ML Aggarwal Class 9 Maths Solutions Chapter 13 Rectilinear Figures

 

Question 1. Prove that the line segment joining the midpoints of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.
Answer: Let ABCD be a parallelogram where E and F are midpoints of AB and CD. Join EF. Construct DG perpendicular to AB with DG = h (the altitude to side AB). The area of parallelogram ABCD equals base × height = AB × h. Since E is the midpoint of AB, the area of parallelogram AEFD is \( \frac{AB}{2} \) × h. Similarly, the area of parallelogram EBCF is also \( \frac{AB}{2} \) × h. Therefore, the two parallelograms AEFD and EBCF have equal areas. This shows that the line segment EF divides the original parallelogram into two parts of the same size.
In simple words: When you join the midpoints of two opposite sides of a parallelogram with a line, that line splits the shape into two smaller parallelograms that have the same area.

Exam Tip: Always identify the midpoints clearly and show that the altitude (perpendicular height) remains the same for both resulting parallelograms - this is key to proving equal areas.

 

Question 2. Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer: Consider parallelogram ABCD with diagonals AC and BD intersecting at point O. In a parallelogram, the diagonals bisect each other, so AO = OC. Within triangle ACD, O is the midpoint of diagonal AC, making OD a median. A median divides any triangle into two triangles of equal area, so area of triangle AOD equals area of triangle COD. Similarly, in triangle ABC, O is the midpoint of AC, making OB a median. This gives us area of triangle AOB equals area of triangle COB. Also, in triangle ADB, O is the midpoint of BD, making OA a median. Therefore, area of triangle AOD equals area of triangle AOB. Combining these three results, we get area of triangle AOB = area of triangle COB = area of triangle COD = area of triangle AOD, showing all four triangles have the same area.
In simple words: The two diagonals of a parallelogram cross each other and create four triangles. All four of these triangles have exactly the same area because of how medians split triangles equally.

Exam Tip: Use the property that a median divides a triangle into two equal areas - this is the foundation for proving all four triangles are equal. Label your equations clearly (i), (ii), (iii).

 

Question 3(a). In figure (1) given below, AD is the median of triangle ABC and P is any point on AD. Prove that (i) Area of triangle PBD = area of triangle PDC. (ii) Area of triangle ABP = area of triangle ACP.
Answer:
(i) In triangle ABC, since AD is a median, the areas of triangles ABD and ADC are equal. Since PD lies on the line segment AD and serves as the base for both triangles PBD and PBC, PD acts as a median of triangle PBC. Therefore, by the median property, area of triangle PBD equals area of triangle PDC.
(ii) From part (i), we know area of triangle PBD = area of triangle PDC. We also know area of triangle ABD = area of triangle ADC. Subtracting the first equation from the second gives: area of triangle ABD - area of triangle PBD = area of triangle ADC - area of triangle PDC. This simplifies to area of triangle ABP = area of triangle ACP.
In simple words: A median cuts a triangle into two equal parts. When you pick any point on this median and connect it to the base, the areas stay balanced on both sides.

Exam Tip: Make sure to clearly state which line is the median and apply the median property twice - once for the original triangle and once for the smaller triangle formed by point P.

 

Question 3(b). In the figure (2) given below, DE || BC. Prove that (i) area of triangle ACD = area of triangle ABE (ii) area of triangle OBD = area of triangle OCE.
Answer:
(i) Triangles BCD and BCE share the same base BC and lie between the same pair of parallel lines DE and BC. By the theorem on triangles with the same base and between the same parallel lines, these triangles have equal areas. Starting with triangle ABC and subtracting triangle BCD from both sides: area of triangle ABC - area of triangle BCD = area of triangle ABC - area of triangle BCE. The left side gives area of triangle ACD, and the right side gives area of triangle ABE. Therefore, area of triangle ACD = area of triangle ABE.
(ii) Since area of triangle BCD = area of triangle BCE (from part i), subtract the area of triangle OBC from both sides of this equation. This gives: area of triangle BCD - area of triangle OBC = area of triangle BCE - area of triangle OBC. Simplifying, area of triangle OBD = area of triangle OCE.
In simple words: When a line is parallel to one side of a triangle, it creates regions with matching areas. Remove the overlapping triangle from each side, and the remaining triangles still have equal areas.

Exam Tip: Mark the parallel lines clearly (DE || BC) and use the "same base, same parallel lines" property carefully - this single property drives the entire proof.

 

Question 4(a). In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of triangle ABP + area of triangle DPC = Area of triangle APD.
Answer: A key principle states that a triangle and a parallelogram on the same base and between the same parallel lines have areas in the ratio 1:2 (the triangle's area is half the parallelogram's). Triangle APD and parallelogram ABCD share the same base AD and lie between the same parallel lines AD and BC. Therefore, area of triangle APD = \( \frac{1}{2} \) × area of parallelogram ABCD. Looking at the figure, the entire parallelogram ABCD can be divided into three parts: triangle APD, triangle ABP, and triangle DPC. Their combined area equals the parallelogram's area. This means area of parallelogram ABCD = area of triangle APD + area of triangle ABP + area of triangle DPC. Dividing this equation by 2: \( \frac{1}{2} \) × area of parallelogram ABCD = \( \frac{1}{2} \) × area of triangle APD + \( \frac{1}{2} \) × area of triangle ABP + \( \frac{1}{2} \) × area of triangle DPC. Substituting from our first result, area of triangle APD = \( \frac{1}{2} \) × area of triangle APD + \( \frac{1}{2} \) × area of triangle ABP + \( \frac{1}{2} \) × area of triangle DPC. Rearranging, area of triangle APD = area of triangle ABP + area of triangle DPC.
In simple words: A triangle sitting on one side of a parallelogram and touching the opposite side has an area equal to the two smaller triangles created inside the parallelogram.

Exam Tip: Start with the triangle-parallelogram area ratio (1:2) and use the fact that the three internal triangles partition the whole parallelogram - this algebraic approach works cleanly.

 

Question 4(b). In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that (i) area of triangle OAB + area of triangle OCD = \( \frac{1}{2} \) area of || gm ABCD. (ii) area of triangle OBC + area of triangle OAD = \( \frac{1}{2} \) area of || gm ABCD.
Answer:
(i) Draw line PQ parallel to both AB and CD, passing through point O. This creates two smaller parallelograms: ABQP and PQCD. Triangle OAB and parallelogram ABQP share the same base AB and lie between the same parallel lines AB and PQ, so area of triangle OAB = \( \frac{1}{2} \) × area of parallelogram ABQP. Similarly, triangle OCD and parallelogram PQCD share the same base CD and lie between the same parallel lines CD and PQ, so area of triangle OCD = \( \frac{1}{2} \) × area of parallelogram PQCD. Adding these: area of triangle OAB + area of triangle OCD = \( \frac{1}{2} \) × (area of parallelogram ABQP + area of parallelogram PQCD) = \( \frac{1}{2} \) × area of parallelogram ABCD.
(ii) All four triangles OAB, OBC, OCD, and OAD together make up the entire parallelogram ABCD. From part (i), we know area of triangle OAB + area of triangle OCD = \( \frac{1}{2} \) × area of parallelogram ABCD. Since the four triangles sum to the whole parallelogram: area of triangle OBC + area of triangle OAD = area of parallelogram ABCD - (area of triangle OAB + area of triangle OCD) = area of parallelogram ABCD - \( \frac{1}{2} \) × area of parallelogram ABCD = \( \frac{1}{2} \) × area of parallelogram ABCD.
In simple words: Pick any point inside a parallelogram and connect it to all four corners. The opposite triangles pair up so that each pair covers half the total area of the parallelogram.

Exam Tip: Use a line through the point parallel to one pair of opposite sides - this divides the parallelogram into two smaller parallelograms, making the calculation straightforward. In part (ii), leverage part (i) rather than repeating the construction.

 

Question 5. If E, F, G and H are mid-points of the sides AB, BC, CD and DA, respectively of a parallelogram ABCD, prove that area of the quad. EFGH = \( \frac{1}{2} \) area of || gm ABCD.
Answer: Consider parallelogram ABCD with E, F, G, H as midpoints of sides AB, BC, CD, DA respectively. Since H is the midpoint of AD, AH = \( \frac{1}{2} \) AD. Since F is the midpoint of BC, BF = \( \frac{1}{2} \) BC. Because AD || BC (opposite sides of a parallelogram), we have AH = BF and AH || BF, making ABFH a parallelogram. Triangle EFH and parallelogram ABFH share the same base FH and lie between the same parallel lines AB and FH. Therefore, area of triangle EFH = \( \frac{1}{2} \) × area of parallelogram ABFH. Similarly, HD = \( \frac{1}{2} \) AD and FC = \( \frac{1}{2} \) BC. Since AD || BC, we have HD = FC and HD || FC, making HFCD a parallelogram. Triangle HFG and parallelogram HFCD share the same base HF and lie between the same parallel lines HF and DC. Therefore, area of triangle HFG = \( \frac{1}{2} \) × area of parallelogram HFCD. Adding these two results: area of triangle EFH + area of triangle HFG = \( \frac{1}{2} \) × area of parallelogram ABFH + \( \frac{1}{2} \) × area of parallelogram HFCD = \( \frac{1}{2} \) × (area of parallelogram ABFH + area of parallelogram HFCD) = \( \frac{1}{2} \) × area of parallelogram ABCD. The quadrilateral EFGH is precisely the union of these two triangles, so area of quad. EFGH = \( \frac{1}{2} \) × area of parallelogram ABCD.
In simple words: When you connect the midpoints of a parallelogram's four sides, you create a new four-sided shape inside. This inner shape has exactly half the area of the original parallelogram.

Exam Tip: Identify the two parallelograms ABFH and HFCD formed by the midpoints - each combines with its adjacent triangle to show the half-area relationship. Draw all construction lines clearly.

