ML Aggarwal Class 10 Maths Solutions Chapter 13 Similarity

Question 1. State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):
Answer: Looking at the two sets of triangles, we need to compare the ratios of their corresponding sides and check for equal angles.

For triangles ABC and PQR: The side ratios are \( \frac{AB}{PQ} = \frac{3.2}{4} = \frac{4}{5} \), \( \frac{AC}{PR} = \frac{3.6}{4.5} = \frac{4}{5} \), and \( \frac{BC}{QR} = \frac{3}{5.4} = \frac{5}{9} \). Since the ratios are not all equal, these triangles are not similar.

For triangles DEF and LMN: Both have an angle of 40°, and the side ratios are \( \frac{DE}{LN} = \frac{4}{2} = \frac{1}{2} \) and \( \frac{EF}{MN} = \frac{4.8}{2.4} = \frac{1}{2} \). Using the Side-Angle-Side (SAS) rule of similarity, we conclude that △DEF ~ △LMN.
In simple words: To check if triangles are similar, compare the ratios of matching sides - if all ratios are equal, they are similar. Also check if matching angles are equal.

Exam Tip: Always find all three side ratios and check if they're equal - even two matching ratios don't guarantee similarity. For SAS similarity, confirm both the side ratio and the included angle match.

Question 2. If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles are similar? Why?
Answer: Yes, the two right triangles will definitely be similar. Here's why: In both triangles, one angle is always 90°. If one acute angle of the first triangle equals one acute angle of the second triangle (let's call it a°), then the third angle of each triangle must also be equal because the sum of all angles in a triangle is 180°. This third angle would be 180° - (90° + a°). Since all three pairs of corresponding angles are equal, the triangles satisfy the Angle-Angle (AA) rule of similarity, making them similar.
In simple words: When two right triangles have one equal acute angle, the third angles must also be equal, so the triangles are similar by AA rule.

Exam Tip: Remember that in right triangles, one angle is always fixed at 90°, so you only need one more equal angle to prove similarity using AA rule.

Question 3. It is given that △ABC ~ △EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.
Answer: Since △ABC ~ △EDF, the corresponding sides are proportional: \( \frac{AB}{DE} = \frac{AC}{EF} = \frac{BC}{DF} \).

Using \( \frac{AB}{DE} = \frac{AC}{EF} \): \( \frac{5}{12} = \frac{7}{EF} \) gives \( EF = \frac{7 \times 12}{5} = \frac{84}{5} = 16.8 \) cm.

Using \( \frac{AB}{DE} = \frac{BC}{DF} \): \( \frac{5}{12} = \frac{BC}{15} \) gives \( BC = \frac{5 \times 15}{12} = \frac{75}{12} = 6.25 \) cm.

Therefore, EF = 16.8 cm and BC = 6.25 cm.
In simple words: When two triangles are similar, you can find missing sides by setting up equal ratios and cross-multiplying to solve.

Exam Tip: Always identify which sides correspond to each other in similar triangles - matching the vertices in the similarity statement tells you which sides pair together.

Question 4(a). If △ABC ~ △DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, then find the perimeter of △ABC.
Answer: Since △ABC ~ △DEF, all corresponding sides have the same ratio. We can find this ratio from the given information: \( \frac{AB}{DE} = \frac{4}{6} = \frac{2}{3} \).

Using this ratio, we calculate the other sides: \( \frac{AC}{DF} = \frac{2}{3} \) gives \( AC = \frac{2}{3} \times 12 = 8 \) cm. And \( \frac{BC}{EF} = \frac{2}{3} \) gives \( BC = \frac{2}{3} \times 9 = 6 \) cm.

Therefore, the perimeter of △ABC = AB + BC + AC = 4 + 6 + 8 = 18 cm.
In simple words: Find the scale factor from one pair of corresponding sides, then use it to find all other sides, then add them to get the perimeter.

Exam Tip: The ratio of perimeters of similar triangles equals the ratio of their corresponding sides - use this as a quick check for your answer.

Question 4(b). If △ABC ~ △PQR, perimeter of △ABC = 32 cm, perimeter of △PQR = 48 cm and PR = 6 cm, then find the length of AC.
Answer: For similar triangles, the ratio of their perimeters equals the ratio of their corresponding sides. We have \( \frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle PQR} = \frac{AC}{PR} \). Substituting the values: \( \frac{32}{48} = \frac{AC}{6} \). Simplifying \( \frac{32}{48} = \frac{2}{3} \), we get \( AC = \frac{2}{3} \times 6 = 4 \) cm.
In simple words: When triangles are similar, compare their perimeters the same way you compare individual sides - the ratios must be equal.

Exam Tip: This approach saves time - you don't need to find all the sides individually when you know the perimeters and one side length.

Question 5. Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.
Answer: We are told that △ABC ~ △DEF, where the shortest side of △ABC is BC = 6 cm, and the sides of △DEF are DE = 8 cm, EF = 4 cm, and DF = 7 cm. The shortest side of △DEF is EF = 4 cm.

The scale factor is \( \frac{6}{4} = \frac{3}{2} \). Using this ratio to find the other sides: \( AB = \frac{3}{2} \times 8 = 12 \) cm and \( AC = \frac{3}{2} \times 7 = 10.5 \) cm.

Therefore, the other sides of the triangle are 12 cm and 10.5 cm.
In simple words: Match the shortest sides of both triangles to find the scale factor, then multiply all other sides of the known triangle by this factor.

Exam Tip: Always identify the shortest (or longest) sides first - this helps you match the triangles correctly and find the right scale factor.

Question 6(a). In the figure (1) given below, AB ∥ DE, AC = 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.
Answer: From the diagram, we identify two important angle relationships: ∠ACB = ∠DCE (vertically opposite angles) and ∠BAC = ∠CED (alternate angles, since AB ∥ DE). By the AA rule of similarity, △ABC ~ △CDE.

From this similarity, we have \( \frac{AC}{CE} = \frac{BC}{CD} \), which gives \( \frac{3}{7.5} = \frac{BC}{CD} \), or \( 7.5 \times BC = 3 \times CD \).

Since BD = 14 cm, we know BC + CD = 14. Let BC = x cm, then CD = (14 - x) cm. Substituting: \( 7.5x = 3(14 - x) \). Solving: \( 7.5x = 42 - 3x \), so \( 10.5x = 42 \), giving \( x = 4 \).

Therefore, CB = 4 cm and DC = 10 cm.
In simple words: When lines are parallel, they create similar triangles with vertically opposite angles. Use the property that matching sides of similar triangles are in the same ratio.

Exam Tip: Always mark equal angles clearly and check for both vertically opposite angles and alternate angles created by parallel lines - these guarantee similarity.

Question 6(b). In the figure (2) given below, CA ∥ BD, the lines AB and CD meet at O.
(i) Prove that △ACO ~ △BDO.
(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.
Answer: (i) In △ACO and △BDO: ∠AOC = ∠BOD (vertically opposite angles) and ∠CAO = ∠DBO (alternate angles, since CA ∥ BD). By the AA rule of similarity, △ACO ~ △BDO.

(ii) From the similarity, we have \( \frac{AC}{BD} = \frac{OA}{OB} = \frac{OC}{OD} \).

Using \( \frac{AC}{BD} = \frac{OA}{OB} \): \( \frac{3.6}{2.4} = \frac{OA}{3.2} \) gives \( OA = \frac{3.6 \times 3.2}{2.4} = \frac{11.52}{2.4} = 4.8 \) cm.

Using \( \frac{AC}{BD} = \frac{OC}{OD} \): \( \frac{3.6}{2.4} = \frac{OC}{4} \) gives \( OC = \frac{3.6 \times 4}{2.4} = \frac{14.4}{2.4} = 6 \) cm.
In simple words: When two lines intersect and create parallel segments, the triangles formed are similar. Use the ratios of corresponding sides to find unknown lengths.

Exam Tip: In problems with intersecting lines and parallel segments, set up the proportion carefully - always match corresponding sides from the similarity statement.

Question 7(a). In the figure (i) given below, ∠P = ∠RTS. Prove that △RPQ ~ △RTS.
Answer: In △RPQ and △RTS: ∠R = ∠R (common to both triangles) and ∠P = ∠RTS (given). By the AA rule of similarity, △RPQ ~ △RTS.
In simple words: When two angles of one triangle match two angles of another triangle, the triangles must be similar.

Exam Tip: Always identify the common angle first - a shared angle is often the key to proving similarity quickly.

Question 7(b). In the figure (ii) given below, ∠ADC = ∠BAC. Prove that CA² = DC × BC.
Answer: In △ABC and △ADC: ∠C = ∠C (common angle to both triangles) and ∠BAC = ∠ADC (given). By the AA rule of similarity, △BAC ~ △ADC.

From the similarity property: \( \frac{CA}{DC} = \frac{BC}{CA} \). Cross-multiplying gives \( CA^2 = DC \times BC \). Hence proved.
In simple words: When triangles are similar, you can set up equal ratios between corresponding sides, then cross-multiply to find relationships between the lengths.

Exam Tip: Always cross-multiply carefully when working with proportions from similar triangles - this often leads to useful relationships like squared terms.

Question 8(a). In the figure (1) given below, AP = 2PB and CP = 2PD.
(i) Prove that △ACP is similar to △BDP and AC ∥ BD.
(ii) If AC = 4.5 cm, calculate the length of BD.
Answer: (i) Given AP = 2PB and CP = 2PD, we can write \( \frac{AP}{PB} = \frac{2}{1} \) and \( \frac{CP}{PD} = \frac{2}{1} \). Since ∠APC = ∠BPD (vertically opposite angles), by the SAS rule of similarity, △ACP ~ △BDP.

Because the triangles are similar, ∠CAP = ∠PBD. Since these are alternate angles with respect to line AB and points C and D, we conclude that AC ∥ BD. Hence proved.

(ii) From the similarity, \( \frac{AP}{PB} = \frac{AC}{BD} \), which gives \( \frac{2}{1} = \frac{AC}{BD} \). Therefore, \( BD = \frac{AC}{2} = \frac{4.5}{2} = 2.25 \) cm.
In simple words: When sides from a common vertex are in the same ratio and the included angle is shared, the triangles are similar. Parallel lines follow from matching alternate angles.

Exam Tip: For SAS similarity, you need two pairs of sides in equal ratios AND the included angle to be the same - all three conditions must hold.

Question 8(b). In the figure (2) given below, ∠ADE = ∠ACB.
(i) Prove that △ABC and △AED are similar.
(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.
Answer: (i) In △ABC and △AED: ∠A = ∠A (common to both triangles) and ∠ACB = ∠ADE (given). By the AA rule of similarity, △ABC ~ △AED.

(ii) Since D lies on AB, we have AD = AB - BD = 6 - 1 = 5 cm. From the similarity, \( \frac{AB}{AE} = \frac{AC}{AD} \), which gives \( \frac{6}{3} = \frac{AC}{5} \). Therefore, \( AC = 2 \times 5 = 10 \) cm.
In simple words: A common angle at one vertex plus one more matching angle makes the triangles similar. Then use the ratio of matching sides from the common vertex to find unknowns.

Exam Tip: Always identify shared angles and angles formed on the same line - these provide quick paths to proving similarity without finding all three angle pairs.

Question 8(c). In the figure (3) given below, ∠PQR = ∠PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.
Answer: Given that ∠PQR = ∠PRS. Since ∠P is shared by both triangles PQR and PRS, we can write ∠P = ∠P. By the AA rule of similarity, we conclude that △PQR ~ △PRS. When two triangles are similar, their corresponding sides are in proportion. This gives us the relation:
\( \frac{PQ}{PR} = \frac{PR}{PS} \)

Substituting the known values PR = 8 cm and PS = 4 cm:
\( \frac{PQ}{8} = \frac{8}{4} \)

\( \frac{PQ}{8} = 2 \)

\( PQ = 16 \) cm
In simple words: When two triangles have equal angles, they are similar. This means their sides match up in the same ratio. We use this to find the missing side PQ.

Exam Tip: Always check for equal angles first - if two angles match, the triangles are similar by the AA rule. Use the proportional sides to find unknown lengths.

Question 9. In the adjoining figure, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. Prove that BM × NP = CN × MP.
Answer: Consider triangle ABC where AB = AC is given. By the property of isosceles triangles, the angles opposite equal sides are equal, so ∠B = ∠C. Now look at triangles BMP and CNP. In triangle BMP, we have ∠M = 90° (since PM ⊥ AB), and in triangle CNP, we have ∠N = 90° (since PN ⊥ AC). Since both triangles have right angles at M and N respectively, and we know ∠B = ∠C, by the AA rule of similarity we have △BMP ~ △CNP. When triangles are similar, their corresponding sides are proportional. This means:
\( \frac{BM}{CN} = \frac{MP}{NP} \)

Cross-multiplying gives us BM × NP = CN × MP, which is what we needed to prove.
In simple words: Equal sides in a triangle make equal angles at the base. When we drop perpendiculars from a point to these sides, the resulting smaller triangles become similar. Similar triangles have proportional sides, which leads to the required product relationship.

Exam Tip: Recognize isosceles triangle properties first, then identify right angles from the perpendiculars. Use AA similarity and cross-multiplication to convert the ratio into a product form.

Question 10. Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
Answer: Let two similar triangles be △ABC and △PQR. When triangles are similar, the ratios of their matching sides are equal. This means:
\( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \)

By a property of ratios, if we have equal ratios, we can write:
\( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{AB + BC + AC}{PQ + QR + PR} \)

Now, AB + BC + AC is the perimeter of △ABC, and PQ + QR + PR is the perimeter of △PQR. Therefore:
\( \frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle PQR} = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \)

This shows that the ratio of perimeters equals the ratio of corresponding sides, which completes the proof.
In simple words: When shapes are similar, all their matching sides have the same ratio. If you add up all three sides (the perimeter) from each triangle, that ratio stays the same.

Exam Tip: Remember the "property of equal ratios" - if multiple fractions are equal, their sum over sum of denominators keeps the ratio. Apply this principle to connect individual side ratios to the perimeter ratio.

Question 11. In the adjoining figure, ABCD is a trapezium in which AB ∥ DC. The diagonals AC and BD intersect at O. Prove that \( \frac{AO}{OC} = \frac{BO}{OD} \). Using the above result, find the value(s) of x if OA = 3x - 19, OB = x - 4, OC = x - 3 and OD = 4.
Answer: Consider triangles AOB and COD. The angle ∠AOB equals ∠COD because they are vertically opposite angles. Since AB is parallel to DC, the angle ∠OAB equals ∠OCD as they are alternate angles. By the AA rule of similarity, △AOB ~ △COD. When triangles are similar, their corresponding sides are in proportion, giving us:
\( \frac{AO}{OC} = \frac{BO}{OD} \)

Now, substituting the given values:
\( \frac{3x - 19}{x - 3} = \frac{x - 4}{4} \)

Cross-multiplying:
\( 4(3x - 19) = (x - 3)(x - 4) \)
\( 12x - 76 = x^2 - 7x + 12 \)
\( 0 = x^2 - 19x + 88 \)
\( 0 = (x - 8)(x - 11) \)

Therefore, x = 8 or x = 11.
In simple words: In a trapezium, when diagonals cross, they create two pairs of similar triangles. This means the segments of one diagonal split in the same ratio as the segments of the other diagonal. We solve a quadratic to find which values of x work.

Exam Tip: Identify alternate angles formed by parallel lines and transversals, and recognize vertically opposite angles. Factor the quadratic carefully - both solutions are typically valid unless constraints from the figure eliminate one.

Question 12(a). In the figure (1) given below, AB, EF and CD are parallel lines. Given that AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate (i) EF (ii) AC
Answer:
(i) Consider triangles EFG and CGD. The angle ∠EGF equals ∠CGD since they are vertically opposite. The angle ∠FEG equals ∠GCD because these are alternate angles (formed by parallel lines EF and CD). By AA similarity, △EFG ~ △CGD. From the proportional sides:
\( \frac{EG}{GC} = \frac{EF}{CD} \)
\( \frac{5}{10} = \frac{EF}{18} \)
\( EF = \frac{5 \times 18}{10} = 9 \) cm

(ii) Now consider triangles ABC and EFC. Angle ∠C is common to both. Angle ∠ABC equals ∠EFC because they are alternate angles (AB and EF are parallel). By AA similarity, △ABC ~ △EFC. Using the proportional relationship:
\( \frac{AC}{EC} = \frac{AB}{EF} \)
\( \frac{AC}{EG + GC} = \frac{15}{9} \)
\( \frac{AC}{5 + 10} = \frac{15}{9} \)
\( \frac{AC}{15} = \frac{15}{9} \)
\( AC = \frac{15 \times 15}{9} = 25 \) cm
In simple words: When parallel lines are cut by two intersecting lines, they create similar triangles. The matching sides of these triangles are in the same ratio, allowing us to find missing lengths.

Exam Tip: Always identify which sides are parallel first. Recognize vertically opposite angles and alternate angles from parallel lines. Use the correct correspondence between sides when setting up ratios.

Question 12(b). In the figure (2) given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm and BE = x and AE = y. Find the values of x and y.
Answer: Consider triangles AEF and CED. The angle ∠AEF equals ∠CED since they are vertically opposite. The angle ∠F equals ∠C because these are alternate angles formed by parallel lines AF and CD. By AA similarity, △AEF ~ △CED. From the proportional sides:
\( \frac{AF}{CD} = \frac{AE}{ED} \)
\( \frac{7.5}{4.5} = \frac{y}{3} \)
\( y = \frac{7.5 \times 3}{4.5} = 5 \) cm

Now consider triangles ABE and ACD. Angle ∠A is common to both. Angle ∠ABE equals ∠ACD because they are alternate angles (BE and CD are parallel). By AA similarity, △ABE ~ △ACD. From proportional sides:
\( \frac{BE}{CD} = \frac{AE}{AD} \)
\( \frac{x}{4.5} = \frac{y}{y + ED} \)
\( \frac{x}{4.5} = \frac{5}{5 + 3} \)
\( \frac{x}{4.5} = \frac{5}{8} \)
\( x = \frac{4.5 \times 5}{8} = \frac{22.5}{8} = 2\frac{13}{16} \) cm
In simple words: Parallel lines cut by two transversals form similar triangles at each intersection point. We find the missing lengths by using the proportional sides from these similar triangles.

Exam Tip: Work through the figure systematically, identifying one pair of similar triangles first, then use the result to find additional unknowns. Check that parallel line conditions create the necessary angle equalities.

Question 13. In the given figure, ∠A = 90° and AD ⊥ BC. If BD = 2 cm and CD = 8 cm, find AD.
Answer: Since ∠A = 90°, we have ∠BAD + ∠DAC = 90° ... (i). Now consider triangle ADC where ∠ADC = 90° (since AD is perpendicular to BC). This means ∠DCA + ∠DAC = 90° ... (ii). Comparing equations (i) and (ii), we get ∠BAD = ∠DCA ... (iii). Consider triangles BDA and ADC. Both have ∠BDA = ∠ADC = 90°. From equation (iii), ∠BAD = ∠DCA. By the AA rule of similarity, △BDA ~ △ADC. Since corresponding sides of similar triangles are proportional:
\( \frac{BD}{AD} = \frac{AD}{DC} \)
\( AD^2 = BD \times CD \)
\( AD^2 = 2 \times 8 \)
\( AD^2 = 16 \)
\( AD = 4 \) cm (taking the positive value)
In simple words: When a line drops perpendicularly from a right angle to the opposite side, it creates two smaller similar triangles. The perpendicular's length squared equals the product of the two segments it creates on the opposite side.

Exam Tip: Recognize the geometric mean relationship in right triangles - the altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse. Always reject negative solutions for length.

Question 14. A 15 meters high tower casts a shadow of 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 meters long. Find the height of the telephone pole.
Answer: Let AB be the tower with height 15 m and CD be the telephone pole with unknown height. The shadow of the tower is BE = 24 m, and the shadow of the pole is DE = 16 m. Consider triangles ABE and CDE. Both have ∠ABE = ∠CDE = 90° (angles at the base where objects meet the ground). The angle ∠AEB = ∠CED is common to both triangles (same sun angle). By the AA rule of similarity, △ABE ~ △CDE. Corresponding sides are proportional:
\( \frac{AB}{CD} = \frac{BE}{DE} \)
\( \frac{15}{CD} = \frac{24}{16} \)
\( CD = \frac{15 \times 16}{24} = 10 \) m
In simple words: Objects casting shadows at the same time create similar triangles because the sun's rays hit at the same angle. The ratio of heights equals the ratio of shadow lengths.

Exam Tip: Identify that objects and their shadows form similar triangles when cast at the same time. The height-to-shadow ratio is identical for all objects, making this a direct application of similar triangles.

Question 15. A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height 1.5 m casts a shadow of 3 m, find how far she is away from the base of the pole?
Answer: Let AB be the pole with height 6 m and DE be the woman with height 1.5 m. The woman's shadow is EF = 3 m. Let BE (the distance of the woman from the pole) = x meters. Consider triangles ABF and EFD. Both have ∠ABF = ∠DEF = 90° (right angles where the pole and woman meet the ground). The angle ∠F = ∠F is common to both triangles (at the end of the shadow). By the AA rule of similarity, △ABF ~ △EFD. Corresponding sides are proportional:
\( \frac{BF}{EF} = \frac{AB}{DE} \)
\( \frac{3 + x}{3} = \frac{6}{1.5} \)
\( \frac{3 + x}{3} = 4 \)
\( 3 + x = 12 \)
\( x = 9 \) m
In simple words: Light rays from a point above the ground cast shadows. The pole and woman's heights compared to their shadow lengths form similar triangles. Using this similarity, we find the distance separating them.

Exam Tip: Draw a clear diagram showing the light source, the pole, the woman, and the shadow endpoint. Identify that the light ray is a common side/angle, creating the similarity needed to set up the proportion.

Exercise 13.2

Question 1(a). In the figure (i) given below, if DE ∥ BC, AD = 3 cm, BD = 4 cm and BC = 5 cm, find (i) AE : EC (ii) DE.
Answer: When triangles ABC and ADE are compared, angle A matches in both (it is a common angle). Also, angle ADE equals angle ABC because they are alternate angles. Using the AA similarity rule, triangles ABC and ADE are similar.

Since the triangles are similar, their corresponding sides are in the same ratio.

\( \frac{AE}{AC} = \frac{AD}{AB} \)

\( \Rightarrow \frac{AE}{AE + EC} = \frac{AD}{AD + BD} \)

\( \Rightarrow \frac{AE}{AE + EC} = \frac{3}{3 + 4} \)

\( \Rightarrow 7AE = 3(AE + EC) \)

\( \Rightarrow 7AE = 3AE + 3EC \)

\( \Rightarrow 7AE - 3AE = 3EC \)

\( \Rightarrow 4AE = 3EC \)

\( \Rightarrow \frac{AE}{EC} = \frac{3}{4} \)

Therefore, AE : EC = 3 : 4.

For part (ii), since triangles ABC and ADE are similar:

\( \frac{DE}{BC} = \frac{AD}{AB} \)

\( \Rightarrow \frac{DE}{5} = \frac{3}{3 + 4} \)

\( \Rightarrow \frac{DE}{5} = \frac{3}{7} \)

\( \Rightarrow DE = \frac{15}{7} = 2\frac{1}{7} \) cm
In simple words: When two triangles are similar, their matching sides have the same ratio. By finding these ratios, we get AE : EC = 3 : 4 and DE = 2\(\frac{1}{7}\) cm.

Exam Tip: Always verify that two triangles are similar using the AA rule before applying the ratio of corresponding sides - this is what makes the calculation valid.

Question 1(b). In the figure (ii) given below, PQ ∥ AC, AP = 4 cm, PB = 6 cm and BC = 8 cm, find CQ and BQ.
Answer: When triangles ABC and PBQ are examined, angle B is the same in both (common angle). Since PQ is parallel to AC, angle BPQ equals angle BAC (corresponding angles are equal). By the AA similarity rule, triangles ABC and PBQ are similar.

Since the triangles are similar, the ratio of corresponding sides is constant:

\( \frac{BQ}{BC} = \frac{BP}{BA} \)

\( \Rightarrow \frac{BQ}{8} = \frac{6}{AP + PB} \)

\( \Rightarrow \frac{BQ}{8} = \frac{6}{4 + 6} \)

\( \Rightarrow \frac{BQ}{8} = \frac{6}{10} \)

\( \Rightarrow BQ = \frac{48}{10} = 4.8 \) cm

Now we find CQ:

\( CQ = BC - BQ = 8 - 4.8 = 3.2 \) cm

Therefore, BQ = 4.8 cm and CQ = 3.2 cm.
In simple words: Using the property that similar triangles have proportional sides, we calculate BQ and then subtract from BC to get CQ.

Exam Tip: Always label which sides are corresponding sides in similar triangles - this prevents mixing up the ratios.

Question 1(c). In the figure (iii) given below, if XY ∥ QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.
Answer: Triangles PQR and PXY are examined. Angle P is present in both triangles (common angle). Since XY is parallel to QR, angle PXY equals angle PQR (corresponding angles). Using the AA similarity rule, triangles PQR and PXY are similar.

Since the triangles are similar, the ratio of corresponding sides is constant:

\( \frac{PX}{QX} = \frac{PY}{YR} \)

\( \Rightarrow \frac{1}{3} = \frac{PY}{4.5} \)

\( \Rightarrow PY = \frac{4.5}{3} = 1.5 \) cm

For XY, using the same similar triangles:

\( \frac{XY}{QR} = \frac{PX}{PQ} \)

\( \Rightarrow \frac{XY}{9} = \frac{1}{1 + 3} \)

\( \Rightarrow \frac{XY}{9} = \frac{1}{4} \)

\( \Rightarrow XY = \frac{9}{4} = 2.25 \) cm

Therefore, PY = 1.5 cm and XY = 2.25 cm.
In simple words: Because the triangles are similar, the lengths of sides in the smaller triangle follow the same proportions as the larger triangle.

Exam Tip: When parallel lines are involved, always look for similar triangles - the parallel line creates the similarity by producing equal corresponding angles.

Question 2. In the adjoining figure, DE ∥ BC. (i) If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value of x. (ii) If DB = x - 3, AB = 2x, EC = x - 2 and AC = 2x + 3, find the value of x.
Answer:
(i) Triangles ABC and ADE are compared. Angle A is the same (common angle), and angle ADE equals angle ABC (corresponding angles). By the AA similarity rule, triangles ABC and ADE are similar.

Since the triangles are similar, the ratio of corresponding sides is the same:

\( \frac{AD}{AB} = \frac{AE}{AC} \)

\( \Rightarrow \frac{AD}{AD + DB} = \frac{AE}{AE + EC} \)

\( \Rightarrow \frac{x}{x + x - 2} = \frac{x + 2}{x + 2 + x - 1} \)

\( \Rightarrow \frac{x}{2x - 2} = \frac{x + 2}{2x + 1} \)

\( \Rightarrow x(2x + 1) = (x + 2)(2x - 2) \)

\( \Rightarrow 2x^2 + x = 2x^2 - 2x + 4x - 4 \)

\( \Rightarrow 2x^2 - 2x^2 + x + 2x - 4x = -4 \)

\( \Rightarrow -x = -4 \)

\( \Rightarrow x = 4 \)

(ii) Since triangles are similar, the ratio of corresponding sides remains constant:

\( \frac{AD}{AB} = \frac{AE}{AC} \)

\( \Rightarrow \frac{AB - DB}{AB} = \frac{AC - EC}{AC} \)

\( \Rightarrow \frac{2x - (x - 3)}{2x} = \frac{2x + 3 - (x - 2)}{2x + 3} \)

\( \Rightarrow \frac{x + 3}{2x} = \frac{x + 5}{2x + 3} \)

\( \Rightarrow (x + 3)(2x + 3) = 2x(x + 5) \)

\( \Rightarrow 2x^2 + 3x + 6x + 9 = 2x^2 + 10x \)

\( \Rightarrow 2x^2 - 2x^2 + 9 = 10x - 9x \)

\( \Rightarrow x = 9 \)
In simple words: When a line is drawn parallel to one side of a triangle, it cuts the other two sides proportionally. This allows us to write an equation and solve for the unknown length.

Exam Tip: Always set up the proportionality statement correctly - AD/AB = AE/AC, not mixed ratios - to avoid sign errors when solving.

Question 3. E and F are points on the sides PQ and PR respectively of a △PQR. For each of the following cases, state whether EF ∥ QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm. (ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Answer:
(i) For EF to be parallel to QR, the sides must be cut proportionally. We check whether the ratios of the segments are equal:

\( \frac{PE}{PQ} \text{ and } \frac{PF}{PR} \)

\( \Rightarrow \frac{PE}{PE + EQ} \text{ and } \frac{PF}{PF + FR} \)

\( \Rightarrow \frac{3.9}{3.9 + 3} \text{ and } \frac{8}{8 + 9} \)

\( \Rightarrow \frac{3.9}{6.9} \text{ and } \frac{8}{17} \)

\( \Rightarrow \frac{39}{69} \text{ and } \frac{8}{17} \)

Since \( \frac{39}{69} \neq \frac{8}{17} \), the ratios are different. Therefore, EF and QR are not parallel.

(ii) We check whether the ratios are equal:

\( \frac{PE}{PQ} \text{ and } \frac{PF}{PR} \)

\( \Rightarrow \frac{0.18}{1.28} \text{ and } \frac{0.36}{2.56} \)

\( \Rightarrow \frac{18}{128} \text{ and } \frac{36}{256} \)

\( \Rightarrow \frac{9}{64} \text{ and } \frac{9}{64} \)

Since both ratios are equal, by the converse of the basic proportionality theorem, EF and QR are parallel.
In simple words: If a line cuts two sides of a triangle in the same ratio, then that line is parallel to the third side. Check the ratios to see if they match.

Exam Tip: Simplify fractions to lowest terms before comparing - this makes it easier to spot whether they are equal or different.

Question 4. A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB ∥ QR? Give reasons for your answer.
Answer: To determine if AB is parallel to QR, we check whether the sides are cut proportionally by examining triangles PAB and PQR.

First, we find the relevant ratios:

\( \frac{PA}{PQ} \text{ and } \frac{PB}{PR} \)

\( \Rightarrow \frac{PA}{PQ} \text{ and } \frac{PB}{PB + BR} \)

\( \Rightarrow \frac{5}{12.5} \text{ and } \frac{4}{4 + 6} \)

\( \Rightarrow \frac{50}{125} \text{ and } \frac{4}{10} \)

\( \Rightarrow \frac{2}{5} \text{ and } \frac{2}{5} \)

Since both ratios are equal, by the converse of the basic proportionality theorem, triangles PAB and PQR are similar. Therefore, AB and QR are parallel.
In simple words: When the ratio PA/PQ equals PB/PR, the line AB divides the triangle in a way that makes it parallel to the opposite side QR.

Exam Tip: Always check that you are comparing the correct ratios - the ratio from P to the point on one side should match the ratio from P to the point on the other side.

Question 5(a). In the figure (i) given below, CD ∥ LA and DE ∥ AC. Find the length of CL if BE = 4 cm and EC = 2 cm.
Answer: Triangles ABC and BDE are examined. Angle B appears in both (common angle). Since DE is parallel to AC, angle DEB equals angle ACB (corresponding angles are equal). Using the AA similarity rule, triangles ABC and BDE are similar.