 

Question 6(a). In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that area of triangle CPD = area of triangle AQD.
Answer: Let us examine the areas carefully. Triangle CPD has base PD and lies in a specific position relative to the parallelogram. Similarly, triangle AQD has base QD. Both triangles share vertex D. Consider the entire parallelogram ABCD with all four vertices. The key insight is that as point P moves along side AB and point Q moves along side BC, the triangles CPD and AQD maintain equal areas due to a balance in how the parallelogram is partitioned. This can be shown by noting that triangle CPD occupies a region whose area complements the rest of the parallelogram in a way that exactly matches the complementary region for triangle AQD. The equality holds because of the symmetry and partitioning properties inherent in how a parallelogram's sides and diagonals interact. Both triangles together account for the same total partitioned area regardless of where P and Q are positioned on their respective sides.
In simple words: When you pick any point on one side of a parallelogram and any point on an adjacent side, the two triangles you can draw to the opposite corner have the same area.

Exam Tip: Draw the parallelogram carefully with both triangles CPD and AQD marked. Use the properties of parallel sides and how triangles partition the overall space to support your reasoning about equal areas.

 

Question 6(b). ∆CPD and || gm ABCD are on the same base CD and between the same parallel lines AB and CD. ∆AQD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC. Prove that area of ∆CPD = area of ∆AQD.
Answer: Since ∆CPD and || gm ABCD share the same base CD and lie between parallel lines AB and CD, the area of ∆CPD equals half the area of || gm ABCD (i). Similarly, ∆AQD and || gm ABCD have the same base AD and lie between parallel lines AD and BC, so the area of ∆AQD also equals half the area of || gm ABCD (ii). From statements (i) and (ii), it follows that the area of ∆CPD equals the area of ∆AQD, which completes the proof.
In simple words: Both triangles have areas equal to half the parallelogram's area, so they must have the same area as each other.

Exam Tip: Key mark-getter: recognizing that a triangle and a parallelogram on the same base between parallel lines have a fixed area ratio of 1:2. Clearly state "same base" and "between the same parallel lines" before invoking the relationship.

 

Question 6(b) [Continued]. In the figure (2) given below, PQRS and ABRS are parallelograms, and X is any point on the side BR. Show that area of ∆AXS = area of || gm PQRS.
Answer: Since PB is a straight line and PQ || SR, it follows that PB || SR. The || gm PQRS and ABRS both rest on the same base SR and lie between the parallel lines PB and SR, therefore the area of || gm PQRS equals the area of || gm ABRS (i). Next, ∆AXS and || gm ABRS share the same base AS and lie between parallel lines AS and BR, so the area of ∆AXS equals half the area of || gm ABRS (ii). Combining (i) and (ii), the area of ∆AXS equals half the area of || gm PQRS, which is what we needed to prove.
In simple words: Two parallelograms on the same base between parallel lines have equal areas. A triangle on the same base as one of them, between the same parallels, has half that area.

Exam Tip: Examiners reward explicit mention of "same base," "between the same parallel lines," and correct application of the parallelogram-to-triangle area ratio. State each relationship with reference numbers (i), (ii) for clarity.

 

Question 7. D, E and F are mid-points of the sides BC, CA and AB respectively of a ∆ABC. Prove that (i) FDCE is a parallelogram (ii) area of ∆DEF = area of ∆ABC (iii) area of || gm FDCE = area of ∆ABC.
Answer:
(i) Since F and E are the midpoints of AB and AC respectively, by the mid-point theorem, FE || BC and FE = ½BC. Because D is the midpoint of BC, CD = ½BC. From these two facts, FE || BC implies FE || CD, and FE = CD (statement 3). Similarly, D and F are the midpoints of BC and AB. By the mid-point theorem, DF || AC and DF = ½AC. Since E is the midpoint of AC, EC = ½AC. This gives DF || AC, so DF || EC, and DF = EC (statement 6). Since the opposite sides of FDCE are both parallel and equal in length, FDCE is a parallelogram.
(ii) In || gm FDCE, the diagonal DE divides it into two triangles of equal area, so area of ∆DEF = area of ∆DEC (statement 1). In || gm BDEF, the diagonal FD divides it into two triangles of equal area, so area of ∆DEF = area of ∆FBD (statement 2). In || gm AFDE, the diagonal FE divides it into two triangles of equal area, so area of ∆DEF = area of ∆AFE (statement 3). From all three statements, the four triangles DEF, DEC, FBD, and AFE all have equal areas (statement 4). Looking at the figure, ∆ABC is made up of these four triangles: area of ∆ABC = area of ∆DEF + area of ∆DEC + area of ∆FBD + area of ∆AFE = 4 × area of ∆DEF. Therefore, area of ∆DEF = ¼ area of ∆ABC.
(iii) From the figure, the area of || gm FDCE equals the sum of ∆DEF and ∆DEC. Since ∆DEC has the same area as ∆DEF (from statement 1 in part ii), the area of || gm FDCE = area of ∆DEF + area of ∆DEF = 2 × area of ∆DEF. Substituting the result from part (ii), area of || gm FDCE = 2 × ¼ area of ∆ABC = ½ area of ∆ABC.
In simple words: When you connect the midpoints of a triangle's sides, you create a parallelogram and four equal smaller triangles. The central triangle is ¼ the original, and the parallelogram made by three of these small triangles is ½ the original.

Exam Tip: For part (i), clearly show that both pairs of opposite sides are parallel AND equal. For parts (ii) and (iii), use the diagonal property of parallelograms and carefully organize the area relationships with numbered statements to avoid losing marks for incomplete reasoning.

 

Question 8. In the adjoining figure, D, E and F are midpoints of the sides BC, CA and AB respectively of ∆ABC. Prove that BCEF is a trapezium and area of the trap. BCEF = area of ∆ABC.
Answer: Since D and E are the midpoints of BC and CA respectively, by the mid-point theorem, DE || AB and DE = ½AB. Since F is the midpoint of AB, BF = ½AB. Therefore, BF || DE and BF = DE, which means BDEF is a || gm. Similarly, F and E are the midpoints of AB and CA. By the mid-point theorem, EF || BC and EF = ½BC. Since D is the midpoint of BC, DC = ½BC. Thus, EF || DC and EF = DC, making EFDC a || gm. Since FE || BC but FB and EC are not parallel to each other, EFBC (or BCEF) is a trapezium. Additionally, F and D are the midpoints of AB and BC. By the mid-point theorem, FD || AC and FD = ½AC. Since E is the midpoint of AC, AE = ½AC. Therefore, FD || AE and FD = AE, so AFDE is a || gm. From the diagonal property of parallelograms, in || gm BDEF with diagonal FD: area of ∆DEF = area of ∆BDF (i); in || gm EFDC with diagonal DE: area of ∆DEF = area of ∆EDC (ii); and in || gm AFDE with diagonal FE: area of ∆DEF = area of ∆AFE (iii). From (i), (ii), and (iii), all four triangles have equal areas (iv). Since ∆ABC is composed of these four triangles, area of ∆ABC = 4 × area of ∆DEF, so area of ∆DEF = ¼ area of ∆ABC (v). The trapezium BCEF consists of three of these equal triangles: area of trapezium BCEF = area of ∆DEF + area of ∆BDF + area of ∆EDC = 3 × area of ∆DEF (vi). Substituting the result from (v): area of trapezium BCEF = 3 × ¼ area of ∆ABC = ¾ area of ∆ABC.
In simple words: The trapezium is made of three small equal triangles, and together they make up ¾ of the original triangle's area.

Exam Tip: Clearly prove the trapezium property by showing one pair of parallel sides (FE || BC) and that the other pair (FB and EC) are not parallel. Use numbered diagonal relationships to organize the area calculation, and always reference the figure when decomposing the original triangle into smaller regions.

 

Question 9(a). In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m - n. Prove that area of ∆ABD - area of ∆ADC = m - n.
Answer: Since point D divides BC in the ratio m - n, we have BD - DC = m - n. The area of ∆ABD equals ½ × BD × AE, where AE is the perpendicular distance from A to BC (i). The area of ∆ADC equals ½ × DC × AE (ii). Dividing (i) by (ii): (area of ∆ABD) ÷ (area of ∆ADC) = (½ × BD × AE) ÷ (½ × DC × AE) = BD ÷ DC = m ÷ n. Therefore, area of ∆ABD - area of ∆ADC = m - n, which completes the proof.
In simple words: Two triangles sharing the same height and having bases in a certain ratio will have their areas in that same ratio.

Exam Tip: The key insight is recognizing that both triangles have the same height (the perpendicular from A to line BC). When bases are in ratio m - n, so are the areas. Always write the ratio of areas = ratio of bases clearly to earn full marks.

 

Question 9(b). In the figure (2) given below, P is a point on the side BC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5PQ, find area of ∆AQC - area of ∆ABC.
Answer: We are given that PC = 2BP. From the figure, BC = BP + PC = BP + 2BP = 3BP, so BP - BC = 1 - 3. By Question 9(a), triangles ABP and APC have areas in the ratio BP - PC = 1 - 2, therefore area of ∆ABP = ⅓ area of ∆ABC and area of ∆APC = ⅔ area of ∆ABC. Next, Q lies on AP such that QA = 5PQ. This means AP = AQ + QP = 5PQ + PQ = 6PQ, so AQ - AP = 5 - 6. Since triangles AQC and APC share the same base AC, and Q lies on AP, they have areas in the ratio AQ - AP = 5 - 6, therefore area of ∆AQC = (5/6) × area of ∆APC = (5/6) × ⅔ area of ∆ABC = (5/9) area of ∆ABC. Thus, area of ∆AQC - area of ∆ABC = 5 - 9.
In simple words: First, find where P divides the base to get the area of ∆APC. Then use the fact that Q divides AP to find what fraction of ∆APC the smaller triangle ∆AQC represents, and multiply to get the final answer.

Exam Tip: Break this into two steps: (1) use the base ratio to find area of ∆APC as a fraction of ∆ABC; (2) use the height ratio to find area of ∆AQC as a fraction of ∆APC. Combine the fractions carefully at the end. Writing intermediate ratios clearly prevents arithmetic errors and shows logical flow to the examiner.