Since the triangles are similar, corresponding sides are in the same ratio:

\( \frac{BE}{BC} = \frac{BD}{BA} \)

\( \Rightarrow \frac{BE}{BE + EC} = \frac{BD}{BA} \)

\( \Rightarrow \frac{4}{4 + 2} = \frac{BD}{AB} \)

\( \Rightarrow \frac{4}{6} = \frac{BD}{AB} \) ... (Eq 1)

Next, triangles ABL and BDC are examined. Angle B is the same (common angle). Since CD is parallel to LA, angle BDC equals angle BAL (corresponding angles). By the AA similarity rule, triangles ABL and BDC are similar.

Since these triangles are similar:

\( \frac{BA}{BD} = \frac{BL}{BC} \)

\( \Rightarrow \frac{BA}{BD} = \frac{BE + EC + CL}{BE + EC} \)

\( \Rightarrow \frac{BA}{BD} = \frac{4 + 2 + CL}{4 + 2} \)

\( \Rightarrow \frac{BA}{BD} = \frac{6 + CL}{6} \) ... (Eq 2)

From Eq 1, we have \( \frac{BA}{BD} = \frac{6}{4} \). Substituting this into Eq 2:

\( \frac{6}{4} = \frac{6 + CL}{6} \)

\( \Rightarrow 36 = 4(6 + CL) \)

\( \Rightarrow 36 = 24 + 4CL \)

\( \Rightarrow 4CL = 36 - 24 \)

\( \Rightarrow 4CL = 12 \)

\( \Rightarrow CL = 3 \) cm
In simple words: By finding two pairs of similar triangles and using their proportional sides, we create equations that can be solved to find the unknown length CL.

Exam Tip: When two pairs of parallel lines are given, look for multiple pairs of similar triangles - solving two equations together often reveals the answer.

Question 5(b). In the figure (ii) given below, ∠D = ∠E and \( \frac{AD}{DB} = \frac{AE}{EC} \). Prove that ABC is an isosceles triangle.
Answer: Since angle D equals angle E, the sides opposite to these angles must also be equal. Therefore, AD = AE.

We are given that \( \frac{AD}{DB} = \frac{AE}{EC} \) ... (Eq 1)

By the basic proportionality theorem (converse), this condition means that DE is parallel to BC.

Since AD = AE (from the equal angles) and we need to satisfy Eq 1, this means DB = EC must hold true.

Now, we can express the full sides:

\( AB = AD + DB = AE + EC \)

\( AC = AE + EC \)

Since AB = AC, triangle ABC is an isosceles triangle (a triangle with two equal sides).
In simple words: When two angles at the base of a configuration are equal, the segments they face are equal. Combined with the given ratio, this forces two sides of the triangle to be equal, making it isosceles.

Exam Tip: In proving geometry statements, always identify what the given conditions guarantee - here, equal angles guarantee equal opposite sides, which is key to the proof.

Question 6. In the adjoining figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.
Answer: Consider triangle POQ. Since AB is parallel to PQ (given), by the basic proportionality theorem:

\( \frac{OA}{AP} = \frac{OB}{BQ} \) ... (Eq 1)

Now consider triangle OPR. Since AC is parallel to PR (given), by the basic proportionality theorem:

\( \frac{OA}{AP} = \frac{OC}{CR} \) ... (Eq 2)

From Eq 1 and Eq 2, we have:

\( \frac{OB}{BQ} = \frac{OC}{CR} \)

This means the sides OQ and OR are divided by points B and C in the same ratio. By the converse of the basic proportionality theorem, BC is parallel to QR.
In simple words: When two different lines are each parallel to a side of a triangle, they create matching ratios on the third side, which guarantees that a line connecting their endpoints is parallel to the base.

Exam Tip: Use the basic proportionality theorem on two different triangles that share a common vertex - combining the results shows the parallelism of the line you need to prove.

Question 7. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem prove that \( \frac{AO}{BO} = \frac{CO}{DO} \).
Answer: Trapezium ABCD has its diagonals intersecting at point O. When we examine triangles OAB and OCD, we observe that \( \angle AOB = \angle COD \) [Vertically opposite angles are equal], \( \angle OBA = \angle ODC \) [Alternate angles are equal since AB || DC], and \( \angle OAB = \angle OCD \) [Alternate angles are equal since AB || DC]. Therefore, by the AA rule of similarity, \( \triangle OAB \sim \triangle OCD \). Since these triangles are similar, their corresponding sides are proportional:
\( \implies \frac{AO}{CO} = \frac{BO}{DO} \)

Cross-multiplying gives us \( \frac{AO}{BO} = \frac{CO}{DO} \), which proves the required result.
In simple words: When two lines cross each other inside a shape, and the sides are parallel, the pieces on one side of the crossing point match up the same way as the pieces on the other side.

Exam Tip: Recognise that AB || DC makes the alternate angles equal. Clearly name both triangles and state the similarity criterion (AA) before using the property of similar triangles.

Question 8. In the adjoining figure, AD is bisector of \( \angle BAC \). If AB = 6 cm, AC = 4 cm and BD = 3 cm, find BC.
Answer: By the angle bisector theorem, when a line divides an angle of a triangle, it splits the opposite side internally in the ratio of the two sides forming that angle:
\( \frac{BD}{DC} = \frac{AB}{AC} \)

Substituting the given values:
\( \frac{3}{DC} = \frac{6}{4} \)

\( \implies DC = \frac{3 \times 4}{6} = \frac{12}{6} = 2 \text{ cm} \)

Since B, D, and C lie on the same line with D between B and C:
\( BC = BD + DC = 3 + 2 = 5 \text{ cm} \)
In simple words: When a line cuts one angle of a triangle in half, it divides the opposite side into two parts that match the ratio of the other two sides of the triangle.

Exam Tip: Always verify your answer by checking that BD + DC = BC. Clearly state the angle bisector theorem by name before applying it.

Exercise 13.3

Question 1. Given that triangles ABC and PQR are similar. Find: (i) the ratio of the area of triangle ABC to the area of triangle PQR if their corresponding sides are in the ratio 1:3. (ii) the ratio of their corresponding sides if area of triangle ABC : area of triangle PQR = 25:36.
Answer:
(i) For two similar triangles, the ratio of their areas equals the ratio of the squares of their corresponding sides. If the sides are in the ratio 1:3, then:
\( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle PQR} = \frac{1^2}{3^2} = \frac{1}{9} \)

Therefore, the area ratio is 1:9.

(ii) Let the corresponding sides be in ratio x:y. We know that for similar triangles:
\( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle PQR} = \frac{x^2}{y^2} \)

Given that this area ratio is 25:36:
\( \frac{x^2}{y^2} = \frac{25}{36} \)

\( \implies \left(\frac{x}{y}\right)^2 = \left(\frac{5}{6}\right)^2 \)

\( \implies \frac{x}{y} = \frac{5}{6} \)

Thus, x:y = 5:6, so the sides are in the ratio 5:6.
In simple words: When shapes are the same but different sizes, their areas grow much faster than their side lengths. If sides double, areas become four times bigger.

Exam Tip: Remember that the area ratio equals the square of the side ratio. When finding sides from area ratios, always take the square root of the area ratio.

Question 2. Triangle ABC ~ triangle DEF. If area of triangle ABC = 9 sq. cm, area of triangle DEF = 16 sq. cm and BC = 2.1 cm, find the length of EF.
Answer: Let the length of EF be x cm. Since the triangles are similar, the ratio of their areas equals the ratio of the squares of corresponding sides:
\( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{BC^2}{EF^2} \)

\( \frac{9}{16} = \frac{(2.1)^2}{x^2} = \frac{4.41}{x^2} \)

\( \implies x^2 = \frac{4.41 \times 16}{9} = \frac{70.56}{9} = 7.84 \)

\( \implies x = \sqrt{7.84} = 2.8 \text{ cm} \)
In simple words: The smaller triangle has a smaller area and smaller sides. Use the area relationship to work backwards and find the missing side length.

Exam Tip: Always check that your final answer makes sense - the side of the larger triangle should be bigger than the corresponding side of the smaller triangle.

Question 3. Triangle ABC ~ triangle DEF. If BC = 3 cm, EF = 4 cm and area of triangle ABC = 54 sq. cm, determine the area of triangle DEF.
Answer: Let the area of triangle DEF be x sq. cm. Using the property that the ratio of areas of similar triangles equals the ratio of squares of corresponding sides:
\( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{BC^2}{EF^2} \)

\( \frac{54}{x} = \frac{3^2}{4^2} = \frac{9}{16} \)

\( \implies x = \frac{54 \times 16}{9} = 6 \times 16 = 96 \text{ sq. cm} \)
In simple words: The triangle with the longer side must have the bigger area. Set up the proportion and solve for the unknown area.

Exam Tip: Verify your work by checking: if sides are in ratio 3:4, then areas should be in ratio 9:16. Here 54:96 = 9:16 ✓

Question 4. The areas of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.
Answer: Let the altitude of the second triangle be x cm. For similar triangles, the ratio of their areas equals the ratio of the squares of their corresponding altitudes:
\( \frac{\text{Area of first} \triangle}{\text{Area of second} \triangle} = \left(\frac{\text{Altitude of first} \triangle}{\text{Altitude of second} \triangle}\right)^2 \)

\( \frac{36}{25} = \left(\frac{2.4}{x}\right)^2 \)

\( \implies \left(\frac{2.4}{x}\right)^2 = \frac{36}{25} \)

\( \implies \frac{2.4}{x} = \frac{6}{5} \)

\( \implies x = \frac{2.4 \times 5}{6} = \frac{12}{6} = 2 \text{ cm} \)
In simple words: Just as sides and areas follow a pattern in similar shapes, heights follow the same pattern too. If areas are in ratio 36:25, then heights are in ratio 6:5.

Exam Tip: Remember that when finding the ratio of altitudes from area ratios, take the square root. The altitude of the smaller triangle will be shorter than the altitude of the larger triangle.

Question 5(a). In the figure (i) given below, PB and QA are perpendiculars to line segment AB. If PO = 6 cm, OQ = 9 cm and the area of triangle POB = 120 cm², find the area of triangle QOA.
Answer: Looking at triangles QOA and POB:
\( \angle QOA = \angle POB \) [Vertically opposite angles are equal]
\( \angle QAO = \angle PBO \) [Both equal 90° since PB and QA are perpendiculars]

By the AA criterion, \( \triangle QOA \sim \triangle POB \). For similar triangles, the ratio of areas equals the ratio of squares of corresponding sides:
\( \frac{\text{Area of } \triangle QOA}{\text{Area of } \triangle POB} = \frac{QO^2}{PO^2} \)

Let the area of triangle QOA be x cm²:
\( \frac{x}{120} = \frac{9^2}{6^2} = \frac{81}{36} \)

\( \implies x = \frac{120 \times 81}{36} = 120 \times \frac{81}{36} = 120 \times 2.25 = 270 \text{ cm}^2 \)
In simple words: These two triangles point in opposite directions but have the same angles, so they are similar. The triangle with the longer side from the point O has a much bigger area.

Exam Tip: Identify vertically opposite angles and right angles carefully. Use the property that in similar figures, area scales with the square of the linear dimensions.

Question 5(b). In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5 cm, AB = 6.5 cm and OD = 2.8 cm. (i) Prove that triangle OAB ~ triangle OCD. (ii) Find CD and OB. (iii) Find the ratio of areas of triangle OAB and triangle OCD.
Answer:
(i) Examining triangles OAB and OCD:
\( \angle AOB = \angle COD \) [Vertically opposite angles are equal]
\( \angle BAO = \angle OCD \) [Alternate angles are equal because AB || DC]

By the AA criterion, \( \triangle OAB \sim \triangle OCD \).

(ii) Since the triangles are similar, corresponding sides are proportional:
\( \frac{AO}{OC} = \frac{AB}{CD} \)

\( \frac{10}{5} = \frac{6.5}{CD} \)

\( \implies CD = \frac{6.5 \times 5}{10} = \frac{32.5}{10} = 3.25 \text{ cm} \)

Similarly:
\( \frac{AO}{OC} = \frac{OB}{OD} \)

\( \frac{10}{5} = \frac{OB}{2.8} \)

\( \implies OB = \frac{2.8 \times 10}{5} = \frac{28}{5} = 5.6 \text{ cm} \)

(iii) The ratio of areas equals the ratio of squares of corresponding sides:
\( \frac{\text{Area of } \triangle OAB}{\text{Area of } \triangle OCD} = \frac{AO^2}{OC^2} = \frac{10^2}{5^2} = \frac{100}{25} = 4:1 \)
In simple words: The two triangles share the same angles because of parallel lines. The larger triangle's sides are exactly twice as long, so its area is four times bigger.

Exam Tip: When proving similarity with parallel lines, state the alternate angle property clearly. For finding the area ratio, use the squares of the ratios of sides or the ratio of corresponding heights.

Question 6(a). In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of triangle ADE = 28 sq. cm, find the area of triangle ABC.
Answer: Looking at triangles ADE and ABC:
\( \angle A = \angle A \) [Common angle]
\( \angle ADE = \angle ABC \) [Corresponding angles are equal because DE || BC]

By the AA criterion, \( \triangle ADE \sim \triangle ABC \). Using the property that areas of similar triangles are proportional to the squares of corresponding sides:
\( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle ADE} = \frac{BC^2}{DE^2} \)

Let the area of triangle ABC be x cm²:
\( \frac{x}{28} = \frac{9^2}{6^2} = \frac{81}{36} \)

\( \implies x = \frac{28 \times 81}{36} = \frac{2268}{36} = 63 \text{ sq. cm} \)
In simple words: The smaller triangle sits inside the larger one. The sides of the big triangle are 1.5 times longer, so the area is 2.25 times bigger.

Exam Tip: Always state which triangles are similar and the similarity criterion. Cross-multiply carefully when setting up the proportion.

Question 6(b). In the figure (ii) given below, DE || BC and AD:DB = 1:2, find the ratio of the areas of triangle ADE and trapezium DBCE.
Answer: Comparing triangles ADE and ABC:
\( \angle A = \angle A \) [Common angle]
\( \angle ADE = \angle ABC \) [Corresponding angles are equal because DE || BC]

By the AA criterion, \( \triangle ADE \sim \triangle ABC \).

Given that AD:DB = 1:2, we can write:
\( \frac{AD}{AB - AD} = \frac{1}{2} \)

\( \implies 2AD = AB - AD \)
\( \implies 3AD = AB \)
\( \implies AD:AB = 1:3 \)

Using the area ratio property for similar triangles:
\( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle ADE} = \frac{AB^2}{AD^2} = \frac{3^2}{1^2} = \frac{9}{1} = 9 \)

This means Area of triangle ABC = 9 × Area of triangle ADE.

Since triangle ABC is divided into triangle ADE and trapezium DBCE:
\( \text{Area of } \triangle ABC = \text{Area of } \triangle ADE + \text{Area of trapezium DBCE} \)
\( 9 \times \text{Area of } \triangle ADE = \text{Area of } \triangle ADE + \text{Area of trapezium DBCE} \)
\( \text{Area of trapezium DBCE} = 8 \times \text{Area of } \triangle ADE \)

Therefore, the ratio of areas is:
\( \frac{\text{Area of } \triangle ADE}{\text{Area of trapezium DBCE}} = \frac{1}{8} \)
In simple words: Point D divides the side into 1 part and 2 parts, making the total 3 parts. The small triangle at the top is 1/9 of the whole triangle, and the trapezium below is 8/9 of the whole.

Exam Tip: Convert the given ratio AD:DB into the ratio AD:AB by adding the parts. Remember that the trapezium area equals the big triangle minus the small triangle.

Question 7. In the given figure, DE || BC.
(i) Prove that △ADE and △ABC are similar.
(ii) Given that AD = BD, calculate DE, if BC = 4.5 cm.
(iii) If area of △ABC = 18 cm², find area of trapezium DBCE.
Answer:
(i) Looking at △ADE and △ABC:

\( \angle A = \angle A \) (Common angles)

\( \angle ADE = \angle ABC \) (Corresponding angles are equal)

By the AA axiom, △ADE ~ △ABC.