 

Question 9(c). In the figure (3) given below, AD is a median of △ABC and P is a point in AC such that area of △ADP : area of △ABD = 2 : 3. Find
(i) AP : PC
(ii) area of △PDC : area of △ABC
Answer:
(i) Looking at the figure, let DE serve as the altitude measured perpendicular to base AC.

A median splits a triangle into two parts that have identical areas.

Since AD functions as the median of triangle ABC,
\[ \text{Area of } \triangle ABD = \text{Area of } \triangle ADC = \frac{1}{2} \text{ Area of } \triangle ABC \text{ ......(1)} \]

We are told that,
\[ \Rightarrow \text{area of } \triangle ADP : \text{area of } \triangle ABD = 2 : 3 \]
\[ \Rightarrow \text{area of } \triangle ADP : \text{area of } \triangle ADC = 2 : 3 \]

Using the area ratio formula for triangles sharing the same height,
\[ \Rightarrow \frac{\text{Area of } \triangle ADP}{\text{Area of } \triangle ADC} = \frac{\frac{1}{2} \times AP \times DE}{\frac{1}{2} \times AC \times DE} = \frac{2}{3} \]
\[ \Rightarrow \frac{AP}{AC} = \frac{2}{3} \]

Setting AP = 2x and AC = 3x,

From the figure,
\[ PC = AC - AP = 3x - 2x = x \]
\[ \frac{AP}{PC} = \frac{2x}{x} = \frac{2}{1} \]

Therefore, AP : PC = 2 : 1

(ii) We recognize that,
\[ PC : AC = x : 3x = 1 : 3 \]

Now,
\[ \Rightarrow \frac{\text{Area of } \triangle PDC}{\text{Area of } \triangle ADC} = \frac{\frac{1}{2} \times PC \times DE}{\frac{1}{2} \times AC \times DE} \]
\[ \Rightarrow \frac{\text{Area of } \triangle PDC}{\text{Area of } \triangle ADC} = \frac{PC}{AC} \]
\[ \Rightarrow \frac{\text{Area of } \triangle PDC}{\text{Area of } \triangle ADC} = \frac{x}{3x} \]
\[ \Rightarrow \frac{\text{Area of } \triangle PDC}{\text{Area of } \triangle ADC} = \frac{1}{3} \text{ .......(1)} \]

Since AD serves as a median of triangle ABC,
\[ \text{area of } \triangle ADC = \frac{1}{2} \text{ area of } \triangle ABC \]

Putting this result into equation (1),
\[ \Rightarrow \frac{\text{Area of } \triangle PDC}{\text{Area of } \triangle ADC} = \frac{1}{3} \]
\[ \Rightarrow \frac{\text{Area of } \triangle PDC}{\frac{1}{2}\text{ Area of } \triangle ABC} = \frac{1}{3} \]
\[ \Rightarrow \frac{\text{Area of } \triangle PDC}{\text{Area of } \triangle ABC} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \]

Therefore, area of △PDC : area of △ABC = 1 : 6.

In simple words: A median divides the triangle into two equal parts. By comparing base lengths and using the area formula, we find that AP is twice as long as PC, and triangle PDC is one-sixth the size of triangle ABC.

Exam Tip: When triangles share the same height, their areas are proportional to their bases. Remember that medians always create two triangles of equal area - this is a key property to use.

 

Question 10(a). In figure (1) given below, area of parallelogram ABCD is 29 cm². Calculate the height of parallelogram ABEF if AB = 5.8 cm.
Answer: Parallelograms ABCD and ABEF lie on the same base (AB) and sit between the same pair of parallel lines AB and DE. This means they must have the same area.

\[ \text{Area of } \parallel\text{gm ABEF} = \text{Area of } \parallel\text{gm ABCD} = 29 \text{ cm}^2 \]

Using the parallelogram area formula,
\[ \text{Area of } \parallel\text{gm ABEF} = \text{base} \times \text{height} \]
\[ \Rightarrow 29 = AB \times \text{height} \]
\[ \Rightarrow 29 = 5.8 \times \text{height} \]
\[ \Rightarrow \text{Height} = \frac{29}{5.8} = 5 \text{ cm} \]

The height of parallelogram ABEF is 5 cm.

In simple words: When two parallelograms share the same base and sit between parallel lines, they have equal areas. Divide the area by the base length to get the height.

Exam Tip: Always identify figures sharing the same base and parallel lines - they have equal areas. Use this property before doing calculations to save time.

 

Question 10(b). In figure (2) given below, area of △ABD is 24 sq. units. If AB = 8 units, find the height of △ABC.
Answer: Given information,
\[ \text{Area of } \triangle ABD = 24 \text{ sq. units} \]

Since triangle ABD and triangle ABC lie on the same base AB and sit between the same parallel lines AB and CD,
\[ \text{Area of } \triangle ABC = \text{Area of } \triangle ABD = 24 \text{ sq. units} \]

Using the triangle area formula,
\[ \Rightarrow \frac{1}{2} \times AB \times \text{height} = 24 \]
\[ \Rightarrow \frac{1}{2} \times 8 \times \text{height} = 24 \]
\[ \Rightarrow 4 \times \text{height} = 24 \]
\[ \Rightarrow \text{Height} = \frac{24}{4} = 6 \text{ units} \]

Therefore, height of △ABC = 6 units.

In simple words: Triangles with the same base and between the same parallel lines have equal areas. Use the triangle area formula with the given base to find the height.

Exam Tip: Recognize when figures occupy the same base and parallel lines - this immediately tells you they have equal areas, making the problem much simpler.

 

Question 10(c). In figure (3) given below, E and F are midpoints of sides AB and CD, respectively, of parallelogram ABCD. If the area of parallelogram ABCD is 36 cm²,
(i) state the area of △APD.
(ii) Name the parallelogram whose area is equal to the area of △APD.
Answer:
(i) Triangle APD and parallelogram ABCD sit on the same base AD and lie between the same pair of parallel lines AD and BC.

\[ \text{Area of } \triangle APD = \frac{1}{2} \times \text{Area of } \parallel\text{gm ABCD} \]
\[ = \frac{1}{2} \times 36 = 18 \text{ cm}^2 \]

Therefore, area of △APD = 18 cm².

(ii) Consider that diagonals AC and BD meet at point O.

In triangle ABC,

Since O marks the midpoint of AC (because parallelogram diagonals bisect one another) and E marks the midpoint of AB, by applying the midpoint theorem,
\[ EO \parallel BC \]

This means,
\[ EF \parallel BC \]

Since BC is parallel to AD,
\[ \Rightarrow EF \parallel AD \]

In the parallelogram ABCD, side AB is parallel to side DC,
\[ \Rightarrow AE \parallel DF \]

Since EF is parallel to AD and AE is parallel to DF,
\[ AEFD \text{ forms a parallelogram} \]

The line segment EF divides the original parallelogram ABCD into two equal halves. This is because E and F are located at the midpoints of AB and CD respectively, and EF is parallel to both BC and AD.

\[ \text{Area of } \parallel\text{gm AEFD} = \frac{1}{2} \times \text{Area of } \parallel\text{gm ABCD} = \frac{1}{2} \times 36 = 18 \text{ cm}^2 \]

We have shown that,
\[ \text{Area of } \triangle APD = \text{Area of } \parallel\text{gm AEFD} \]

Therefore, parallelogram AEFD is the required parallelogram that has area equal to the area of △APD.

In simple words: The line EF connecting the midpoints of opposite sides divides the parallelogram into two equal parts. The triangle APD has the same area as parallelogram AEFD, which is half of the original parallelogram.

Exam Tip: When midpoints are involved, always apply the midpoint theorem and the property that diagonals of a parallelogram bisect each other. This helps identify equal areas quickly.

 

Question 11(a). In figure (1) given below, ABCD is a parallelogram. Points P and Q on BC trisect BC into three equal parts. Prove that: area of △APQ = area of △DPQ = \(\frac{1}{6}\) (area of ∥gm ABCD)
Answer: Construction: Through points P and Q, draw PR and QS parallel to sides AB and CD.

First, triangles APD and AQD both sit on the same base AD and lie between the same pair of parallel lines AD and BC. Therefore,
\[ \text{Area of } \triangle APD = \text{Area of } \triangle AQD \]

Subtracting the area of triangle AOD from both sides,
\[ \text{Area of } \triangle APD - \text{Area of } \triangle AOD = \text{Area of } \triangle AQD - \text{Area of } \triangle AOD \]
\[ \text{Area of } \triangle APO = \text{Area of } \triangle OQD \text{ .......(i)} \]

Adding the area of triangle OPQ to both sides,
\[ \text{Area of } \triangle APO + \text{Area of } \triangle OPQ = \text{Area of } \triangle OQD + \text{Area of } \triangle OPQ \]
\[ \text{Area of } \triangle APQ = \text{Area of } \triangle DPQ \text{ .......(ii)} \]

Next, triangle APQ and parallelogram PQSR share the same base PQ and lie between the same pair of parallel lines PQ and AD. Therefore,
\[ \text{Area of } \triangle APQ = \frac{1}{2} \times \text{Area of } \parallel\text{gm PQRS \text{ ......(iii)}} \]

From the diagram,
\[ \text{Height of } \parallel\text{gm ABCD} = \text{Height of } \parallel\text{gm PQRS} = AE \]

Since P and Q trisect BC,
\[ \Rightarrow PQ = \frac{BC}{3} \]
\[ \Rightarrow BC = 3PQ \]

Calculating the area ratio,
\[ \Rightarrow \frac{\text{Area of } \parallel\text{gm PQRS}}{\text{Area of } \parallel\text{gm ABCD}} = \frac{PQ \times AE}{BC \times AE} \]
\[ \Rightarrow \frac{\text{Area of } \parallel\text{gm PQRS}}{\text{Area of } \parallel\text{gm ABCD}} = \frac{PQ \times AE}{3PQ \times AE} \]
\[ \Rightarrow \frac{\text{Area of } \parallel\text{gm PQRS}}{\text{Area of } \parallel\text{gm ABCD}} = \frac{1}{3} \]

Therefore,
\[ \Rightarrow \text{Area of } \parallel\text{gm PQRS} = \frac{1}{3} \times \text{Area of } \parallel\text{gm ABCD} \]

Substituting this into equation (iii),
\[ \Rightarrow \text{Area of } \triangle APQ = \frac{1}{2} \times \frac{1}{3} \times \text{Area of } \parallel\text{gm ABCD} \]
\[ \Rightarrow \text{Area of } \triangle APQ = \frac{1}{6} \times \text{Area of } \parallel\text{gm ABCD} \text{ ......(iv)} \]

From equations (ii) and (iv),
\[ \text{area of } \triangle APQ = \text{area of } \triangle DPQ = \frac{1}{6} \text{ (area of } \parallel\text{gm ABCD)} \]

Hence, proved that area of △APQ = area of △DPQ = \(\frac{1}{6}\) (area of ∥gm ABCD).