(ii) Given that AD = BD, we know that:

\( AD = \frac{1}{2}BD \)

\( \Rightarrow AD = \frac{1}{2}(AB - AD) \)

\( \Rightarrow 2AD = AB - AD \)

\( \Rightarrow 3AD = AB \)

\( \Rightarrow \frac{AD}{AB} = \frac{1}{3} \)

Since triangles ADE and ABC are similar, their corresponding sides are proportional:

\( \frac{AD}{AB} = \frac{DE}{BC} \)

\( \Rightarrow \frac{1}{3} = \frac{DE}{4.5} \)

\( \Rightarrow DE = \frac{4.5}{3} = 1.5 \) cm

(iii) For similar triangles, the ratio of their areas equals the ratio of the squares of their corresponding sides:

\( \frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC} = \frac{AD^2}{AB^2} = \frac{1^2}{3^2} = \frac{1}{9} \)

\( \Rightarrow \text{Area of } \triangle ADE = \frac{1}{9} \times 18 = 2 \) cm²

Area of trapezium DBCE = Area of △ABC - Area of △ADE = 18 - 2 = 16 cm²

In simple words: When a line parallel to one side of a triangle cuts the other two sides, it creates a smaller triangle similar to the original. We use this to find the side length and then calculate the area left over after removing the smaller triangle.

Exam Tip: For similar triangles, always use the property that the area ratio equals the square of the side ratio - this is a common marking point in geometry questions.

Question 8. In the given figure, AB and DE are perpendiculars to BC.
(i) Prove that △ABC ~ △DEC.
(ii) If AB = 6 cm, DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of △ABC : area of △DEC.
Answer:
(i) Looking at △DEC and △ABC:

\( \angle C = \angle C \) (Common angles)

\( \angle ABC = \angle DEC = 90° \) (Both angles are right angles)

By the AA axiom, △DEC ~ △ABC.

(ii) Since △DEC ~ △ABC, their corresponding sides are proportional:

\( \frac{DE}{AB} = \frac{CD}{AC} \)

\( \Rightarrow \frac{4}{6} = \frac{CD}{15} \)

\( \Rightarrow CD = \frac{15 \times 4}{6} = 10 \) cm

(iii) For similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEC} = \frac{AB^2}{DE^2} = \frac{6^2}{4^2} = \frac{36}{16} = \frac{9}{4} \)

The ratio of area of △ABC : area of △DEC = 9 : 4

In simple words: Two triangles that have a common angle and both have a right angle are similar. When triangles are similar, we can find missing sides using proportions, and compare their sizes using the square of side ratios.

Exam Tip: When given perpendiculars from different points to the same line, look for a common angle to prove similarity using the AA axiom rather than trying to find equal corresponding angles.

Question 9. In the adjoining figure, ABC is a triangle. DE is parallel to BC and \( \frac{AD}{DB} = \frac{3}{2} \).
(i) Determine the ratio \( \frac{AD}{AB}, \frac{DE}{BC} \).
(ii) Prove that △DEF is similar to △CBF. Hence, find \( \frac{EF}{FB} \).
(iii) What is the ratio of the areas of △DEF and △CBF?
Answer:
(i) We need to find \( \frac{AD}{AB} \).

Given: \( \frac{AD}{DB} = \frac{3}{2} \)

\( \Rightarrow \frac{AD}{AB - AD} = \frac{3}{2} \)

\( \Rightarrow 2AD = 3(AB - AD) \)

\( \Rightarrow 2AD = 3AB - 3AD \)

\( \Rightarrow 5AD = 3AB \)

\( \Rightarrow \frac{AD}{AB} = \frac{3}{5} \)

Looking at △ADE and △ABC:

\( \angle A = \angle A \) (Common angles)

\( \angle ADE = \angle ABC \) (Corresponding angles are equal)

By the AA axiom, △ADE ~ △ABC.

Since the triangles are similar, their corresponding sides are proportional:

\( \frac{DE}{BC} = \frac{AD}{AB} = \frac{3}{5} \)

(ii) Looking at △DEF and △CBF:

\( \angle DFE = \angle BFC \) (Vertically opposite angles)

\( \angle EDF = \angle FCB \) (Alternate angles are equal)

By the AA axiom, △DEF ~ △CBF.

Since the triangles are similar, their corresponding sides are proportional:

\( \frac{EF}{FB} = \frac{DE}{BC} = \frac{3}{5} \)

(iii) For similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of } \triangle DEF}{\text{Area of } \triangle CBF} = \frac{EF^2}{FB^2} = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \)

The ratio of the area of △DEF : area of △CBF = 9 : 25

In simple words: When a line is drawn parallel to the base of a triangle, it creates a smaller similar triangle. The sides of this smaller triangle have the same ratio as the points that divide the original triangle's sides.

Exam Tip: In problems involving parallel lines and similar triangles, always identify vertically opposite angles and alternate angles - these are key to proving similarity without calculating all sides.

Question 10. In △PQR, MN is parallel to QR and \( \frac{PM}{MQ} = \frac{2}{3} \).
(i) Find \( \frac{MN}{QR} \).
(ii) Prove that △OMN and △ORQ are similar.
(iii) Find area of △OMN : area of △ORQ.
Answer:
(i) Looking at △PMN and △PQR:

\( \angle P = \angle P \) (Common angles)

\( \angle PMN = \angle PQR \) (Corresponding angles are equal)

By the AA axiom, △PMN ~ △PQR.

Given: \( \frac{PM}{MQ} = \frac{2}{3} \)

\( \Rightarrow \frac{PM}{PQ - PM} = \frac{2}{3} \)

\( \Rightarrow 3PM = 2(PQ - PM) \)

\( \Rightarrow 3PM = 2PQ - 2PM \)

\( \Rightarrow 5PM = 2PQ \)

\( \Rightarrow \frac{PM}{PQ} = \frac{2}{5} \)

Since the triangles are similar, their corresponding sides are proportional:

\( \frac{MN}{QR} = \frac{PM}{PQ} = \frac{2}{5} \)

(ii) Looking at △OMN and △ORQ:

\( \angle MON = \angle QOR \) (Vertically opposite angles are equal)

\( \angle OMN = \angle ORQ \) (Alternate angles are equal)

By the AA axiom, △OMN ~ △ORQ.

(iii) For similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of } \triangle OMN}{\text{Area of } \triangle ORQ} = \frac{MN^2}{QR^2} = \left(\frac{2}{5}\right)^2 = \frac{4}{25} \)

The ratio of the Area of △OMN : Area of △ORQ = 4 : 25

In simple words: A line parallel to one side of a triangle divides the other sides in the same ratio. This creates similar triangles, and we can compare their areas by squaring the ratio of their sides.

Exam Tip: When working with parallel lines in a triangle, remember that the ratio of division on one side is the same as the ratio of the parallel segments to the original sides - use this to avoid repeated calculation.

Question 11. In △ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find :
(i) area △APO : area △ABC
(ii) area △APO : area △CQO.
Answer:
(i) Given: \( \frac{AP}{PB} = \frac{2}{3} \)

\( \Rightarrow \frac{AP}{AB - AP} = \frac{2}{3} \)

\( \Rightarrow 3AP = 2(AB - AP) \)

\( \Rightarrow 3AP = 2AB - 2AP \)

\( \Rightarrow 5AP = 2AB \)

\( \Rightarrow \frac{AP}{AB} = \frac{2}{5} \)

Looking at △APO and △ABC:

\( \angle A = \angle A \) (Common angles)

\( \angle APO = \angle ABC \) (Corresponding angles are equal)

By the AA axiom, △APO ~ △ABC.

For similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of } \triangle APO}{\text{Area of } \triangle ABC} = \frac{AP^2}{AB^2} = \left(\frac{2}{5}\right)^2 = \frac{4}{25} \)

The ratio of the area of △APO : area of △ABC = 4 : 25

(ii) In parallelogram PBCQ, opposite sides are equal:

PB = QC. Therefore: \( \frac{AP}{QC} = \frac{AP}{PB} = \frac{2}{3} \)

Looking at △APO and △CQO:

\( \angle AOP = \angle QOC \) (Vertically opposite angles)

\( \angle OAP = \angle OCQ \) (Alternate angles are equal)

By the AA axiom, △APO ~ △CQO.

For similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of } \triangle APO}{\text{Area of } \triangle CQO} = \frac{AP^2}{QC^2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)

The ratio of the area of △APO : area of △CQO = 4 : 9

In simple words: When you draw a line parallel to the base of a triangle from a point on one side, it cuts off a smaller similar triangle. We can find what fraction of the original triangle's area this smaller triangle is by squaring the ratio of the division point.

Exam Tip: In questions involving parallel lines and areas, always identify the similar triangles first and use the area-ratio property (square of side ratio) rather than calculating individual areas.

Question 12(a). In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of △AOB and △COD.
Answer: Given AB || DC and AB = 2CD:

\( \frac{AB}{CD} = \frac{2}{1} \)

Looking at △AOB and △COD:

\( \angle AOB = \angle COD \) (Vertically opposite angles)

\( \angle OCD = \angle OAB \) (Alternate angles are equal)

By the AA axiom, △AOB ~ △COD.

For similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle COD} = \frac{AB^2}{CD^2} = \frac{2^2}{1^2} = \frac{4}{1} \)

The ratio of the area of △AOB : area of △COD = 4 : 1

In simple words: In a trapezium, the diagonals divide each other and create two pairs of triangles. The triangles at the ends (with the parallel sides as their bases) are similar, and their area ratio is the square of the ratio of those parallel sides.

Exam Tip: In trapezium problems, always look for similar triangles formed by the diagonals - the vertically opposite angles at the intersection are equal, and alternate angles help establish similarity.

Question 12(b). In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find
(i) AB
(ii) BC
(iii) area of △ADM : area of △ANB.
Answer:
(i) Given: AM = 6 cm, AN = 10 cm, and area of parallelogram ABCD = 45 cm².

The area of a parallelogram = base × height = CD × AM = BC × AN

\( AM \times CD = 45 \)

\( 6 \times CD = 45 \)

\( CD = 7.5 \) cm

Since opposite sides of a parallelogram are equal, AB = CD = 7.5 cm

(ii) Similarly, using the area formula with the second base and height:

\( BC \times AN = 45 \)

\( BC \times 10 = 45 \)

\( BC = 4.5 \) cm

(iii) In right triangle ADM, the base is DM and height is AM = 6 cm.

Area of △ADM = \( \frac{1}{2} \times DM \times AM \)

In right triangle ANB, the base is NB and height is AN = 10 cm.

Area of △ANB = \( \frac{1}{2} \times NB \times AN \)

To find DM and NB, we use the fact that in a parallelogram with side length 7.5 cm and height 6 cm from DC:

\( DM = \sqrt{AD^2 - AM^2} \)

And in a parallelogram with side length 4.5 cm and height 10 cm from CB, this is impossible since the height cannot exceed the side length. Let me recalculate:

Since AD = BC = 4.5 cm and AN = 10 cm, and AN is the perpendicular distance from A to CB extended, we have:

For △ADM: Area = \( \frac{1}{2} \times DM \times 6 \)

Since AD = 4.5 cm and AM = 6 cm would mean AM > AD (impossible for a right triangle). Let me reconsider: AM is perpendicular to DC, where DC = AB = 7.5 cm. So in the right triangle formed, DM can be found using Pythagoras if we know AD.

Since ABCD is a parallelogram, AD = BC = 4.5 cm.

\( AD^2 = AM^2 + DM^2 \)

\( 4.5^2 = 6^2 + DM^2 \) - This gives a negative value, which is impossible.

This suggests the given measurements may have an inconsistency in the original problem. However, proceeding with the area ratio approach:

\( \frac{\text{Area of } \triangle ADM}{\text{Area of } \triangle ANB} = \frac{\frac{1}{2} \times DM \times AM}{\frac{1}{2} \times NB \times AN} = \frac{DM \times 6}{NB \times 10} \)

Given the parallelogram properties and the measurements, the area ratio would be determined by the specific positions of M and N on the respective sides.

In simple words: The area of a parallelogram can be calculated using any side as the base and the perpendicular distance to the opposite side as the height. Different base-height pairs give the same area, which lets us find missing side lengths.

Exam Tip: When given a parallelogram with two different perpendicular heights, use the area formula with both base-height pairs to find the side lengths - the area remains constant regardless of which base-height pair you choose.

Question 12(c). In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(i) EF : AD
(ii) area of △BEF : area of △ABD
(iii) area of △ABD : area of trap. AEFD
(iv) area of △FEO : area of △OBC.
Answer:
(i) Looking at △ADB and △EFB, angle B is the same in both triangles. Since angle DAB and angle FEB are corresponding angles (made when a line crosses parallel lines), they are equal. Using the AA axiom, we find that △ADB and △EFB are similar figures.

For similar triangles, the ratio of matching sides stays constant. Since the ratio AE : EB = 2 : 3, we can write:

\( \frac{AE}{EB} = \frac{2}{3} \)

This means:

\( \frac{AB - EB}{EB} = \frac{2}{3} \)

Cross-multiplying: \( 3(AB - EB) = 2EB \)

\( 3AB - 3EB = 2EB \)

\( 3AB = 5EB \)

\( \frac{EB}{AB} = \frac{3}{5} \)

Since the triangles are similar:

\( \frac{EF}{AD} = \frac{EB}{AB} = \frac{3}{5} \)

Hence, EF : AD = 3 : 5.

(ii) In △ABD and △BEF, angle B appears in both. Angle DAB matches angle FEB as corresponding angles. By the AA axiom, △ABD ~ △BEF.

The ratio of areas for similar figures equals the ratio of squared sides:

\( \frac{\text{Area of } \triangle BEF}{\text{Area of } \triangle ABD} = \frac{EF^2}{AD^2} = \frac{3^2}{5^2} = \frac{9}{25} \)

Hence, area of △BEF : area of △ABD = 9 : 25.

(iii) From part (ii), the area of △ABD to the area of △BEF is in the ratio 25 : 9.

\( \frac{\text{Area of } \triangle ABD}{\text{Area of } \triangle BEF} = \frac{25}{9} \)

The trapezium AEFD is what remains when we take away △BEF from △ABD:

\( 9 \times \text{Area of } \triangle ABD = 25 \times (\text{Area of } \triangle ABD - \text{Area of trapezium AEFD}) \)

\( 9 \times \text{Area of } \triangle ABD = 25 \times \text{Area of } \triangle ABD - 25 \times \text{Area of trapezium AEFD} \)

\( 16 \times \text{Area of } \triangle ABD = 25 \times \text{Area of trapezium AEFD} \)

Hence, area of △ABD : area of trapezium AEFD = 25 : 16.

(iv) Looking at △FEO and △OBC, angle FOE matches angle BOC because they are vertically opposite angles. Angle FEO equals angle OCB as they are alternate angles. Using the AA axiom, △FEO ~ △OBC.

For similar triangles, the area ratio equals the square of the side ratio. Since opposite sides of a parallelogram are equal, BC = AD:

\( \frac{\text{Area of } \triangle FEO}{\text{Area of } \triangle OBC} = \left(\frac{EF}{BC}\right)^2 = \left(\frac{EF}{AD}\right)^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \)

Hence, area of △FEO : area of △OBC = 9 : 25.
In simple words: When lines are parallel and create similar triangles, the sides keep the same ratio. The areas of similar shapes follow the ratio of their squared sides, so if the side ratio is 3 : 5, the area ratio is 9 : 25.

Exam Tip: Always verify similarity using the AA axiom before calculating area ratios. Remember that area ratio = (side ratio)², which is the key relationship in similarity problems. Show the calculation steps clearly to earn full marks.

Question 13. In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of △CPQ = 20 cm², find
(i) area of △BPQ.
(ii) area of △CDP.
(iii) area of ||gm ABCD.
Answer:
(i) Draw QN perpendicular to CB. Both triangles △BPQ and △CPQ share the same height QN from vertex Q.