In simple words: When two points divide a side into three equal parts, triangles formed with vertices on the opposite side have areas that are one-sixth of the parallelogram. This happens because the smaller segments are one-third the original length, and triangles are half of parallelograms with the same base and height.

Exam Tip: Break the proof into stages: first show the two triangles are equal using the property of same base and parallel lines, then relate them to the parallelogram using trisection. Write clear equations at each step.

 

Question 11(b). In figure (2) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at point E. Prove that area of quad. ABCD = area of △ABE.
Answer: We need to show that the quadrilateral and the triangle have identical areas.

Since line DE is drawn parallel to diagonal AC,
\[ DE \parallel AC \]

Consider triangles DAC and DAE. These two triangles share the common base DA. Also, since DE is parallel to AC, the perpendicular distances from C to line DA and from E to line DA are equal. Therefore, both triangles have the same height measured from their respective opposite vertices to the common base DA.

\[ \text{Area of } \triangle DAC = \text{Area of } \triangle DAE \]

Now, observe that quadrilateral ABCD can be written as the sum of two triangles,
\[ \text{Area of quad. ABCD} = \text{Area of } \triangle ABC + \text{Area of } \triangle DAC \]

Since Area of △DAC = Area of △DAE,
\[ \text{Area of quad. ABCD} = \text{Area of } \triangle ABC + \text{Area of } \triangle DAE \]

Notice that triangle ABE can be split into two parts: triangle ABC and triangle ACE. However, since E lies on BC produced and DE is parallel to AC, we can express,
\[ \text{Area of } \triangle ABE = \text{Area of } \triangle ABC + \text{Area of } \triangle ACE \]

From the parallel condition DE || AC, triangles DAE and CAE share the same base AE. Since DE || AC, these two triangles lie between the same parallel lines, meaning,
\[ \text{Area of } \triangle CAE = \text{Area of } \triangle DAE \]

Wait, let me reconsider. Since DE || AC, consider triangle DEC. Point E is positioned such that DE || AC. This means triangles CAD and EAD are between parallel lines through DE and AC.

Actually, since DE || AC,
\[ \text{Area of } \triangle ACD = \text{Area of } \triangle ADE \]
(Same base AD, with C and E lying on parallel lines equidistant from AD.)

Therefore,
\[ \text{Area of quad. ABCD} = \text{Area of } \triangle ABC + \text{Area of } \triangle ACD \]
\[ = \text{Area of } \triangle ABC + \text{Area of } \triangle ADE \]
\[ = \text{Area of } \triangle ABE \]

Hence, proved that area of quad. ABCD = area of △ABE.

In simple words: When a line through one vertex is drawn parallel to a diagonal, the triangle formed with this parallel line and two other vertices has the same area as the original quadrilateral. The parallel line rearranges the area without changing the total.

Exam Tip: Identify which triangles can be formed by dividing the quadrilateral. Use the parallel line property to show equal areas between triangles that share a base and have vertices on parallel lines. This substitution is the key to the proof.

 

Question 11(c). In the figure (3) given below, ABCD is a parallelogram. O is any point on the diagonal AC of the parallelogram. Show that the area of ∆AOB is equal to the area of ∆AOD.
Answer: Draw line BD, which intersects AC at point P. Since the diagonals of a parallelogram bisect each other, P is the midpoint of BD. In triangle ABD, AP becomes a median (a line from a vertex to the midpoint of the opposite side). A property of medians tells us that any median divides a triangle into two equal areas, so Area of ∆ABP = Area of ∆ADP .....(i). Similarly, in triangle BOD, PO acts as a median since P is the midpoint of BD. This means Area of ∆BOP = Area of ∆POD .....(ii). Adding equations (i) and (ii) together: Area of ∆ABP + Area of ∆BOP = Area of ∆ADP + Area of ∆POD. The left side simplifies to Area of ∆AOB, and the right side simplifies to Area of ∆AOD. Therefore, Area of ∆AOB = Area of ∆AOD.
In simple words: When you draw a line from B to D inside the parallelogram and it crosses AC at P, the point P splits BD into two equal parts. This makes certain triangles equal in area, which proves that triangles AOB and AOD have the same size.

Exam Tip: The key is recognizing that the diagonals of a parallelogram bisect each other, and that a median of any triangle divides it into two parts of equal area. State both properties clearly to earn full marks.

 

Question 12(a). In the figure (1) given, ABCD and AEFG are two parallelograms. Prove that area of || gm ABCD = area of || gm AEFG.
Answer: Draw a line from B to G. Triangles and parallelograms sharing the same base and lying between identical parallel lines have equal areas. Since triangle ABG and parallelogram ABCD share base AB and lie between the same parallel lines AB and CD, we get: Area of ∆ABG = \( \frac{1}{2} \) Area of || gm ABCD .....(i). Similarly, triangle ABG and parallelogram AEFG share base AG and lie between the same parallel lines AG and EF, giving us: Area of ∆ABG = \( \frac{1}{2} \) Area of || gm AEFG .....(ii). From statements (i) and (ii), we conclude: Area of || gm ABCD = Area of || gm AEFG.
In simple words: A triangle with one side as its base has half the area of a parallelogram with the same base between the same parallel lines. Since both parallelograms give the same triangle area, they must be equal to each other.

Exam Tip: Always state the condition "same base and between same parallel lines" explicitly - this is the property that makes the areas equal, and examiners expect to see it named.

 

Question 12(b). In figure (2) given below, the side AB of the parallelogram ABCD is produced to E. A straight line through A is drawn parallel to CE to meet CB produced at F and parallelogram BFGE is completed. Prove that area of || gm BFGE = area of || gm ABCD.
Answer: Connect segments AC and EF. Triangles AFC and AFE both rest on the same base AF and lie between the parallel lines AF and CE. By the base-parallel property, Area of ∆AFC = Area of ∆AFE. Subtracting the area of triangle ABF from both sides: Area of (∆AFC - ∆ABF) = Area of (∆AFE - ∆ABF), which gives us Area of ∆ABC = Area of ∆BEF. Doubling both sides: 2 Area of ∆ABC = 2 Area of ∆BEF .....(i). From the diagram, parallelogram ABCD and triangle ABC share base AB and lie between parallel lines AB and DC, so Area of ∆ABC = \( \frac{1}{2} \) Area of || gm ABCD, or Area of || gm ABCD = 2 Area of ∆ABC .....(ii). Likewise, parallelogram BFGE and triangle BEF share base BE and lie between parallel lines FG and BE, giving Area of ∆BEF = \( \frac{1}{2} \) Area of || gm BFGE, or Area of || gm BFGE = 2 Area of ∆BEF .....(iii). Substituting (ii) and (iii) into (i): Area of || gm ABCD = Area of || gm BFGE.
In simple words: The two triangles ABC and BEF turn out to have the same area because of the parallel lines in the construction. Since each parallelogram is exactly double the area of its own triangle, the two parallelograms must be equal.

Exam Tip: The critical step is showing that ∆ABC = ∆BEF through the base-parallel property, then using the fact that a parallelogram is always twice the area of a triangle on the same base.

 

Question 12(c). In figure (3) given below, AB || DC || EF, AD || BE and DE || AF. Prove that the area of DEFH is equal to the area of ABCD.
Answer: From the given conditions: AD || BE means AD || EG (since G lies on the extension of E), and DE || FA means DE || GA (since A is on the extension). Since opposite sides are parallel, ADEG forms a parallelogram. Parallelograms ABCD and ADEG rest on the same base AD and lie between the parallel lines AD and EB, so: Area of || gm ABCD = Area of || gm ADEG .....(i). Next, from DE || FA, we get DE || FH (since H is positioned accordingly), and from the configuration, DH || EF (as opposite sides of the figure). This makes DEFH a parallelogram. Parallelograms DEFH and ADEG both sit on the same base DE and lie between the parallel lines DE and FA, giving us: Area of || gm DEFH = Area of || gm ADEG .....(ii). Combining (i) and (ii): Area of || gm ABCD = Area of || gm DEFH.
In simple words: By recognizing that ADEG and DEFH are both parallelograms with the right base-parallel arrangement, each can be shown equal to the original parallelogram ABCD.

Exam Tip: Carefully identify which pairs of opposite sides are parallel to confirm that both ADEG and DEFH are indeed parallelograms - this is essential to apply the equal-area theorem.

 

Question 13. Any point D is taken on the side BC of a ∆ABC and AD is produced to E such that AD = DE, prove that area of ∆BCE = area of ∆ABC.
Answer: Consider triangle ABE, where AD = DE by the given condition. This means D is the midpoint of AE, making BD a median of triangle ABE (a line from vertex B to the midpoint of the opposite side). A fundamental property states that a median splits any triangle into two equal-area parts, so: Area of ∆ABD = Area of ∆BED .....(i). Similarly, in triangle ACE, since AD = DE (given), D acts as the midpoint of AE. This makes CD a median of triangle ACE, so: Area of ∆ACD = Area of ∆CED .....(ii). Adding equations (i) and (ii): Area of ∆ABD + Area of ∆ACD = Area of ∆BED + Area of ∆CED. The left side combines into Area of ∆ABC, and the right side combines into Area of ∆BCE. Therefore: Area of ∆ABC = Area of ∆BCE.
In simple words: When you extend AD to E such that AD equals DE, point D becomes the midpoint of line AE. This makes BD and CD both medians, and medians always cut triangles into equal halves, so the final triangle BCE must match the original triangle ABC in area.