The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).

\( \frac{\text{Area of } \triangle BPQ}{\text{Area of } \triangle CPQ} = \frac{\frac{1}{2} \times BP \times QN}{\frac{1}{2} \times PC \times QN} = \frac{BP}{PC} = \frac{1}{2} \)

Since the area of △CPQ = 20 cm²:

\( \text{Area of } \triangle BPQ = \frac{1}{2} \times 20 = 10 \text{ cm}^2 \)

Hence, the area of △BPQ = 10 cm².

(ii) Now consider △CDP and △BQP. Angle CPD and angle QPB are vertically opposite, so they equal each other. Also, angle PDC and angle PQB are alternate angles (formed by line DQ cutting parallel lines DC and AB), so they are equal too. Using the AA axiom, △CDP ~ △BQP.

For similar triangles, the area ratio equals the square of the side ratio:

\( \frac{\text{Area of } \triangle BQP}{\text{Area of } \triangle CDP} = \frac{BP^2}{PC^2} = \frac{1^2}{2^2} = \frac{1}{4} \)

This means:

\( \text{Area of } \triangle CDP = 4 \times \text{Area of } \triangle BQP = 4 \times 10 = 40 \text{ cm}^2 \)

Hence, the area of △CDP = 40 cm².

(iii) The diagonal BD divides parallelogram ABCD into two equal triangles. When we extend DP to Q, triangle DCQ shares the same base DC and lies between the same pair of parallel lines (AB extended and DC). So the parallelogram and triangle DCQ together have a special relationship: the area of parallelogram ABCD equals twice the area of △DCQ.

Triangle DCQ is made up of △CDP and △CPQ:

\( \text{Area of } \triangle DCQ = \text{Area of } \triangle CDP + \text{Area of } \triangle CPQ = 40 + 20 = 60 \text{ cm}^2 \)

\( \text{Area of } \parallel\text{gm ABCD} = 2 \times 60 = 120 \text{ cm}^2 \)

Hence, the area of ||gm ABCD = 120 cm².
In simple words: Triangles with the same height can be compared by their bases. Similar triangles have areas in the ratio of their squared sides. Parallelograms can be broken into triangles using diagonals and parallel line properties.

Exam Tip: Always use the property that triangles with equal height have areas proportional to their bases. For similar triangles, the square relationship between area ratio and side ratio is essential. Show clearly how triangle areas add up to form the parallelogram area.

Question 14(a). In the figure (i) given below, DE || BC and the ratio of the areas of △ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.
Answer: Given that the ratio of areas is \( \frac{\text{Area of } \triangle ADE}{\text{Area of trapezium DBCE}} = \frac{4}{5} \).

The trapezium DBCE is the region between the two parallel lines (from △ABC minus △ADE):

\( \frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC - \text{Area of } \triangle ADE} = \frac{4}{5} \)

Cross-multiplying: \( 5 \times \text{Area of } \triangle ADE = 4(\text{Area of } \triangle ABC - \text{Area of } \triangle ADE) \)

\( 5 \times \text{Area of } \triangle ADE = 4 \times \text{Area of } \triangle ABC - 4 \times \text{Area of } \triangle ADE \)

\( 9 \times \text{Area of } \triangle ADE = 4 \times \text{Area of } \triangle ABC \)

\( \frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC} = \frac{4}{9} \)

Since DE || BC, triangles ADE and ABC share angle A. The angles at D and B are corresponding angles (on the same side of line AB cut by parallel lines), making them equal. Using the AA axiom, △ADE ~ △ABC.

For similar triangles, the area ratio is the square of the side ratio:

\( \frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC} = \frac{DE^2}{BC^2} = \frac{4}{9} \)

Taking square roots:

\( \frac{DE}{BC} = \sqrt{\frac{4}{9}} = \frac{2}{3} \)

Hence, the ratio of DE : BC is 2 : 3.
In simple words: When a line inside a triangle runs parallel to the base, it creates a smaller similar triangle. If areas are in ratio 4 : 9, then the sides are in ratio 2 : 3 (the square root of the area ratio).

Exam Tip: For parallel lines cutting through a triangle, always recognize the similar triangles formed. The area ratio of similar figures is the square of their side ratio - so always take the square root when going from area ratio to side ratio.

Question 14(b). In the figure (ii) given below, AB || DC and AB = 2DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find
(i) ED
(ii) BE
(iii) area of △EDC : area of trapezium ABCD.
Answer:
(i) Given AB = 2DC, or \( \frac{AB}{DC} = \frac{2}{1} \).

Looking at △AEB and △EDC, they share angle E. Since AB || DC, angle EAB and angle EDC are corresponding angles (cut by line AD extended), so they match. Using the AA axiom, △AEB ~ △EDC.

When two triangles are similar, their matching sides are proportional:

\( \frac{AE}{ED} = \frac{AB}{DC} = \frac{2}{1} \)

Since AE = AD + ED = 3 + ED:

\( \frac{3 + ED}{ED} = \frac{2}{1} \)

\( 3 + ED = 2 \times ED \)

\( 3 = ED \)

Hence, the length of ED = 3 cm.

(ii) Similarly, since △AEB ~ △EDC:

\( \frac{BE}{EC} = \frac{AB}{DC} = \frac{2}{1} \)

Since BE = BC + EC = 4 + EC:

\( \frac{4 + EC}{EC} = \frac{2}{1} \)

\( 4 + EC = 2 \times EC \)

\( 4 = EC \)

\( BE = BC + EC = 4 + 4 = 8 \text{ cm} \)

Hence, the length of BE = 8 cm.

(iii) For similar triangles, the area ratio is the square of the side ratio:

\( \frac{\text{Area of } \triangle EDC}{\text{Area of } \triangle AEB} = \frac{DC^2}{AB^2} = \frac{1^2}{2^2} = \frac{1}{4} \)

This means the area of △AEB is 4 times the area of △EDC. The trapezium ABCD fills the space between these two triangles:

\( \frac{\text{Area of } \triangle EDC}{\text{Area of } \triangle EDC + \text{Area of trapezium ABCD}} = \frac{1}{4} \)

Let Area of △EDC = x. Then:

\( \frac{x}{x + \text{Area of trapezium}} = \frac{1}{4} \)

\( 4x = x + \text{Area of trapezium} \)

\( 3x = \text{Area of trapezium} \)

\( \frac{\text{Area of } \triangle EDC}{\text{Area of trapezium ABCD}} = \frac{1}{3} \)

Hence, the ratio of area of △EDC : area of trapezium ABCD = 1 : 3.
In simple words: When two parallel lines are cut by two intersecting lines, the triangles formed are similar. Use the similarity ratio to find unknown lengths and area ratios using squared ratios.

Exam Tip: Recognize similar triangles formed by parallel lines immediately. For part (iii), set up the relationship between the small triangle and the trapezium carefully - the trapezium is the larger triangle minus the smaller one. Show all algebra steps for finding the final ratio.

Question 15(a). In the figure (i) given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BD = 12 cm, find
(i) BP
(ii) the ratio of areas of △APB and △DPC.
Answer:
(i) In △APB and △CPD, the angles APB and CPD are vertically opposite (where the diagonals meet at P), so they are equal. Since DC || AB, angles PAB and PCD are alternate angles (cut by line AC extended), making them equal. By the AA axiom, △APB ~ △CPD.

For similar triangles, matching sides are proportional:

\( \frac{BP}{PD} = \frac{AB}{CD} = \frac{9}{6} = \frac{3}{2} \)

Since BD = BP + PD = 12:

\( \frac{BP}{12 - BP} = \frac{9}{6} \)

Cross-multiplying: \( 6 \times BP = 9(12 - BP) \)

\( 6 \times BP = 108 - 9 \times BP \)

\( 15 \times BP = 108 \)

\( BP = \frac{108}{15} = 7.2 \text{ cm} \)

Hence, the length of BP = 7.2 cm.

(ii) For similar triangles, the area ratio equals the square of the side ratio:

\( \frac{\text{Area of } \triangle APB}{\text{Area of } \triangle DPC} = \frac{AB^2}{CD^2} = \frac{9^2}{6^2} = \frac{81}{36} = \frac{9}{4} \)

Hence, area of △APB : area of △DPC = 9 : 4.
In simple words: When diagonals of a trapezium cross, they form similar triangles. The sides of similar triangles stay in the same ratio, and their areas follow the square of that ratio.

Exam Tip: For trapezium diagonal problems, always identify the two similar triangles formed at the intersection point. Use side ratios to find unknown lengths on the diagonal, then square those ratios to get area relationships. Label the similar triangles clearly in your working.

Question 1. In the figure (ii) given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm. (i) Prove that △ACD is similar to △BCA. (ii) Find BC and CD. (iii) Find area of △ACD : area of △ABC.
Answer:
(i) Examine △ACD and △BCA. Angle C appears in both triangles (Common angles). We are told that ∠ABC = ∠DAC (Given). Therefore, by the AA axiom, △ACD ~ △BCA.

(ii) When triangles are similar, their matching sides are proportional.

\( \Rightarrow \frac{AC}{BC} = \frac{AD}{AB} \)

\( \Rightarrow \frac{4}{BC} = \frac{5}{8} \)

\( \Rightarrow BC = \frac{4 \times 8}{5} = \frac{32}{5} = 6.4 \) cm

Similarly,

\( \Rightarrow \frac{CA}{CD} = \frac{AB}{AD} \)

\( \Rightarrow \frac{4}{CD} = \frac{8}{5} \)

\( \Rightarrow CD = \frac{4 \times 5}{8} = \frac{20}{8} = 2.5 \) cm

Therefore, BC measures 6.4 cm and CD measures 2.5 cm.

(iii) For similar triangles, the ratio of areas equals the ratio of the squared lengths of matching sides.

\( \therefore \frac{\text{Area of } \triangle ACD}{\text{Area of } \triangle ABC} = \frac{AD^2}{AB^2} \)

\( \Rightarrow \frac{\text{Area of } \triangle ACD}{\text{Area of } \triangle ABC} = \frac{5^2}{8^2} = \frac{25}{64} \)

Therefore, area of △ACD : area of △ABC = 25 : 64.
In simple words: When two triangles have matching angles, their sides grow at the same rate. Use matching sides to find unknown lengths. For areas, square the side ratio.

Exam Tip: Always establish which angles match before applying the AA axiom. When finding the area ratio of similar triangles, remember to square the linear ratio - this is a frequent error point.

Question 2. In the given figure, ∠PQR = ∠PST = 90°, PQ = 5 cm and PS = 2 cm. (i) Prove that △PQR ~ △PST. (ii) Find area of △PQR : area of quadrilateral SRQT.
Answer:
(i) Examine △PQR and △PST. Angle P is shared between them (Common angles). We know that ∠PQR = ∠PST (both are 90°). Therefore, by the AA axiom, △PQR ~ △PST.

(ii) Since the triangles are similar, the ratio of their areas matches the ratio of squared matching sides.

\( \therefore \frac{\text{Area of } \triangle PQR}{\text{Area of } \triangle PST} = \frac{PQ^2}{PS^2} = \frac{5^2}{2^2} = \frac{25}{4} \)

This gives us:

\( \Rightarrow \frac{\text{Area of } \triangle PQR}{\text{Area of } \triangle PQR - \text{Area of SRQT}} = \frac{25}{4} \)

\( \Rightarrow 4 \times \text{Area of } \triangle PQR = 25 \times \text{Area of } \triangle PQR - 25 \times \text{Area of SRQT} \)

\( \Rightarrow 25 \times \text{Area of SRQT} = 25 \times \text{Area of } \triangle PQR - 4 \times \text{Area of } \triangle PQR \)

\( \Rightarrow 25 \times \text{Area of SRQT} = 21 \times \text{Area of } \triangle PQR \)

Therefore, area of △PQR : area of quadrilateral SRQT is 25 : 21.
In simple words: The big triangle is 25/4 times larger than the small triangle. The leftover space (the quadrilateral) is what remains when you remove the small triangle from the big one.

Exam Tip: When asked for the ratio of a triangle to a quadrilateral region, remember that the quadrilateral is the difference between two triangular areas - set up the subtraction equation carefully.

Question 3. ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. (i) Prove that △ADE ~ △ACB. (ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD. (iii) Find, area of △ADE : area of quadrilateral BCED.
Answer:
(i) Examine △ADE and △ACB. Angle A belongs to both triangles (Common angles). We are told that ∠AED = ∠ABC (both equal 90°). Therefore, by the AA axiom, △ADE ~ △ACB.

(ii) Since △ABC is a right angled triangle, apply the Pythagorean theorem:

\( AC^2 = AB^2 + BC^2 \)

\( AB^2 = AC^2 - BC^2 = 13^2 - 5^2 = 169 - 25 = 144 \)

\( AB = \sqrt{144} = 12 \) cm

Since the triangles are similar, their matching sides are proportional:

\( \therefore \frac{AE}{AB} = \frac{AD}{AC} \)

\( \Rightarrow \frac{4}{12} = \frac{AD}{13} \)

\( \Rightarrow AD = \frac{4 \times 13}{12} = \frac{13}{3} = 4\frac{1}{3} \) cm

Also,

\( \therefore \frac{AE}{AB} = \frac{DE}{BC} \)

\( \Rightarrow \frac{4}{12} = \frac{DE}{5} \)

\( \Rightarrow DE = \frac{4 \times 5}{12} = \frac{5}{3} = 1\frac{2}{3} \) cm

Therefore, AD = 4\(\frac{1}{3}\) cm and DE = 1\(\frac{2}{3}\) cm.

(iii) For a right angled triangle, Area = \( \frac{1}{2} \times \) Base \( \times \) Height.

\( \text{Area of } \triangle ADE = \frac{1}{2} \times AE \times DE = \frac{1}{2} \times 4 \times \frac{5}{3} = \frac{10}{3} \) cm²

\( \text{Area of } \triangle ABC = \frac{1}{2} \times BC \times AB = \frac{1}{2} \times 5 \times 12 = 30 \) cm²

\( \text{Area of quadrilateral BCED} = \text{Area of } \triangle ABC - \text{Area of } \triangle ADE = 30 - \frac{10}{3} = \frac{90 - 10}{3} = \frac{80}{3} \) cm²

\( \therefore \frac{\text{Area of } \triangle ADE}{\text{Area of quadrilateral BCED}} = \frac{\frac{10}{3}}{\frac{80}{3}} = \frac{10}{80} = \frac{1}{8} \)

Therefore, area of △ADE : area of quadrilateral BCED is 1 : 8.
In simple words: The small triangle is one-eighth the size of the four-sided shape left over. Finding one area and subtracting it from the total gives you the other area.

Exam Tip: Use the Pythagorean theorem to find the missing side first. Always check that your side ratios match the given similarity statement - the order of vertices matters.

Question 4. Two isosceles triangles have equal vertical angles and their areas are in the ratio 7 : 16. Find the ratio of their corresponding heights.
Answer: Let the two isosceles triangles be ABC and DEF. We know that ∠A = ∠D (Given, as vertical angles are equal).

Since the triangles are isosceles:

\( \angle B = \angle C = \frac{180 - \angle A}{2} \) and \( \angle E = \angle F = \frac{180 - \angle D}{2} \)

Since ∠A = ∠D, it follows that ∠B = ∠E. Therefore, by the AA axiom, △ABC ~ △DEF.

For similar triangles, the ratio of areas equals the ratio of squared matching heights:

\( \therefore \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{(\text{Height of } \triangle ABC)^2}{(\text{Height of } \triangle DEF)^2} \)

\( \Rightarrow \frac{7}{16} = \left( \frac{\text{Height of } \triangle ABC}{\text{Height of } \triangle DEF} \right)^2 \)

\( \Rightarrow \frac{\text{Height of } \triangle ABC}{\text{Height of } \triangle DEF} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \)

Therefore, the ratio of their matching heights is \( \sqrt{7} : 4 \).
In simple words: Equal vertex angles guarantee that isosceles triangles have matching angles everywhere. Take the square root of the area ratio to get the height ratio.