Exam Tip: The key insight is recognizing that D is the midpoint of AE (from AD = DE) and that this creates two medians (BD and CD) in triangles ABE and ACE respectively.

 

Question 14. ABCD is a rectangle and P is the mid-point of AB. DP is produced to meet CB at Q. Prove that the area of rectangle ABCD = area of ∆DQC.
Answer: Examine triangles APD and PQB. Since P is the midpoint of AB, we have AP = BP. Both ABCD being a rectangle means all corners are right angles, so \( \angle \)DAP = \( \angle \)QBP = 90°. Additionally, angles APD and BPQ are vertically opposite angles (formed where lines cross), so \( \angle \)APD = \( \angle \)BPQ. By the ASA (Angle-Side-Angle) rule, triangles APD and PQB are congruent: \( \triangle \)APD \( \cong \) \( \triangle \)PQB. Since the triangles are congruent, their areas are equal: Area of ∆APD = Area of ∆PQB .....(i). Looking at the rectangle: Area of rectangle ABCD = Area of ∆APD + Area of quad. PBCD. Substituting equation (i): Area of rectangle ABCD = Area of ∆PQB + Area of quad. PBCD. The right side equals Area of ∆DQC. Therefore: Area of rectangle ABCD = Area of ∆DQC.
In simple words: The triangles APD and PQB are the same size because they have matching angle measures and equal sides. Swapping one triangle for the other in the rectangle's breakdown shows that the rectangle has the same area as the larger triangle DQC.

Exam Tip: Use the ASA congruence criterion correctly by identifying AP = BP (midpoint condition), the two right angles, and the vertically opposite angles - all three are needed for the proof.

 

Question 15(a). In figure (1) given below, the perimeter of the parallelogram is 42 cm. Calculate the lengths of the sides of the parallelogram.
Answer: Let AB = p. In a parallelogram, opposite sides are always equal in length. The perimeter is the sum of all four sides: Perimeter of || gm ABCD = 2(AB + BC) = 42. Dividing by 2: p + BC = 21, so BC = 21 - p. The area of a parallelogram can be calculated in two ways, depending on which side we choose as the base. Using AB as the base with perpendicular height DM = 6 cm: Area = AB \( \times \) DM = p \( \times \) 6 = 6p .....(i). Using BC as the base with perpendicular height DN = 8 cm: Area = BC \( \times \) DN = (21 - p) \( \times \) 8 = 8(21 - p) .....(ii). Since both expressions represent the same area: 6p = 8(21 - p). Expanding: 6p = 168 - 8p. Adding 8p to both sides: 14p = 168. Dividing: p = 12 cm. Therefore: BC = 21 - 12 = 9 cm. The sides of the parallelogram are AB = 12 cm and BC = 9 cm (with CD = 12 cm and DA = 9 cm as opposites).
In simple words: The perimeter tells you that two opposite sides add up to 21 cm. Using area formulas with both height measurements lets you write two equations for the same area, which you can solve to find each side.

Exam Tip: The critical step is setting two area expressions equal - one using each perpendicular height. Make sure to accurately substitute BC = 21 - p into the second area equation.

 

Question 15(b). In the figure (2) given below, the perimeter of ∆ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, calculate the lengths of the sides of ∆ABC.
Answer: Let BC = p and CA = q. From the perimeter condition: AB + BC + CA = 37, so AB = 37 - (p + q). The area of a triangle can be computed using any side as the base paired with its corresponding altitude. Using BC as the base and AM as the altitude: Area = \( \frac{1}{2} \times \) BC \( \times \) AM = \( \frac{1}{2} \times \) p \( \times \) 5x = 2.5px .....(i). Using CA as the base and BN as the altitude: Area = \( \frac{1}{2} \times \) CA \( \times \) BN = \( \frac{1}{2} \times \) q \( \times \) 6x = 3qx .....(ii). Using AB as the base and CL as the altitude: Area = \( \frac{1}{2} \times \) AB \( \times \) CL = \( \frac{1}{2} \times \) [37 - (p + q)] \( \times \) 4x = 2x[37 - (p + q)] .....(iii). Since all three expressions give the same area, set (i) equal to (ii): 2.5px = 3qx. Dividing by x (assuming x \( \neq \) 0): 2.5p = 3q, or 5p = 6q, which gives q = \( \frac{5p}{6} \). Set (i) equal to (iii): 2.5px = 2x[37 - (p + q)]. Dividing by 2x: 1.25p = 37 - (p + q). Substituting q = \( \frac{5p}{6} \): 1.25p = 37 - p - \( \frac{5p}{6} \). Combining like terms on the right: 1.25p = 37 - p - 0.833p, so 1.25p = 37 - 1.833p. Adding 1.833p: 3.083p = 37, so p = 12 cm. Then q = \( \frac{5 \times 12}{6} \) = 10 cm. And AB = 37 - (12 + 10) = 15 cm. The sides of ∆ABC are: BC = 12 cm, CA = 10 cm, AB = 15 cm.
In simple words: Each altitude gives you a different way to calculate the triangle's area. All three ways must give the same result, so you can set them equal to each other and solve for the unknown side lengths.

Exam Tip: Remember that a triangle's area is constant regardless of which base-altitude pair you use. Setting two area expressions equal and solving the resulting equation is the main strategy for finding the sides.

 

Question 15(c). In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find (i) area of || gm ABCD (ii) DP : PC.
Answer:
(i) Since triangle APB and parallelogram ABCD share the same base AB and lie between the same two parallel lines AB and DC, triangle APB has half the area of the parallelogram. Looking at the figure, the parallelogram's area equals the sum of triangles DAP and BCP doubled:

\( \frac{1}{2} \) area of || gm ABCD = area of (∆DAP + ∆BCP)

\( \frac{1}{2} \) area of || gm ABCD = 25 + 15 = 40

area of || gm ABCD = 2 × 40 = 80 cm²

(ii) Triangles DAP and BCP share the same base CD and lie between the same pair of parallel lines CD and AB. When two triangles are on the same base and between the same parallel lines, the ratio of their areas equals the ratio of their heights from the opposite vertices. Since both triangles sit on line CD with heights measured to the parallel line AB, the ratio of their areas gives us the ratio of segments DP to PC:

\( \frac{\text{Area of } \triangle\text{DAP}}{\text{Area of } \triangle\text{BCP}} = \frac{\text{DP}}{\text{PC}} \)

\( \frac{25}{15} = \frac{\text{DP}}{\text{PC}} \)

\( \frac{\text{DP}}{\text{PC}} = \frac{5}{3} \)

Therefore, DP : PC = 5 : 3
In simple words: The parallelogram breaks into two triangles. Use the triangle areas to find how point P divides side DC.

Exam Tip: Remember that triangles on the same base and between the same parallel lines divide in the ratio of their areas along that base - this is a key property to apply here.

 

Question 16. In the adjoining figure, E is the midpoint of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of ∆ ABC is 25 sq. units, find the area of || gm EBCF.
Answer:
In triangle ABC, since E is the midpoint of AB and EF is parallel to BC, the line EF cuts AC at its midpoint too (by the midpoint theorem). Call this midpoint G, so AG = GC.

Now look at triangles AEG and CFG. These triangles have:
\( \angle \)EAG = \( \angle \)GCF (alternate angles, since EF || BC)
\( \angle \)EGA = \( \angle \)CGF (vertically opposite angles)
AG = GC (proved above)

By the ASA rule, triangle AEG is congruent to triangle CFG. This means they have equal areas:
area of ∆AEG = area of ∆CFG .........(i)

From the figure, the parallelogram EBCF breaks down into quadrilateral BCGE plus triangle CFG:
area of || gm EBCF = area of quad. BCGE + area of ∆CFG

Substituting equation (i):
area of || gm EBCF = area of quad. BCGE + area of ∆AEG = area of ∆ABC = 25 sq. units
In simple words: The parallelogram covers exactly the same area as the whole triangle because the two small triangles at the corners balance out.

Exam Tip: The key insight is showing triangle AEG is congruent to triangle CFG - once you prove this, the area equality follows directly from the fact that the quadrilateral plus one triangle equals the full original triangle.

 

Question 17(a). In the figure (1) given below, BC || AE and CD || BE. Prove that area of ∆ABC = area of ∆EBD.
Answer:
Draw lines CE and AC to help with the proof.

From the figure, triangles ABC and EBC rest on the same base BC and are sandwiched between the same two parallel lines BC and AE. By the property of triangles with the same base and parallel boundaries:
area of ∆ABC = area of ∆EBC .........(i)

Next, triangles EBC and EBD sit on the same base BE and lie between the same parallel lines CD and BE. Applying the same property again:
area of ∆EBC = area of ∆EBD .........(ii)

From statements (i) and (ii) combined, we get:
area of ∆ABC = area of ∆EBD

Hence proved.
In simple words: Two different pairs of triangles share bases with parallel lines, so we can chain two area equalities together to reach the final result.

Exam Tip: The trick is identifying which triangles to compare first - look for shared bases and parallel lines that let you apply the equal-area theorem step by step.

 

Question 17(b). In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that (i) ∆BCF ≅ ∆ABE. (ii) area of square ABFG = area of rectangle BENM.
Answer:
(i) From the figure, angle FBC is made up of angle FBA plus angle ABC. Since FBAG is a square, each interior angle = 90°, so:
\( \angle \)FBC = 90° + \( \angle \)ABC .........(1)

Likewise, angle ABE is made up of angle CBE plus angle ABC. Since BCDE is a square, each interior angle = 90°, so:
\( \angle \)ABE = 90° + \( \angle \)ABC .........(2)

From equations (1) and (2):
\( \angle \)FBC = \( \angle \)ABE .........(3)

Now compare triangles BCF and ABE. They have:
BF = AB (both are sides of square FBAG)
\( \angle \)FBC = \( \angle \)ABE (from equation 3)
BC = BE (both are sides of square BCDE)

By the SAS rule of congruency, ∆BCF ≅ ∆ABE, hence proved.

(ii) From part (i), we know the two triangles are congruent, so:
area of ∆BCF = area of ∆ABE .........(4)

Triangle BCF and square AGFB share the same base FB and lie between the same pair of parallel lines FB and GC. Therefore, the triangle has half the area of the square:
\( \frac{1}{2} \) area of square AGFB = area of ∆BCF .........(5)

Triangle ABE and rectangle BENM sit on the same base BE and lie between the same pair of parallel lines BE and AN. Therefore, the triangle has half the area of the rectangle:
\( \frac{1}{2} \) area of rectangle BENM = area of ∆ABE .........(6)

From equations (4), (5) and (6):
\( \frac{1}{2} \) area of square AGFB = \( \frac{1}{2} \) area of rectangle BENM

Therefore, area of square AGFB = area of rectangle BENM, hence proved.
In simple words: Congruent triangles have the same area. Each triangle is exactly half of its paired four-sided shape, so the squares and rectangles must be equal too.

Exam Tip: Part (i) hinges on recognizing that both angles equal 90° plus angle ABC - once you have equal angles plus equal adjacent sides, SAS congruency is immediate. Part (ii) then follows from the "same base, same parallels" property applied twice.

 

Multiple Choice Questions

 

Question 1. In the adjoining figure, if l || m, AF || BE, FC ⊥ m and ED ⊥ m, then the correct statement is
(a) area of || ABEF = area of rect. CDEF
(b) area of || ABEF = area of quad. CBEF
(c) area of || ABEF = 2 area of △ACF
(d) area of || ABEF = 2 area of △EBD
Answer: (a) area of || ABEF = area of rect. CDEF
In simple words: A parallelogram and a rectangle built on the same base and between the same parallel lines must have the same area.

Exam Tip: This tests the fundamental property that shapes with the same base and parallel boundaries are equal in area - verify by identifying base EF and the two parallel lines l and m.

 

Question 2. Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 3 : 1
Answer: (b) 1 : 1
In simple words: When two parallelograms have equal bases and sit between the same pair of parallel lines, they cover the same amount of space.

Exam Tip: The ratio is always 1 : 1 regardless of how different the parallelograms look - the base and parallel-line constraint is what matters for area.

 

Question 3. If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of area of the triangle to the area of parallelogram is
(a) 1 : 3
(b) 1 : 2
(c) 3 : 1
(d) 1 : 4
Answer: (b) 1 : 2
In simple words: A triangle always takes up exactly half the space of a parallelogram when they share a base and sit between the same parallel lines.

Exam Tip: The ratio is always 1 : 2 because a triangle's area formula is \( \frac{1}{2} \) × base × height, while a parallelogram's is base × height - the factor of \( \frac{1}{2} \) always appears.

 

Question 4. A median of a triangle divides it into two
(a) triangles of equal area
(b) congruent triangles
(c) right triangles
(d) isosceles triangles
Answer: (a) triangles of equal area
In simple words: When you draw a median from a vertex to the midpoint of the opposite side, it splits the triangle into two parts with matching areas.

Exam Tip: The median creates two triangles with the same base (the midpoint divides the original base in half) and the same height (both reach the opposite vertex), so their areas must be equal.

 

Question 5. In the adjoining figure, area of parallelogram ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL
(d) AD × DL
Answer: (c) DC × DL
In simple words: Area of a parallelogram equals the base times the perpendicular distance from the opposite side to it.

Exam Tip: Identify which side is the base (DC) and which perpendicular line represents the height (DL measured at right angles to DC) - the product of these two gives the area.

 

Question 6. The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to
(a) \( \frac{1}{2} \) area of △ABC
(b) \( \frac{1}{3} \) area of △ABC
(c) \( \frac{1}{4} \) area of △ABC
(d) area of △ABC
Answer: (a) \( \frac{1}{2} \) area of △ABC
In simple words: When you use the three midpoints of a triangle's sides plus any one vertex to form a four-sided shape, that shape covers exactly half the triangle's area.

Exam Tip: The parallelogram formed is sometimes called a "midpoint parallelogram" - it always measures half the original triangle regardless of which vertex you pair with the three midpoints.

 

Question. Area of a parallelogram formed by points E, D, C, F is equal to half the area of triangle ABC.
Answer: Consider parallelogram CDEF. Its diagonal DF splits it into two triangles of equal size. Therefore, area of triangle CDF equals area of triangle EDF (equation i). Since E and F serve as midpoints of sides AB and AC respectively, by applying the midpoint theorem, we find that EF equals BC/2, which also equals BD (since D is the midpoint of BC), and EF is parallel to BC. Because EF is parallel to BC, we can observe from the figure that EF is parallel to BD and EF equals BD. This means EBDF forms a parallelogram whose diagonal ED splits it into two triangles of equal size. So area of triangle EBD equals area of triangle EDF (equation ii). Similarly, since E and D are midpoints of sides AB and BC respectively, the midpoint theorem tells us that ED equals AC/2, which also equals AF (since F is the midpoint of AC), and ED is parallel to AC. From the figure, ED is parallel to AF. Therefore, AEDF forms a parallelogram whose diagonal EF divides it into two triangles of equal size. Hence, area of triangle AEF equals area of triangle EDF (equation iii). Combining equations (i), (ii), and (iii), all four triangles - EDF, CDF, EBD, and AEF - have the same area x. From the figure, triangle ABC comprises all four of these triangles, so its total area equals 4x. Parallelogram EDCF consists of triangles EDF and CDF, so its area equals 2x. Therefore, the area of parallelogram EDCF equals (2x)/(4x) = 1/2 of the area of triangle ABC.
In simple words: When you divide a triangle using two midpoints, you create a parallelogram in the middle that takes up exactly half the space of the original triangle.

Exam Tip: Always identify and apply the midpoint theorem correctly, and track how triangles combine to form the larger shape - counting each smaller triangle and showing their equal areas is key to obtaining full marks.

 

Question 7. In the adjoining figure, ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are mid-points of the non-parallel sides. The ratio of area of ABFE and area of EFCD is
(1) a : b
(2) (3a + b) : (a + 3b)
(3) (a + 3b) : (3a + b)
(4) (2a + b) : (3a + b)
Answer: Let us take h as the distance between the parallel sides AB, CD, and the line EF. Draw line BD, which meets EF at point M. In triangle ABD, E is the midpoint of AD and EM is parallel to AB. Using the midpoint theorem, M becomes the midpoint of BD, and EM equals AB/2 (equation 1). In triangle CBD, F is the midpoint of BC and M is the midpoint of BD. By the midpoint theorem, MF equals CD/2 (equation 2). Adding equations (1) and (2): EM + MF equals (AB + CD)/2, which gives us EF equals (AB + CD)/2 = (a + b)/2. Using the trapezium area formula: Area of ABFE equals (1/2) times [a + (a + b)/2] times h = (1/2) times [(2a + a + b)/2] times h = (h/4)(3a + b). Similarly, Area of EFCD equals (1/2) times [b + (a + b)/2] times h = (1/2) times [(2b + a + b)/2] times h = (h/4)(3b + a). The required ratio is [(h/4)(3a + b)] divided by [(h/4)(3b + a)] = (3a + b)/(3b + a) = (3a + b) : (3b + a).
In simple words: Using the midpoint theorem, the line connecting the midpoints divides the trapezium into two smaller trapeziums. Calculate each area by adding the parallel sides and dividing by two, then find their ratio by canceling common factors.

Exam Tip: Apply the midpoint theorem systematically to find EF, then use the trapezium area formula carefully with correct substitutions - examiners award marks for showing each step clearly.

 

Question 8. In the adjoining figure, AB || DC and AB ≠ DC. If the diagonals AC and BD of the trapezium ABCD intersect at O, then which of the following statements is not true?
(1) area of triangle ABC = area of triangle ABD
(2) area of triangle ACD = area of triangle BCD
(3) area of triangle OAB = area of triangle OCD
(4) area of triangle OAD = area of triangle OBC
Answer: We use the fact that triangles standing on the same base and lying between the same pair of parallel lines have equal area. Since triangles ABC and ABD both rest on base AB and lie between parallel lines AB and DC, they have equal areas (equation 1). Similarly, triangles ACD and BCD both rest on base CD and lie between the same parallel lines, so they have equal areas (equation 2). From the figure and equation (1), triangle AOB plus triangle OAD equals triangle AOB plus triangle OBC. Subtracting triangle AOB from both sides gives us area of triangle OAD equals area of triangle OBC (equation 3). This leaves statement (3) - that area of triangle OAB equals area of triangle OCD - unproven. In fact, these two triangles do not have equal areas because they do not satisfy the conditions required (same base and parallel lines). Therefore, statement (3) is not true.
In simple words: Statements 1, 2, and 4 all follow from the rule that triangles on the same base between parallel lines have equal area. Statement 3 cannot be proven this way.

Exam Tip: Recognize which geometric conditions (same base, parallel lines) lead to equal areas, and test each statement methodically against these conditions - this approach eliminates guessing.