Exam Tip: When working with isosceles triangles, use the fact that base angles are equal to establish similarity. Always take the square root when converting an area ratio to a length ratio.

Question 5. On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements: AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate: (i) the actual length of AB in km. (ii) the area of the plot in sq. km.
Answer:
(i) The map is made to a scale of 1 : 250000.

\( \therefore K \text{ (Scale factor)} = 250000 \)

Actual length of AB = scale factor \( \times \) (length of AB on map) = 250000 \( \times \) 3 = 750000 cm

Converting to km: 1 cm = 10⁻⁵ km

\( \therefore 750000 \) cm = 750000 \( \times \) 10⁻⁵ km = 7.5 km

Therefore, the actual length of AB is 7.5 km.

(ii) The plot is a right angled triangle. Area of a right angled triangle = \( \frac{1}{2} \times \) base \( \times \) height.

Area of map model = \( \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 3 \times 4 = 6 \) cm²

We know that 1 cm² = 10⁻¹⁰ km²

\( \therefore \) Area of model = 6 \( \times \) 10⁻¹⁰ km²

Actual area of plot = (scale factor)² \( \times \) (area of model)

\( = (250000)^2 \times 6 \times 10^{-10} = 625 \times 10^8 \times 6 \times 10^{-10} = 3750 \times 10^{-2} = 37.5 \) km²

Therefore, the area of the plot is 37.5 km².
In simple words: Multiply the map measurement by the scale factor to get the real length. Square the scale factor when finding actual area.

Exam Tip: Always convert units carefully - remember that 1 cm = 10⁻⁵ km and 1 cm² = 10⁻¹⁰ km². Never forget to square the scale factor for areas and cube it for volumes.

Question 6. On a map drawn to a scale of 1 : 50000, a rectangular plot of land ABCD has the following dimensions. AB = 6 cm; BC = 8 cm. Find: (i) the actual length of the diagonal AC of the plot in km. (ii) the actual area of the plot in sq. km.
Answer:
The map is constructed at a scale of 1 : 50000.

\( \therefore k \text{ (Scale factor)} = 50000 \)

The diagonal of a rectangle can be found using the Pythagorean theorem:

\( AC = \sqrt{AB^2 + BC^2} \)

Substituting values:

\( \Rightarrow AC = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) cm

Actual length of diagonal = scale factor \( \times \) length of diagonal on map

\( = 50000 \times 10 = 500000 \) cm

Converting to km: \( = 500000 \times 10^{-5} \) km \( = 5 \) km

Therefore, actual length of diagonal = 5 km.

(ii) Area of the map model ABCD = AB \( \times \) BC \( = 6 \times 8 = 48 \) cm²

Actual area of plot = (scale factor)² \( \times \) (area of model)

\( = (50000)^2 \times 48 = 25 \times 10^8 \times 48 = 1200 \times 10^8 = 12 \times 10^{10} \) cm²

We know that 1 cm² = 10⁻¹⁰ km²

\( \therefore \) Actual area of plot = 12 \( \times \) 10¹⁰ \( \times \) 10⁻¹⁰ km² \( = 12 \) km²

Therefore, the actual area of the plot is 12 km².
In simple words: Use Pythagoras to find the diagonal on the map, then multiply by the scale factor. For area, square the scale factor before multiplying.

Exam Tip: In rectangular problems, always use the Pythagorean theorem to find diagonals. Remember the conversion: divide centimetres by 10⁵ to get kilometres, and cm² by 10¹⁰ to get km².

Question 7. A map of a square plot of land is drawn to a scale of 1 : 25000. If the area of the plot in the map is 72 cm², find: (i) the actual area of the plot of land. (ii) the length of the diagonal in the actual plot of land.
Answer:
(i) The square plot is created at a scale of 1 : 25000.

\( k \text{ (Scale factor)} = 25000 \)

Actual area of plot = (scale factor)² \( \times \) (area of map model)

Given, area of map model = 72 cm². Inserting this value:

\( = (25000)^2 \times 72 = 625000000 \times 72 = 45000000000 = 45 \times 10^9 \) cm²

We know that 1 cm² = 10⁻¹⁰ km²

\( \therefore \) Actual area of plot = 45 \( \times \) 10⁹ \( \times \) 10⁻¹⁰ km² \( = 4.5 \) km²

Therefore, the actual area of the plot is 4.5 km².

(ii) For a square, the relationship between diagonal length and area is:

\( \frac{1}{2} \times \text{(length of diagonal)}^2 = \text{area of square} \)

Using the actual area = 4.5 km²:

\( \Rightarrow \frac{1}{2} \times \text{(Length of diagonal)}^2 = 4.5 \)

\( \Rightarrow \text{(Length of diagonal)}^2 = 9 \)

\( \Rightarrow \text{Length of diagonal} = \sqrt{9} = 3 \) km

Therefore, the length of the diagonal in the actual plot of land is 3 km.
In simple words: Square the scale factor to find the real area. For a square, the diagonal squared is twice the area - rearrange to find the diagonal length.

Exam Tip: The formula \( \frac{1}{2} \text{(diagonal)}^2 = \text{area} \) comes from dividing a square into two triangles. Know this relationship for quick diagonal calculations in square problems.

Question 8. The model of a building is constructed with the scale factor 1 : 30. (i) If the height of the model is 80 cm, find the actual height of the building in metres. (ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model.
Answer:
(i) The building model is made at a scale of 1 : 30.

\( \therefore k \text{ (Scale factor)} = 30 \)

Actual height of building = scale factor \( \times \) height of model

\( = 30 \times 80 = 2400 \) cm \( = \frac{2400}{100} = 24 \) m

Therefore, the actual height of the building is 24 m.

(ii) For volumes, we must cube the scale factor. The relationship between model and actual volumes is:

\( \text{Actual volume} = (\text{scale factor})^3 \times \text{(model volume)} \)

We know the actual tank volume is 27 m³. Rearranging:

\( \text{Model volume} = \frac{\text{Actual volume}}{(\text{scale factor})^3} = \frac{27}{30^3} = \frac{27}{27000} = \frac{1}{1000} = 0.001 \) m³

Therefore, the volume of the tank on the model is 0.001 m³ or 1000 cm³.
In simple words: For height, use the scale factor once. For volume, cube the scale factor - this makes the volume difference much larger than the height difference.

Exam Tip: Scale problems involve multiplying for actual measurements (1D), squaring for areas (2D), and cubing for volumes (3D). This is a frequent source of errors - always ask yourself: what dimension am I working with?

Question 23. A model of a high rise building is made to a scale of 1 : 50.
(i) If the height of the model is 0.8 m, find the height of the actual building.
(ii) If the floor area of a flat in a building is 20 m², find the floor area of that in the model.
Answer:
(i) The model is built using a scale of 1 : 50, which means the scale factor k = 50.

Height of the actual building = k × Height of the model = 50 × 0.8 = 40 m

Therefore, the building's height is 40 m.

(ii) The floor area of a flat follows the rule: Area of the actual flat = k² × Area of the model flat

Setting up the equation with the given data:

20 = 50² × Floor area of model flat

20 = 2500 × Floor area of model flat

Floor area of model flat = 20 ÷ 2500 = 0.008 m²

Therefore, the floor area in the model is 0.008 m².
In simple words: When a model is built to a scale, you multiply the model's length by the scale factor to get the real length. For area, you multiply by the scale factor squared. For volume, you multiply by the scale factor cubed.

Exam Tip: Remember that linear dimensions (length, height) use the scale factor k directly, areas use k², and volumes use k³. Always identify which type of measurement you're working with before applying the formula.

Question 24. A model of a ship is made to a scale of 1 : 200.
(i) If the length of the model is 4 m, find the length of the ship.
(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.
(iii) If the volume of the model is 200 litres, find the volume of the ship in m³.
Answer:
(i) Given that the model is made to a scale of 1 : 200, the scale factor K = 200.

The actual length of the ship = k × (length of the model) = 200 × 4 = 800 m

Therefore, the ship's length is 800 m.

(ii) The relationship between areas is: Area of the actual deck = k² × (Area of the model deck)

Let the area of the model deck be x m².

Substituting into the formula:

160000 = (200)² × x

160000 = 40000 × x

x = 160000 ÷ 40000 = 4 m²

Therefore, the area of the ship's deck is 4 m².

(iii) The volume relationship is: Volume of the ship = k³ × (volume of the model)

Given: volume of model = 200 litres = 200 ÷ 1000 = 0.2 m³

Substituting the values:

Volume of ship = (200)³ × 0.2

= 8,000,000 × 0.2

= 1,600,000 m³

Therefore, the volume of the ship is 1,600,000 m³.
In simple words: A scale model uses the same rules every time. Lengths multiply by k, areas by k², and volumes by k³. Convert units (like litres to cubic metres) before doing calculations.

Exam Tip: Always convert all measurements to the same unit system before applying scale formulas. Common mistakes include forgetting to convert litres to m³, or using k instead of k² for area calculations.

Question 1. In the adjoining figure, △ABC ~ △QPR. Then ∠R is
(a) 60°
(b) 50°
(c) 70°
(d) 80°
Answer: (a) 60°
In simple words: When two triangles are similar, angles in the same positions match. Find the missing angle in triangle ABC by adding 70° and 50° and subtracting from 180°. That angle will be the same as angle R in triangle QPR.

Exam Tip: In similar triangles, corresponding angles are always equal. Use the angle sum property (angles in a triangle total 180°) to find unknown angles quickly.

Question 2. In the adjoining figure, △ABC ~ △QPR. The value of x is
(a) 2.25 cm
(b) 4 cm
(c) 4.5 cm
(d) 5.25 cm
Answer: (a) 2.25 cm
In simple words: When triangles are similar, the ratios of matching sides are the same. Set up an equation using these equal ratios and solve for the unknown length.

Exam Tip: Always identify which sides correspond to each other in similar triangles. Write the ratio carefully: side from first triangle over matching side from second triangle, keeping the order consistent.

Question 3. In the adjoining figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to
(a) 50°
(b) 30°
(c) 60°
(d) 100°
Answer: (d) 100°
In simple words: Check if the two triangles formed are similar by comparing side ratios and angles. If they match, use the similarity to find missing angles.

Exam Tip: When two chords intersect, check the ratios: PA/PD and PB/PC. If these are equal and there's a vertically opposite angle match, the triangles are similar. Use this to transfer angle information between triangles.

Question 4. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Answer: (b) similar but not congruent
In simple words: Two angles matching means the triangles have the same shape (they are similar). But the sides are not equal in length (AB is three times DE), so they are not congruent (which requires equal size and shape).

Exam Tip: Similarity needs matching angles; congruence needs both matching angles and equal sides. If angles match but sides differ, the triangles are similar but not congruent.

Question 5. The adjoining figure, AB || DE. The length of CD is
(a) 2.5 cm
(b) 2.7 cm
(c) \( \frac{10}{3} \) cm
(d) 3.5 cm
Answer: (b) 2.7 cm
In simple words: When two lines are parallel, they create similar triangles. The sides of similar triangles are in proportion. Set up a ratio using the matching sides to find the unknown length.

Exam Tip: Parallel lines within triangles create proportional segments. Use the basic proportionality theorem: if a line is drawn parallel to one side, it divides the other two sides proportionally.

Question 6. If △PQR ~ △ABC, PQ = 6 cm, AB = 8 cm and perimeter of △ABC is 36 cm, then perimeter of △PQR is
(a) 20.25 cm
(b) 27 cm
(c) 48 cm
(d) 64 cm
Answer: (b) 27 cm
In simple words: When triangles are similar, the ratio of their sides equals the ratio of their perimeters. If one pair of matching sides has a ratio of 6 to 8, then the perimeters also have that same ratio.

Exam Tip: For similar triangles, all corresponding measurements (sides, perimeters, heights) are in the same proportion. Set up the proportion as: side₁/side₂ = perimeter₁/perimeter₂ and solve.

Question 7. In the adjoining figure, DE || BC and all measurements are in centimetres. The length of AE is
(a) 2 cm
(b) 2.25 cm
(c) 3.5 cm
(d) 4 cm
Answer: (b) 2.25 cm
In simple words: A line parallel to one side of a triangle divides the other two sides proportionally. Use this rule to set up an equation and solve for the missing length.

Exam Tip: When DE || BC, the ratio AD/DB equals the ratio AE/EC. Write both segments in terms of known values and solve the resulting equation carefully.

Question 8. In the adjoining figure, PQ || CA and all lengths are given in centimetres. The length of BC is
(a) 6.4 cm
(b) 7.4 cm
(c) 8 cm
(d) 9 cm
Answer: (c) 8 cm
In simple words: Parallel lines create similar triangles. Matching sides of these similar triangles have equal ratios. Use this property to write an equation and find the unknown side.

Exam Tip: When PQ || CA, triangles ABC and PBQ are similar. Set up the proportion: BQ/BC = BP/BA, substituting known values and solving for the unknown.

Question 9. In the adjoining figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is
(a) 4 cm
(b) 3.6 cm
(c) 2 cm
(d) 3 cm
Answer: (d) 3 cm
In simple words: When a line is parallel to the base of a triangle, it divides the two sides proportionally. The ratio of one side's segments equals the ratio of the other side's segments.

Exam Tip: For MN || QR, use the basic proportionality theorem: PM/MQ = PN/NR. Express MQ as (PQ - PM) to create an equation with one unknown, then solve by cross-multiplication.

Question 10. It is given that △ABC ~ △PQR with \( \frac{BC}{QR} = \frac{1}{3} \), then \( \frac{\text{area of } \triangle PQR}{\text{area of } \triangle ABC} \) is equal to
(1) 9
(2) 3
(3) \( \frac{1}{3} \)
(4) \( \frac{1}{9} \)
Answer: (1) 9
In simple words: When two triangles are similar, their area ratio equals the square of their side ratio. Since the side ratio is 1:3, the area ratio is 1²:3² = 1:9, which means the first triangle's area is 1/9 of the second triangle's area.

Exam Tip: Remember that the area ratio of similar triangles always equals the square of their corresponding side ratio - this is a key theorem tested frequently in similarity questions.

Question 11. If the areas of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio
(1) 9 : 4
(2) 3 : 2
(3) 2 : 3
(4) 16 : 81
Answer: (3) 2 : 3
In simple words: To find the side ratio from an area ratio, take the square root. Since the areas are in ratio 4:9, the sides are in ratio \( \sqrt{4} : \sqrt{9} = 2 : 3 \).

Exam Tip: Always remember to reverse the operation when moving from areas to sides - if area ratio is given, find its square root to get the side ratio.

Question 12. If △ABC ~ △PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of △ABC and △PQR is
(1) 8 : 6
(2) 3 : 4
(3) 9 : 16
(4) 16 : 9
Answer: (4) 16 : 9
In simple words: The sides BC and QR are in ratio 8:6, which simplifies to 4:3. When we square this ratio for areas, we get 16:9.

Exam Tip: Make sure to square the entire side ratio - don't just square individual numbers separately, as this can lead to errors.

Question 13. If △ABC ~ △QRP, \( \frac{\text{area of } \triangle ABC}{\text{area of } \triangle PQR} = \frac{9}{4} \), AB = 18 cm and BC = 15 cm, then the length of PR is equal to
(1) 10 cm
(2) 12 cm
(3) \( \frac{20}{3} \) cm
(4) 8 cm
Answer: (1) 10 cm
In simple words: The area ratio 9:4 means the side ratio is 3:2. Since BC corresponds to PR, and BC is 15 cm, we can set up the proportion 15:PR = 3:2, which gives PR = 10 cm.

Exam Tip: Identify which sides correspond in similar triangles carefully - matching the correct pairs is essential for getting the right answer.

Question 14. If △ABC ~ △PQR, area of △ABC = 81 cm², area of △PQR = 144 cm² and QR = 6 cm, then length of BC is
(1) 4 cm
(2) 4.5 cm
(3) 9 cm
(4) 12 cm
Answer: (2) 4.5 cm
In simple words: The area ratio is 81:144, so the side ratio is \( \sqrt{81} : \sqrt{144} = 9 : 12 \), which simplifies to 3:4. Since QR = 6 cm and BC:QR = 3:4, we get BC = 4.5 cm.

Exam Tip: When you have areas and one side length, always extract the side ratio from the area ratio using square roots before setting up your proportion.

Question 15. In the adjoining figure, DE || CA and D is a point on BC such that BD : DC = 2 : 1. The ratio of area of △ABC to area of △BDE is
(1) 4 : 1
(2) 9 : 1
(3) 9 : 4
(4) 3 : 2
Answer: (3) 9 : 4
In simple words: Since BD:DC = 2:1, the total BC = 3 parts, so BD = 2 parts. The triangles BDE and BCA are similar with side ratio BD:BC = 2:3. Therefore, the area ratio is 4:9 (or reversed as 9:4 comparing ABC to BDE).

Exam Tip: For parallel line problems, use the basic proportionality theorem to establish the similarity, then apply the area ratio rule based on corresponding sides.

Question 16. If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is
(1) 2 : 1
(2) 1 : 2
(3) 1 : 4
(4) 4 : 1
Answer: (4) 4 : 1
In simple words: Since D is the midpoint of BC, side BD is half of BC. Both triangles are equilateral and therefore similar, with side ratio 1:2. The area ratio becomes 1²:2² = 1:4, so ABC:BDE = 4:1.

Exam Tip: When dealing with equilateral triangles, remember they are always similar to each other - focus on finding the correct side ratio from the given geometric information.

Question 17. The areas of two similar triangles are 81 cm² and 49 cm² respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is
(1) 9 cm
(2) 7 cm
(3) 6 cm
(4) 4.5 cm
Answer: (4) 4.5 cm
In simple words: The altitude ratio of similar triangles equals the square root of their area ratio. Area ratio is 81:49, so the altitude ratio is \( \sqrt{81} : \sqrt{49} = 9 : 7 \). If the smaller triangle's altitude is 3.5 cm, the larger triangle's altitude is 4.5 cm.

Exam Tip: For altitude problems with similar triangles, the altitude ratio behaves the same way as the side ratio - it is the square root of the area ratio.

Question 18. Given △ABC ~ △PQR, area of △ABC = 54 cm² and area of △PQR = 24 cm². If AD and PM are medians of △'s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is
(1) \( \frac{49}{9} \) cm
(2) \( \frac{20}{3} \) cm
(3) 15 cm
(4) 22.5 cm
Answer: (3) 15 cm
In simple words: For similar triangles, the median ratio equals the side ratio, which is the square root of the area ratio. Area ratio is 54:24, so median ratio is \( \sqrt{54} : \sqrt{24} \), which simplifies to 3:2. If PM = 10 cm, then AD = 15 cm.

Exam Tip: Medians of similar triangles follow the same ratio principle as sides - use the square root of the area ratio to find corresponding linear measurements.

Question 19. In the given diagram, △ABC ~ △PQR and \( \frac{AD}{PS} = \frac{3}{8} \). The value of AB : PQ is :
(1) 8 : 3
(2) 3 : 5
(3) 3 : 8
(4) 5 : 8
Answer: (3) 3 : 8
In simple words: AD and PS are altitudes of the similar triangles ABC and PQR. Since corresponding altitudes in similar triangles are in the same ratio as corresponding sides, AB:PQ = AD:PS = 3:8.

Exam Tip: In similar triangle problems involving altitudes or medians, use the principle that all linear corresponding measurements maintain the same ratio.

Question 20. In the given diagram, △ABC ~ △PQR. If AD and PS are bisectors of ∠BAC and ∠QPR respectively then:
(1) △ABC ~ △PQS
(2) △ABD ~ △PQS
(3) △ABD ~ △PSR
(4) △ABC ~ △PSR
Answer: (2) △ABD ~ △PQS
In simple words: When angle bisectors are drawn in similar triangles, they create new smaller triangles. Since ∠BAD = ∠QPS (half of equal angles) and ∠B = ∠Q (from original similarity), triangles ABD and PQS are similar by AA axiom.

Exam Tip: When angle bisectors are involved in similar triangle problems, identify which new triangles are formed and check for equality of angles using properties of the bisectors.

Question. Assertion (A): In a △ABC, if D is a point on the side BC such that AD divides BC in ratio AB : AC, then AD is the bisector of ∠A.
Reason (R): The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(4) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
In simple words: The reason statement describes the angle bisector theorem, which directly explains why the assertion is true - if a line divides the opposite side in the ratio of the adjacent sides, it must be an angle bisector.

Exam Tip: For assertion-reason questions, check both statements separately first, then determine whether one correctly justifies the other using logical reasoning.

Question. Assertion (A): In a triangle ABC, if a point D exists on side BC such that AD splits BC in the ratio AB : AC, then AD is the angle bisector of angle A.
Reason (R): The internal angle bisector of an angle in a triangle divides the opposite side internally in the ratio of the adjacent sides.
Answer: Both the assertion and the reason are correct, and the reason provides the correct explanation for the assertion. The converse of the angle bisector theorem confirms this relationship - if AD divides BC in the ratio AB : AC, then AD must bisect angle A. Since corresponding sides of similar triangles maintain equal ratios, the reason directly supports and justifies the assertion.
In simple words: If a line splits the opposite side in the same ratio as the two sides that make the angle, then that line cuts the angle in half. This is exactly what the angle bisector theorem tells us.

Exam Tip: For assertion-reason questions, always verify both statements independently first, then check if one logically leads to the other. The angle bisector theorem and its converse are key relationships to memorize.

Question. Assertion (A): When two triangles ABC and PQR are similar with area ratio 16 : 25, the ratio of their perimeters is 4 : 5.
Reason (R): In similar triangles, the ratio of their perimeters equals the ratio of their matching sides.
Answer: Both the assertion and reason are correct, and the reason provides the correct justification for the assertion. When the area ratio is 16 : 25, taking square roots gives us the side ratio as 4 : 5. Since all corresponding sides of similar triangles maintain the same ratio, the perimeter ratio must also be 4 : 5. The sum of all sides in the same proportion yields a perimeter in that same proportion.
In simple words: Areas of similar triangles relate to the square of their side ratio. Perimeters follow the same ratio as individual sides. So if area ratio is 16 : 25, the side and perimeter ratio becomes 4 : 5.

Exam Tip: Always remember the key relationships: side ratio equals perimeter ratio, while area ratio equals the square of the side ratio.

Question. Assertion (A): When triangles PQR and DEF are similar with area ratio 9 : 49, the ratio of their matching medians is 3 : 7.
Reason (R): For similar triangles, the ratio of matching medians equals the ratio of matching sides.
Answer: The reason is correct - in similar triangles, corresponding medians do maintain the same ratio as corresponding sides. However, the assertion is false. When the area ratio is 9 : 49, we take the square root to find the side ratio: \( \sqrt{\frac{9}{49}} = \frac{3}{7} \). Therefore, the median ratio should also be 3 : 7, not 4 : 9. The assertion incorrectly states the median ratio as 4 : 9, making it false even though the underlying reason is true.
In simple words: The area ratio 9 : 49 means sides and medians are in ratio 3 : 7, not 4 : 9. Taking the square root of area ratios always gives you the side and median ratios.

Exam Tip: When dealing with similar triangles, carefully distinguish between area ratios (which involve squaring) and side/median ratios (which don't). The assertion-reason format tests whether you can identify correct statements separately from correct reasoning.

Question. Assertion (A): Given triangles ABC and DEF are similar, with area of ABC = 64 cm², area of DEF = 49 cm², and BC = 4 cm, then EF must be 7 cm.
Reason (R): The ratio of areas of two similar triangles equals the ratio of the squares of their matching sides.
Answer: The reason is correct - this is a fundamental property of similar triangles. However, the assertion is false. Using the area ratio relationship: \( \frac{\text{Area ABC}}{\text{Area DEF}} = \frac{BC^2}{EF^2} \), we get \( \frac{64}{49} = \frac{16}{EF^2} \). Solving: \( EF^2 = \frac{49 \times 16}{64} = \frac{784}{64} = \frac{49}{4} \), so \( EF = \frac{7}{2} = 3.5 \) cm, not 7 cm. The calculation error in the assertion makes it incorrect despite the reason being valid.
In simple words: Even though the rule about area ratios and side ratios is correct, the actual length of EF comes out to be 3.5 cm when you do the math, not 7 cm.

Exam Tip: Always work through the complete calculation when an assertion gives a specific numerical answer. Check your algebra carefully - one arithmetic mistake can flip a true assertion into a false one.

Chapter Test

Question 1. In the adjoining figure, angle 1 = angle 2 and angle 3 = angle 4. Show that PT × QR = PR × ST.
Answer: Given that angle 1 equals angle 2, we add angle QPT to both sides: angle 1 + angle QPT = angle 2 + angle QPT. This simplifies to angle SPT = angle QPR. We also know that angle PST = angle PQR because angle 3 = angle 4. By the AA (Angle-Angle) similarity criterion, triangles PQR and PST are similar. Since the triangles are similar, their corresponding sides are proportional: \( \frac{PR}{PT} = \frac{QR}{ST} \). Cross-multiplying yields PT × QR = PR × ST, which completes the proof.
In simple words: When you have matching angles in two triangles, those triangles have the same shape. When shapes match like this, their sides are proportional. Rearranging the proportion gives you the final result.

Exam Tip: For similarity proofs, identify at least two pairs of equal angles to apply the AA criterion. Always state the similarity clearly before writing the proportion of sides.

Question 2. In the adjoining figure, AB = AC. If PM is perpendicular to AB and PN is perpendicular to AP, show that PM × PC = PN × PB.
Answer: We examine triangles PNC and PMB. Both angle PNC and angle PMB equal 90 degrees (given perpendicularity). Since AB = AC (given), angle NCP = angle PBM. By the AA similarity criterion, triangle PNC is similar to triangle PMB. Since the triangles are similar, their corresponding sides are proportional: \( \frac{PC}{PB} = \frac{PN}{PM} \). Cross-multiplying gives PC × PM = PN × PB, completing the required proof.
In simple words: Two right triangles with one matching angle pair will always have the same shape. When shapes match, side ratios must be equal. Setting up the proportion correctly and cross-multiplying gives the final answer.

Exam Tip: In perpendicularity problems, always identify the right angles first. These are your strongest angle matches for similarity arguments.

Question 3(a). In the figure given below, angle AED = angle ABC. Find the values of x and y.
Answer: We consider triangles ABC and ADE. Since angle AED = angle ABC (given) and angle A = angle A (common angle), by AA similarity, triangle ABC is similar to triangle ADE. From similarity, corresponding sides are proportional: \( \frac{AD}{AC} = \frac{DE}{BC} \). Substituting the known values: \( \frac{3}{4+2} = \frac{y}{10} \), which gives \( \frac{3}{6} = \frac{y}{10} \). Solving yields y = 5. Next, using the other proportion: \( \frac{AB}{AE} = \frac{BC}{DE} \), we get \( \frac{3+x}{4} = \frac{10}{5} \). This simplifies to \( 3 + x = 8 \), giving x = 5.
In simple words: When two triangles are similar, matching sides divide each other in the same proportion. Set up two ratios using the known values, then solve each equation step by step.

Exam Tip: Carefully identify which sides correspond to each other in similar triangles. Labeling the ratios correctly before substituting values prevents many algebra mistakes.

Question 3(b). In the figure given below, medians BE and CF of triangle ABC meet at G. Prove that: (i) Triangle FGE is similar to triangle CGB, and (ii) BG = 2GE.
Answer: (i) Examining triangles FGE and CGB: angle FGE = angle BGC (vertically opposite angles are equal), and angle GFE = angle GCB (alternate angles are equal). By the AA criterion, triangle FGE is similar to triangle CGB. (ii) Next, we look at triangles AFE and ABC. Angle A = angle A (common), and angle AFE = angle ABC (corresponding angles are equal). By AA similarity, triangle AFE is similar to triangle ABC. Since BE is a median of AC, we have AE = EC, so AC = 2AE. From the similarity of AFE and ABC: \( \frac{FE}{BC} = \frac{AE}{AC} = \frac{1}{2} \). From the similarity of FGE and CGB: \( \frac{FE}{BC} = \frac{GE}{BG} = \frac{1}{2} \). Therefore, BG = 2GE, completing the proof.
In simple words: Vertically opposite angles and alternate angles are always equal in geometric figures. When these angle pairs match between triangles, similarity follows. Using known median properties and similarity ratios, you can prove the length relationship.

Exam Tip: Median problems frequently involve the centroid and its property that it divides each median in a 2:1 ratio from vertex to opposite side. Use similarity to establish this fundamental relationship carefully.

Question 4. In the given figure, P is a point on AB such that PB : AP = 3 : 4 and PQ || AC.
(i) Calculate PQ : AC.
(ii) If AR ⊥ CP, QS ⊥ CB and QS = 6 cm, calculate the length of AR.
Answer:
(i) We are given that AP : PB = 4 : 3. Since PQ is parallel to AC, we apply the Basic Proportionality Theorem, which tells us that the sides are cut proportionally.

From the theorem: \( \frac{PB}{AP} = \frac{QB}{CQ} \)

Substituting: \( \frac{QB}{CQ} = \frac{3}{4} \)

This means: \( \frac{BQ}{BC - BQ} = \frac{3}{4} \)

Cross-multiplying: \( 3(BC - BQ) = 4BQ \)

\( 3BC - 3BQ = 4BQ \)

\( 3BC = 7BQ \)

\( \frac{BQ}{BC} = \frac{3}{7} \)

Now, looking at the similar triangles △PBQ and △ABC: they share angle B, and since PQ || AC, corresponding angles are equal (∠QPB = ∠CAB and ∠PQB = ∠ACB). By AA similarity, the triangles are similar, so their corresponding sides are proportional.

\( \frac{PQ}{AC} = \frac{BQ}{BC} = \frac{3}{7} \)

Therefore, PQ : AC = 3 : 7.

(ii) We now examine triangles △ARC and △QSP. We know ∠ARC = ∠QSP = 90° (both are right angles). Also, ∠ACR = ∠SPQ (these are alternate angles, since AB || the relevant lines). By AA similarity, △ARC ~ △QSP.

For similar triangles, corresponding sides are proportional:

\( \frac{AR}{QS} = \frac{AC}{PQ} \)

From part (i), we found \( \frac{AC}{PQ} = \frac{7}{3} \)

So: \( AR = \frac{AC}{PQ} \times QS = \frac{7}{3} \times 6 = 7 \times 2 = 14 \)

Therefore, the length of AR = 14 cm.
In simple words: When a line parallel to one side of a triangle cuts the other sides, it divides them in the same ratio. This helps us find that PQ is 3/7 of AC. Then we use similar triangles with right angles to find that AR equals 14 cm.

Exam Tip: Always identify parallel lines and use the Basic Proportionality Theorem. Look for similar triangles formed by these parallel lines - the ratio of corresponding sides in similar triangles equals the ratio of their squared sides for area comparisons.

Question 5. In a △ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Answer: The triangle ABC has points D on AB and E on AC, with DE parallel to BC.

Since DE is parallel to BC, triangles ABC and ADE satisfy the conditions for similarity. They share angle A, and since the sides are parallel, ∠ADE = ∠ABC and ∠AED = ∠ACB (corresponding angles). By AA similarity, △ABC ~ △ADE.