 

Question 9. Consider the following two statements:
Statement 1: The line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.
Statement 2: Diagonals of a parallelogram divide it into four triangles of equal area.
Which of the following is valid?
(1) Both the statements are true.
(2) Both the statements are false.
(3) Statement 1 is true, and Statement 2 is false.
(4) Statement 1 is false, and Statement 2 is true.
Answer: Consider parallelogram ABCD where E and F are midpoints of sides AB and CD respectively. Draw a perpendicular DG from D to side AB, with length h (the height from AB to CD). The area of parallelogram ABCD equals base AB times height h. For parallelogram AEFD: its area equals AE times h. Since E is the midpoint of AB, AE equals AB/2, so the area of AEFD is (AB/2) times h (equation 1). For parallelogram EBCF: its area equals EB times h. Since E is the midpoint of AB, EB equals AB/2, so the area of EBCF is (AB/2) times h (equation 2). From equations (1) and (2), the two parallelograms have equal areas. Thus, Statement 1 is true. For the diagonals: In a parallelogram, the diagonals bisect each other at point O, so AO equals OC. In triangle ACD, O is the midpoint of AC, making OD a median. A median divides a triangle into two triangles of equal area, so area of triangle AOD equals area of triangle COD (equation 3). In triangle ABC, O is the midpoint of AC, making OB a median, so area of triangle AOB equals area of triangle COB (equation 4). In triangle ADB, O is the midpoint of BD, making OA a median, so area of triangle AOD equals area of triangle AOB (equation 5). From equations (3), (4), and (5): all four triangles have equal areas. Thus, Statement 2 is true. Therefore, both statements are true.
In simple words: A line connecting opposite midpoints splits a parallelogram into two equal parts. The two diagonals of any parallelogram cross to create four triangles that all have the same area.

Exam Tip: For Statement 1, use the area formula (base times height) and the midpoint property to show equal areas. For Statement 2, identify the midpoint of each diagonal and apply the median theorem to each triangle - both paths lead to the correct answer.

 

Assertion Reason Type Questions

 

Question. Assertion (A): Area of trapezium = 1/2 (sum of parallel sides) x height.
Reason (R): A parallelogram and a rectangle on the same base and between the same parallel lines are equal in area.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: By the standard formula, the area of a trapezium is calculated as (1/2) times the sum of the parallel sides times the perpendicular height. Therefore, Assertion (A) is true. The statement about parallelograms and rectangles is also correct: shapes with the same base and lying between identical parallel lines possess equal areas. Since a rectangle is a special case of a parallelogram, this relationship holds. Therefore, Reason (R) is true. However, the reason given in (R) does not explain why the trapezium formula works the way it does. The trapezium area formula arises from dividing the shape into triangles or by averaging the two parallel sides, not from comparing it to a parallelogram and rectangle. While both statements are correct, Reason (R) is not the underlying justification for Assertion (A).
In simple words: Both the trapezium area formula and the statement about equal-area shapes are correct facts, but one does not explain the other.

Exam Tip: When evaluating Assertion-Reason pairs, check not only whether each statement is true, but also whether the reason logically supports the assertion - a true reason that doesn't explain the assertion still earns a partial-credit answer.

 

Question. Assertion (A): In the adjoining figure, the two triangles are equal figures.
Reason (R): Two figures are called equal if and only if they have the same area.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: The term "equal figures" in geometry specifically means two shapes that have the same area, regardless of whether they are congruent (identical in shape and size). Using the triangle area formula (1/2 times base times height): the first triangle with base 8 cm and height 6 cm has area = (1/2) times 8 times 6 = (1/2) times 48 = 24 cm². The second triangle with base 12 cm and height 4 cm has area = (1/2) times 12 times 4 = (1/2) times 48 = 24 cm². Since both triangles have the same area, they are equal figures. Therefore, Assertion (A) is true. The Reason (R) correctly defines "equal figures" - two figures are equal if and only if they have the same area. This definition does not require them to be congruent, only to have equal areas. Therefore, Reason (R) is also true and correctly explains why Assertion (A) is true.
In simple words: Two shapes are called equal if they cover the same amount of space, even if they look different. The reason explains this idea perfectly.

Exam Tip: Remember that "equal figures" means same area only, not congruence - distinguishing this concept from congruence is crucial for assertion-reason problems involving area and shape comparisons.

 

Question. Assertion (A): Two triangles having equal area are congruent.
Reason (R): Two congruent triangles have equal area.
Answer: Reason (R) is true: if two triangles are congruent (meaning they have the same shape and size), then they must have equal areas. This is a fundamental property of congruent figures. However, Assertion (A) claims the reverse - that two triangles with equal areas must be congruent. This is false. Two triangles can have the same area without being congruent. For example, a triangle with base 8 and height 6 has area 24, and a triangle with base 12 and height 4 also has area 24, but these triangles are not congruent because they have different shapes. Therefore, Assertion (A) is false while Reason (R) is true. The correct answer reflects that equal area does not guarantee congruence, only that congruence guarantees equal area (a one-way relationship, not reversible).
In simple words: Two triangles can have the same area but different shapes. However, two triangles that are exactly the same shape and size always have the same area.

Exam Tip: Understand the direction of logical implication: congruence always brings equal area, but equal area does not always bring congruence - testing both directions helps identify false assertions in reasoning problems.

 

Question. Assertion (A) is true, Reason (R) is false.

Exam Tip: Check both statements independently before deciding the relationship between them - a true assertion and false reason is a distinct outcome from other combinations.

 

Question. Assertion (A) is false, Reason (R) is true.

Exam Tip: When the reason is correct but the assertion fails, the explanation itself is sound even though the claim being explained is wrong.

 

Question. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Exam Tip: This is the strongest option - both parts must be true AND logically connected for full marks.

 

Question. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Exam Tip: Even when both statements are accurate, they must be properly linked - a true reason that does not explain the assertion does not earn full credit.

 

Chapter Test

 

Question 1(a). In the figure (1) given below, ABCD is a rectangle (not drawn to scale) with side AB = 4 cm and AD = 6 cm. Find (i) the area of parallelogram DEFC (ii) area of △EFG.

Answer:(i) We know that when a parallelogram and a rectangle share the same base and lie between the same parallel lines, they have equal areas. Looking at the figure, rectangle ABCD and parallelogram DEFC both rest on base DC and lie between parallel lines DG and AF. Therefore, the area of parallelogram DEFC equals the area of rectangle ABCD. Area of rectangle ABCD = AB × AD = 4 × 6 = 24 cm² Hence, area of parallelogram DEFC = 24 cm²
In simple words: The parallelogram DEFC has the same area as rectangle ABCD because they sit on the same base between the same parallel lines. The rectangle's area is length times width, which is 4 times 6 equals 24 square centimetres.(ii) We know that a triangle covering the same base and lying between the same parallel lines as a parallelogram takes up half the parallelogram's area. Triangle GEF and parallelogram DEFC share base EF and lie between parallel lines DG and AF. Therefore, area of triangle EFG = \( \frac{1}{2} \) × area of parallelogram DEFC = \( \frac{1}{2} \) × 24 = 12 cm² Hence, area of triangle EFG = 12 cm²
In simple words: A triangle using the same base as the parallelogram takes up exactly half its area. Half of 24 is 12 square centimetres.

Exam Tip: Remember the key relationships - a parallelogram and rectangle on the same base between same parallels have equal area; a triangle on the same base between same parallels has half the area of the parallelogram.

 

Question 1(b). In the figure (2) given below, PQRS is a parallelogram formed by drawing lines parallel to the diagonals of a quadrilateral ABCD through its corners. Prove that area of || gm PQRS = 2 × area of quad. ABCD.

Answer:Looking at the figure: Triangle ACD and parallelogram ACRS occupy the same base AC and lie between the same parallel lines AC and SR. Therefore, area of triangle ACD = \( \frac{1}{2} \) × area of parallelogram ACRS .......(i) This means: area of parallelogram ACRS = 2 × area of triangle ACD Similarly: Triangle ABC and parallelogram APQC occupy the same base AC and lie between the same parallel lines AC and PQ. Therefore, area of triangle ABC = \( \frac{1}{2} \) × area of parallelogram APQC .......(ii) This means: area of parallelogram APQC = 2 × area of triangle ABC Now adding equations (i) and (ii): Area of parallelogram ACRS + area of parallelogram APQC = 2 × area of triangle ACD + 2 × area of triangle ABC Area of parallelogram PQRS = 2 × (area of triangle ACD + area of triangle ABC) Area of parallelogram PQRS = 2 × (area of quadrilateral ABCD) Hence, proved that area of parallelogram PQRS = 2 × area of quadrilateral ABCD
In simple words: The big parallelogram made from the diagonal lines is exactly twice the size of the original quadrilateral inside it. This works because two smaller parallelograms together add up to double what the two triangles were worth separately.

Exam Tip: The key is recognizing that the parallelogram can be split into two smaller parallelograms, each of which has double the area of a triangle that shares its base and parallel lines.

 

Question 2. In the parallelogram ABCD, P is a point on the side AB and Q is a point on the side BC. Prove that (i) area of △CPD = area of △AQD (ii) area of △ADQ = area of △APD + area of △CPB.

Answer:(i) Triangle CPD and parallelogram ABCD share base CD and lie between the same parallel lines AB and CD. Therefore: area of △CPD = \( \frac{1}{2} \) × area of parallelogram ABCD .......(1) Triangle AQD and parallelogram ABCD share base AD and lie between the same parallel lines AD and BC. Therefore: area of △AQD = \( \frac{1}{2} \) × area of parallelogram ABCD .......(2) From equations (1) and (2): Area of △CPD = area of △AQD Hence, proved that area of △CPD = area of △AQD
In simple words: Both triangles take up the same amount of space because each one is half the parallelogram when measured from its own base and parallel lines.(ii) From part (i): Area of △CPD = \( \frac{1}{2} \) × area of parallelogram ABCD Therefore: area of parallelogram ABCD - area of △CPD = \( \frac{1}{2} \) × area of parallelogram ABCD .......(3) Looking at the figure, the parallelogram minus triangle CPD breaks down into two smaller triangles: Area of parallelogram ABCD - area of △CPD = area of △APD + area of △CPB .......(4) Combining equations (3) and (4): Area of △APD + area of △CPB = \( \frac{1}{2} \) × area of parallelogram ABCD From equation (2) we already found: Area of △ADQ = \( \frac{1}{2} \) × area of parallelogram ABCD Therefore: area of △APD + area of △CPB = area of △ADQ Hence, proved that area of △ADQ = area of △APD + area of △CPB
In simple words: When you take away triangle CPD from the whole parallelogram, you get exactly the same amount as triangle AQD - because both equal half the parallelogram. The leftover pieces are triangles APD and CPB put together.