For similar triangles, the ratios of corresponding sides are equal:

\( \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} \)

First, find AB using the ratio \( \frac{AD}{AB} = \frac{DE}{BC} \):

\( \frac{2.4}{AB} = \frac{2}{5} \)

\( AB = \frac{2.4 \times 5}{2} = \frac{12}{2} = 6 \) cm

Next, find AC using \( \frac{AE}{AC} = \frac{DE}{BC} \):

\( \frac{3.2}{AC} = \frac{2}{5} \)

\( AC = \frac{3.2 \times 5}{2} = \frac{16}{2} = 8 \) cm

Now we can calculate BD and CE:

\( BD = AB - AD = 6 - 2.4 = 3.6 \) cm

\( CE = AC - AE = 8 - 3.2 = 4.8 \) cm

Therefore, BD = 3.6 cm and CE = 4.8 cm.
In simple words: When a line inside a triangle is parallel to one side, it cuts the other two sides into parts that match the same ratio. Using this property, we found the total lengths AB and AC, then subtracted to get BD and CE.

Exam Tip: Always check that the parallel lines create similar triangles - this is the key to solving these problems. Use the Basic Proportionality Theorem to set up your ratio equations before solving for unknowns.

Question 6. In a △ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm. Is DE || BC? Justify your answer.
Answer: First, we find EC by subtracting AE from AC:

\( EC = AC - AE = 8.8 - 3.3 = 5.5 \) cm

To check if DE is parallel to BC, we apply the converse of the Basic Proportionality Theorem: if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

Calculate \( \frac{AD}{DB} \):

\( \frac{AD}{DB} = \frac{5.7}{9.5} = \frac{57}{95} = \frac{3}{5} \)

Calculate \( \frac{AE}{EC} \):

\( \frac{AE}{EC} = \frac{3.3}{5.5} = \frac{33}{55} = \frac{3}{5} \)

Since \( \frac{AD}{DB} = \frac{AE}{EC} = \frac{3}{5} \), the line DE divides sides AB and AC in the same ratio. By the converse of the Basic Proportionality Theorem, DE || BC.
In simple words: If a line cuts two sides of a triangle into equal ratios, then that line must be parallel to the third side. Here both ratios equal 3/5, so DE is parallel to BC.

Exam Tip: When asked to prove or disprove parallelism, always check if the ratios are equal - this is the most direct method. Simplify fractions completely to avoid errors in comparison.

Question 7. If the areas of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.
Answer: Let the corresponding side of the second triangle be x cm. For similar triangles, there is an important relationship: the ratio of their areas equals the ratio of the squares of their corresponding sides.

\( \frac{\text{Area of first triangle}}{\text{Area of second triangle}} = \frac{(\text{Side of first triangle})^2}{(\text{Side of second triangle})^2} \)

Substitute the known values:

\( \frac{360}{250} = \frac{8^2}{x^2} \)

\( \frac{360}{250} = \frac{64}{x^2} \)

Cross-multiply to solve for x²:

\( x^2 = \frac{64 \times 250}{360} = \frac{16000}{360} = \frac{1600}{36} \)

Taking the square root:

\( x = \sqrt{\frac{1600}{36}} = \frac{40}{6} = \frac{20}{3} = 6\frac{2}{3} \) cm

Therefore, the length of the corresponding side of the second triangle is 6⅔ cm (or 20/3 cm).
In simple words: When two triangles have the same shape, their area ratio is the square of their side ratio. So we set up a proportion, square both sides, and solve.

Exam Tip: Remember the key formula: area ratio = (side ratio)². Always square the side ratio when comparing areas, and take the square root when working backwards from areas to sides.

Question 8. In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find (i) BC (ii) DC (iii) area of △ACD : area of △BCA
Answer:
(i) Consider triangles ABC and ACD. Angle C appears in both triangles (common angle). We are also given that ∠ABC = ∠CAD. By AA similarity, △ABC ~ △ACD. Therefore, the ratios of corresponding sides are equal:

\( \frac{AB}{AD} = \frac{BC}{AC} \)

\( \frac{5}{4} = \frac{BC}{3} \)

\( BC = \frac{5 \times 3}{4} = \frac{15}{4} = 3.75 \) cm

(ii) Since △ABC and △ACD are similar, we have another proportion from corresponding sides:

\( \frac{AB}{AD} = \frac{AC}{DC} \)

\( \frac{5}{4} = \frac{3}{DC} \)

\( DC = \frac{3 \times 4}{5} = \frac{12}{5} = 2.4 \) cm

(iii) For similar triangles, the ratio of their areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of △BCA}}{\text{Area of △ACD}} = \frac{AB^2}{AD^2} = \frac{5^2}{4^2} = \frac{25}{16} \)

Therefore, area of △ACD : area of △BCA = 16 : 25.
In simple words: When two triangles share an angle and have one pair of equal angles, they are similar. Their sides match the same ratio, and their areas are in the ratio of the squares of those sides.

Exam Tip: Identify the common angle and the given equal angles carefully - this confirms similarity. Then use the similarity property to find all unknown sides before comparing areas.

Question 9. In the adjoining figure, the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of △AOE : area of ||gm ABCD.
Answer: In parallelogram ABCD, the diagonals AC and BD meet at O. We draw OE parallel to CB to meet AB at E.

Key property: In a parallelogram, the diagonals bisect each other. So O is the midpoint of both diagonals. Also, the diagonals divide the parallelogram into four triangles of equal area, each having area equal to 1/4 of the parallelogram. Therefore, area of △OAB = (1/4) × area of parallelogram ABCD.

Since OE || CB and ABCD is a parallelogram (where opposite sides are parallel), we know that OE || AD as well. This means that E divides AB into two equal parts, making E the midpoint of AB.

The line OE acts as a median of triangle AOB - it connects a vertex (O) to the midpoint of the opposite side (E is midpoint of AB). A median divides a triangle into two triangles of equal area. Therefore:

\( \text{Area of △AOE} = \frac{1}{2} \times \text{Area of △AOB} = \frac{1}{2} \times \frac{1}{4} \times \text{Area of parallelogram ABCD} = \frac{1}{8} \times \text{Area of parallelogram ABCD} \)

Therefore, area of △AOE : area of ||gm ABCD = 1 : 8.
In simple words: The parallelogram is split into 4 equal triangles by its diagonals. Triangle AOB is one of these 4 parts. The parallel line OE cuts AOB in half. So AOE is 1/8 of the whole parallelogram.

Exam Tip: Remember that diagonals of a parallelogram bisect each other and create four equal triangles. A median always divides a triangle into two parts of equal area - use this property when a line connects a vertex to the midpoint of the opposite side.

Question 10. In the adjoining figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of △AOB and △COD.
Answer: We are given that 2AB = 3DC, which means \( \frac{AB}{DC} = \frac{3}{2} \).

Examine triangles AOB and COD. The angles ∠AOB and ∠COD are vertically opposite angles at point O where the diagonals intersect, so they are equal. Also, ∠OAB and ∠OCD are alternate angles (since AB || DC), so they are equal. By AA similarity, △AOB ~ △COD.

For similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of △AOB}}{\text{Area of △COD}} = \frac{AB^2}{DC^2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \)

Therefore, area of △AOB : area of △COD = 9 : 4.
In simple words: The parallel sides of the trapezium create similar triangles in the middle. Since AB is 3/2 times larger than DC, the area of triangle AOB is (3/2)² = 9/4 times larger.

Exam Tip: When a trapezium has its diagonals drawn, the two triangles formed from the parallel sides are similar. The ratio of their areas is always the square of the ratio of the parallel sides.

Question 11. In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that (i) DO : OE = 2 : 1 (ii) area of △OEC : area of △OAD = 1 : 4
Answer:
(i) E is the midpoint of BC, so BC = 2EC. Since ABCD is a parallelogram, opposite sides are equal: BC = AD. Therefore, AD = 2EC.

Now examine triangles AOD and EOC. The angles ∠AOD and ∠EOC are vertically opposite angles where the lines intersect, so they are equal. Also, ∠OAD and ∠OCE are alternate angles (since AD || BC in the parallelogram), so they are equal. By AA similarity, △AOD ~ △EOC.

For similar triangles, corresponding sides are proportional:

\( \frac{DO}{OE} = \frac{AD}{EC} = \frac{2EC}{EC} = \frac{2}{1} \)

Therefore, DO : OE = 2 : 1. (Proved)

(ii) From part (i), we established that △AOD ~ △EOC. For similar triangles, the area ratio equals the ratio of the squares of corresponding sides:

\( \frac{\text{Area of △OEC}}{\text{Area of △OAD}} = \frac{OE^2}{DO^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)

Therefore, area of △OEC : area of △OAD = 1 : 4. (Proved)
In simple words: Since E is the midpoint, EC is half of AD. The similar triangles have sides in ratio 1:2, so their areas are in ratio 1:4 (because we square the ratio).

Exam Tip: When a parallelogram problem involves a midpoint, use the equality of opposite sides carefully. Similar triangles formed by intersecting lines inside the parallelogram follow the standard area ratio = (side ratio)² rule.

Question 12. In the given diagram △ADB and △ACB are two right angled triangles with ∠ADB = ∠BCA = 90°. If AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm (a) Prove that △APD ~ △BPC. (b) Find the length of BD and PB (c) Hence, find the length of PA (d) Find area △APD : area △BPC
Answer:
(a) Both triangles APD and BPC have a right angle: ∠ADP = 90° (given) and ∠BCP = 90° (given). Also, ∠APD and ∠BPC are vertically opposite angles (they share the same vertex at P on line segment AB), so they are equal. By AA similarity, △APD ~ △BPC. (Proved)

(b) In right triangle ADB with ∠ADB = 90°, use the Pythagorean theorem to find BD:

\( AB^2 = AD^2 + BD^2 \)

\( 10^2 = 6^2 + BD^2 \)

\( 100 = 36 + BD^2 \)

\( BD^2 = 64 \)

\( BD = 8 \) cm

In right triangle ACB with ∠BCA = 90°, use the Pythagorean theorem to find AC:

\( AB^2 = AC^2 + BC^2 \)

\( 10^2 = AC^2 + 2.4^2 \)

\( 100 = AC^2 + 5.76 \)

\( AC^2 = 94.24 \)

\( AC = 9.71 \) cm (approximately)

Since triangles APD and BPC are similar, corresponding sides are proportional:

\( \frac{AP}{BP} = \frac{AD}{BC} = \frac{6}{2.4} = \frac{5}{2} \)

Also, AP + BP = AB = 10. Setting AP = 5k and BP = 2k:

\( 5k + 2k = 10 \)

\( k = \frac{10}{7} \)

\( BP = 2k = \frac{20}{7} \approx 2.86 \) cm

(c) From above, \( AP = 5k = \frac{50}{7} \approx 7.14 \) cm

(d) For similar triangles, the area ratio equals the square of the side ratio:

\( \frac{\text{Area of △APD}}{\text{Area of △BPC}} = \left(\frac{AD}{BC}\right)^2 = \left(\frac{6}{2.4}\right)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \)

Therefore, area △APD : area △BPC = 25 : 4.
In simple words: Two right triangles overlap at point P. They share a right angle and have vertically opposite angles, making them similar. Using the Pythagorean theorem and the similarity ratio 5:2, we find all the missing lengths, and the area ratio is the square of this: 25:4.

Exam Tip: When two right triangles share a vertex, always check for vertically opposite angles and other angle equalities. Use the Pythagorean theorem to find missing sides in right triangles, and remember that the area ratio of similar triangles is always the square of the linear ratio.

Question 12(a). In △APD and △BPC, prove the triangles are similar.
Answer: In △APD and △BPC:

∠APD = ∠BPC (vertically opposite angles are always equal)

∠ADP = ∠BCP (both measure 90°)

Since two angles of one triangle match two angles of the other triangle, △APD ∼ △BPC by the AA (Angle-Angle) similarity criterion.
In simple words: When two angles in one triangle are the same as two angles in another triangle, the triangles must have the same shape. That makes them similar.

Exam Tip: Always name corresponding angles clearly and state the similarity criterion (AA, SAS, or SSS) — just listing equal angles without naming the criterion loses marks.

Question 12(b). Find BD and PB using the Pythagorean theorem in △ADB.
Answer: In △ADB, applying the Pythagorean theorem:

\( AB^2 = AD^2 + BD^2 \)

\( 10^2 = 6^2 + BD^2 \)

\( 100 = 36 + BD^2 \)

\( BD^2 = 64 \)

\( BD = \sqrt{64} = 8 \text{ cm} \)

Now, since PB = BD - PD:

\( PB = 8 - 4.5 = 3.5 \text{ cm} \)

Therefore, BD = 8 cm and PB = 3.5 cm.
In simple words: The Pythagorean theorem helps us find BD by squaring and subtracting. Then we find PB by taking away PD from the full length BD.

Exam Tip: Always show each step of the calculation clearly, especially when simplifying square roots — examiners check that you can handle the arithmetic correctly.

Question 12(c). Find the length of AP using the Pythagorean theorem in △APD.
Answer: In △APD, applying the Pythagorean theorem:

\( AP^2 = AD^2 + DP^2 \)

\( AP^2 = 6^2 + (4.5)^2 \)

\( AP^2 = 36 + 20.25 \)

\( AP^2 = 56.25 \)

\( AP = \sqrt{56.25} = 7.5 \text{ cm} \)

Hence, the length of AP is 7.5 cm.
In simple words: We square both the sides AD and DP, add them together, then take the square root to get AP.

Exam Tip: Be careful with decimal calculations when squaring — verify your arithmetic by checking that 7.5² = 56.25 before finalizing your answer.

Question 12(d). Find the ratio of the areas of △APD and △BPC.
Answer: For similar triangles, the ratio of their areas equals the square of the ratio of their matching sides.

\( \frac{\text{Area of } \triangle APD}{\text{Area of } \triangle BPC} = \frac{AD^2}{BC^2} \)

\( = \frac{6^2}{(2.4)^2} \)

\( = \frac{36}{5.76} \)

\( = \frac{6 \times 6}{2.4 \times 2.4} \)

\( = \frac{1 \times 1}{0.4 \times 0.4} \)

\( = \frac{10 \times 10}{4 \times 4} \)

\( = \frac{100}{16} \)

\( = \frac{25}{4} \)

Therefore, Area △APD : Area △BPC = 25 : 4.
In simple words: When triangles are similar, their areas are in the same ratio as the square of any matching pair of sides.

Exam Tip: Always simplify fraction ratios fully before writing the final ratio form — examiners expect the simplest whole number ratio, not a fraction.

Question 13. A model of a ship is made to a scale of 1 : 250. Calculate: (i) the length of the ship, if the length of model is 1.6 m. (ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m². (iii) the volume of the model, if the volume of the ship is 1 km³.
Answer:

(i) The model is made at a scale of 1 : 250, which means the scale factor k = 250.

Actual length of ship = k × (length of model)

\( = 250 \times 1.6 = 400 \text{ m} \)

The ship's length is 400 m.

(ii) For areas, we use the square of the scale factor:

Area of ship's deck = k² × (area of model deck)

\( = (250)^2 \times 2.4 \)

\( = 250 \times 250 \times 2.4 \)

\( = 1,50,000 \text{ m}^2 \)

The ship's deck area is 1,50,000 m².

(iii) For volumes, we use the cube of the scale factor.

Given: Volume of ship = 1 km³ = (1000)³ m³

Let volume of model = x m³

Using the relationship: Volume of ship = k³ × (volume of model)

\( 1000^3 = 250^3 \times x \)

\( x = \frac{1000 \times 1000 \times 1000}{250 \times 250 \times 250} \)

\( x = 4 \times 4 \times 4 \)

\( x = 64 \)

The model's volume is 64 m³.
In simple words: When a model is made at a scale, lengths multiply by the scale factor, areas multiply by the scale factor squared, and volumes multiply by the scale factor cubed.

Exam Tip: Remember the three formulas: length uses k, area uses k², and volume uses k³ - forgetting to square or cube the scale factor is a common mistake that costs marks.