Exam Tip: Draw clear diagrams showing which triangles and regions you are discussing - the relationships between the parts and the whole are easier to track visually. Use the property that triangles on the same base between the same parallels are equal.

 

Question 3. In the adjoining figure, X and Y are points on the side LN of triangle LMN. Through X, a line is drawn parallel to LM to meet MN at Z. Prove that area of △LZY = area of quad. MZYX.

Answer:From the figure: Triangle LZX and triangle MZX share the same base XZ and lie between the same parallel lines LM and XZ. Therefore: area of △LZX = area of △MZX Now add the area of triangle XZY to both sides of this equation: Area of △LZX + area of △XZY = area of △MZX + area of △XZY Looking at the figure, we can see: △LZX + △XZY = △LZY and △MZX + △XZY = quadrilateral MZYX Therefore: area of △LZY = area of quadrilateral MZYX Hence, proved that area of △LZY = area of quadrilateral MZYX
In simple words: Two triangles that share the same base between parallel lines have equal areas. When you add the same triangle to each side, the totals stay equal. On one side you get triangle LZY, on the other side you get the four-sided shape MZYX, and they are the same size.

Exam Tip: The strategy here is to use equal triangles as a stepping stone - establish equality of two triangles, then add a common region to both to reach the final result. This is a useful technique for many area proofs.

 

Question 4. Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.

Answer:Since ABC is an equilateral triangle, let each side have length x cm. PN, PM, and PL are perpendiculars from point P to sides AB, AC, and BC respectively. AD is an altitude from vertex A to side BC. Join PA, PB, and PC. Area of △ABC = \( \frac{1}{2} \) × base × altitude = \( \frac{1}{2} \) × BC × AD = \( \frac{x}{2} \) AD .......(1) Area of △APB = \( \frac{1}{2} \) × AB × NP = \( \frac{1}{2} \) × x × NP .......(2) Area of △APC = \( \frac{1}{2} \) × AC × MP = \( \frac{1}{2} \) × x × MP .......(3) Area of △BPC = \( \frac{1}{2} \) × BC × LP = \( \frac{1}{2} \) × x × LP .......(4) Adding equations (2), (3), and (4): Area of (△APB + △APC + △BPC) = \( \frac{1}{2} \) × (x × NP + x × MP + x × LP) = \( \frac{x}{2} \) × (NP + MP + LP) .......(5) From the figure, the three smaller triangles together make up the whole triangle ABC: Area of △ABC = \( \frac{x}{2} \) × (NP + MP + LP) .......(6) From equations (1) and (6): \( \frac{x}{2} \) × AD = \( \frac{x}{2} \) × (NP + LP + MP) Therefore: AD = NP + LP + MP Hence, proved that the sum of the three perpendiculars is equal to the altitude of the triangle
In simple words: When you add up the three perpendiculars from point P to each side, the total distance equals the height of the triangle. This is because the three smaller triangles formed together cover the whole big triangle.

Exam Tip: The key insight is breaking the main triangle into three smaller triangles using the point inside, then expressing each smaller triangle's area in terms of a perpendicular. When these add up, the perpendiculars combine to give the altitude.

 

Question 5. If each diagonal of a quadrilateral divides it into two triangles of equal areas, then prove that the quadrilateral is a parallelogram.

Answer:Let ABCD be a quadrilateral where each diagonal divides it into triangles of equal areas. Then: Area of △ABC = \( \frac{1}{2} \) × area of ABCD .......(1) Area of △ABD = \( \frac{1}{2} \) × area of ABCD .......(2) Area of △BCD = \( \frac{1}{2} \) × area of ABCD .......(3) We know that triangles on the same base and having equal areas lie between the same parallel lines. From equations (1) and (2): Area of △ABC = area of △ABD Since △ABC and △ABD share base AB and have equal area, they must lie between the same parallel lines. Therefore: AB || CD Similarly, from equations (1) and (3): Area of △ABC = area of △BCD Since △ABC and △BCD share base BC and have equal area, they must lie between the same parallel lines. Therefore: BC || AD Since AB || CD and BC || AD, quadrilateral ABCD is a parallelogram. Hence, proved that ABCD is a parallelogram
In simple words: When both diagonals split the quadrilateral into equal-area triangles, this forces opposite sides to be parallel to each other - which is exactly what makes a parallelogram.

Exam Tip: Use the converse property - if two triangles on the same base have equal area, they lie between parallel lines. Apply this property separately for each diagonal to establish that both pairs of opposite sides are parallel.

 

Question 6. In the adjoining figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of △DFB = 3 cm², find the area of parallelogram ABCD.

Answer:From the figure, BE is a straight line. Since BC || AD (opposite sides of a parallelogram), we have CE || AD. Since both CE and AD are parallel, and they connect with other sides, quadrilateral ACED forms a parallelogram. The diagonals of parallelogram ACED are AE and DC. Diagonals of a parallelogram bisect each other, so F is the midpoint of DC. Since F is the midpoint of DC, line BF is a median of triangle BDC. A median divides a triangle into two triangles of equal area. Therefore: area of △BFC = area of △DFB = 3 cm² So: area of △BDC = area of △BFC + area of △DFB = 3 + 3 = 6 cm² From the figure, triangle BDC and parallelogram ABCD share the same base CD and lie between the same parallel lines AB and CD. Therefore: area of △BDC = \( \frac{1}{2} \) × area of parallelogram ABCD So: 6 = \( \frac{1}{2} \) × area of parallelogram ABCD Therefore: area of parallelogram ABCD = 12 cm² Hence, area of parallelogram ABCD = 12 cm²
In simple words: A median cuts a triangle exactly in half. Since BF is the median of triangle BDC, each piece has area 3 cm². Together they make 6 cm². The triangle is half the parallelogram, so the parallelogram must be 12 cm².

Exam Tip: Recognize when a quadrilateral forms a parallelogram from the given conditions - this allows you to use the property that diagonals bisect each other. Also remember that a median of a triangle creates two equal-area triangles.

 

Question 7. In the adjoining figure, ABCD is a square. E and F are mid-points of sides BC and CD respectively. If R is mid-point of EF, prove that: area of △AER = area of △AFR

Answer:Since ABCD is a square, let each side have length a. E is the midpoint of BC, so BE = EC = a/2. F is the midpoint of CD, so CF = FD = a/2. R is the midpoint of EF. Since R is the midpoint of segment EF, and we need to compare triangles AER and AFR, we observe that both triangles share vertex A and their bases lie on line EF. Triangles that share the same vertex and have bases on the same line segment are equal in area if and only if their bases are equidistant from that vertex. Since R is the midpoint of EF, the segments ER and RF are equal in length. Triangles AER and AFR both have the same height from vertex A to line EF (the perpendicular distance from A to line EF is the same for both). Since ER = RF (R is the midpoint of EF) and both triangles have the same perpendicular height from A to line EF: Area of △AER = \( \frac{1}{2} \) × ER × height = \( \frac{1}{2} \) × RF × height = area of △AFR Hence, proved that area of △AER = area of △AFR
In simple words: R is the exact middle of line EF, so segment ER equals segment RF. Both triangles AER and AFR stand on these equal bases and share the same height from point A. Equal bases and equal heights mean equal areas.

Exam Tip: When a point is the midpoint of a line segment and two triangles share a common vertex but have their bases on that segment, equal bases guarantee equal areas (provided the height is the same). This is a quick way to prove area equality without calculating exact measurements.

 

Question 7. Prove that Area of ∆AER = Area of ∆AFR
Answer: In triangles ABE and ADF, we have AB = AD (sides of a square are equal), angle B = angle D (each interior angle of a square measures 90 degrees), and BE = DF (since E marks the midpoint of BC and F marks the midpoint of DC). Using the SAS criterion, triangle ABE is congruent to triangle ADF. From this congruence, we get AE = AF.

Now consider triangles AER and AFR. We know AE = AF (from the step above), AR is a shared side (common to both triangles), and ER = FR (as R lies at the midpoint of EF). By the SSS criterion, triangle AER is congruent to triangle AFR.

Since the two triangles are congruent, their areas must be equal. Therefore, the area of triangle AER equals the area of triangle AFR.
In simple words: Two triangles are equal in area because they match on all three sides, which means they have the same shape and size.

Exam Tip: Clearly identify the congruent triangles at each step and state which criterion (SAS or SSS) applies - examiners award marks for naming the specific axiom used.

 

Question 8. In the adjoining figure, X and Y are midpoints of the sides AC and AB respectively of ∆ABC. QP || BC and CYQ and BXP are straight lines. Prove that area of ∆ABP = area of ∆ACQ.
Answer: Given that X and Y serve as the midpoints of AC and AB respectively, the midpoint theorem tells us that XY is parallel to BC. Since we are also given that QP is parallel to BC, we can conclude that QP, BC, and XY are all parallel to one another.

Within triangle BAP, Y is the midpoint of AB and the line segment XY is parallel to AP. By the converse of the midpoint theorem, X must be the midpoint of BP. This gives us the relation \( XY = \frac{1}{2}AP \) ... (1).

Using the same reasoning in triangle AQC: X is the midpoint of AC and XY runs parallel to QA. By the converse of the midpoint theorem, Y is the midpoint of QC. We obtain \( XY = \frac{1}{2}QA \) ... (2).

From equations (1) and (2), we can deduce \( \frac{1}{2}QA = \frac{1}{2}AP \), which simplifies to QA = AP.

Since triangles ABP and ACQ have equal bases (QA = AP) and lie between the same pair of parallel lines BC and QP, the triangles must have equal areas. Therefore, the area of triangle ABP equals the area of triangle ACQ.
In simple words: When two triangles sit between the same two parallel lines and their bases are the same length, the triangles cover the same amount of space.

Exam Tip: Always apply the midpoint theorem and its converse systematically - show each step where you identify a midpoint, and confirm parallel relationships before concluding equality of areas.