ML Aggarwal Class 9 Maths Solutions Chapter 12 Pythagoras Theorem

 

Question 1. If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.
Answer: The sum of all angles in a quadrilateral equals 360°. Let the other two angles be 3x and 4x. We can write: 3x + 4x + 40° + 110° = 360°. Simplifying, 7x + 150° = 360°, so 7x = 210°. Dividing both sides by 7 gives x = 30°. Therefore, 3x = 90° and 4x = 120°. The other two angles are 90° and 120°.
In simple words: Add up all four angles. Two are already known, and the other two are multiples of the same number. Solve for that number, then find both unknown angles.

Exam Tip: Always remember the sum of angles in any quadrilateral is 360°. Set up an equation using the given ratio to find the unknown angles quickly.

 

Question 2. If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.
Answer: Let the quadrilateral be ABCD with angles x, 2x, 3x, and 4x in order. Since the sum of angles equals 360°, we have x + 2x + 3x + 4x = 360°, which gives 10x = 360°, so x = 36°. This means ∠A = 36°, ∠B = 72°, ∠C = 108°, and ∠D = 144°. Looking at the figure, angles A and D are co-interior angles, and their sum is 36° + 144° = 180°. When co-interior angles sum to 180°, the sides AB and CD must be parallel. We can also check that opposite angles do not sum to 180° (36° + 108° = 144° and 72° + 144° = 216°), which confirms this is a trapezium and not a parallelogram.
In simple words: If two angles on the same side of a line add up to 180°, those opposite sides are parallel. This makes it a trapezium.

Exam Tip: When proving a quadrilateral is a trapezium, show that one pair of opposite sides is parallel (using co-interior angles) and that opposite angles do not equal.

 

Question 3. If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.
Answer: Let parallelogram ABCD have ∠A = x and ∠B = (2/3)x. Since AD is parallel to BC, the co-interior angles on the same side must sum to 180°. Thus, x + (2/3)x = 180°. Combining the terms, (3x + 2x)/3 = 180°, so (5x)/3 = 180°. Solving for x, we get x = (180° × 3)/5 = 108°. Therefore, ∠B = (2/3) × 108° = 72°. In a parallelogram, opposite angles are equal, so ∠C = ∠A = 108° and ∠D = ∠B = 72°. The angles are 108°, 72°, 108°, and 72°.
In simple words: One angle is two-thirds of the next one. Two adjacent angles in a parallelogram always add to 180°. Use this to find x, then use the opposite angle rule.

Exam Tip: Remember that adjacent angles in a parallelogram are supplementary (sum to 180°), while opposite angles are equal.

 

Question 4(a). In figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.
Answer: Since AD is parallel to BC, co-interior angles sum to 180°. We have ∠A + ∠B = 180°, so ∠B = 180° - 70° = 110°. From the figure, ∠B = ∠DBA + ∠DBC, which gives 110° = ∠DBA + 80°, so ∠DBA = 30°. In triangle DAB, the angles must sum to 180°, so ∠DAB + ∠DBA + ∠ADB = 180°. Substituting, 70° + 30° + ∠ADB = 180°, which gives ∠ADB = 80°. Since opposite angles in a parallelogram are equal, ∠C = ∠A = 70°. In triangle DBC, we have ∠DBC + ∠BCD + ∠CDB = 180°, so 80° + 70° + ∠CDB = 180°, giving ∠CDB = 30°. Therefore, ∠ADB = 80° and ∠CDB = 30°.
In simple words: Split angle B into two parts using the diagonal. Find each part separately using parallel line properties and triangle angle rules.

Exam Tip: When a diagonal divides an angle, use the property that the two parts add up to the original angle. This helps break down the problem into manageable triangles.

 

Question 4(b). In figure (2) given below, ABCD is a parallelogram. Find the angles of △AOD.
Answer: In triangle BOC, the angles sum to 180°. We have ∠BOC + ∠OCB + ∠CBO = 180°, so ∠BOC + 35° + 77° = 180°, which gives ∠BOC = 68°. Since the diagonals of the parallelogram intersect at O, angles ∠AOD and ∠BOC are vertically opposite, so ∠AOD = 68°. Also using the property of alternate angles formed by the diagonals and parallel sides, ∠OAD = ∠OCB = 35° and ∠ADO = ∠CBO = 77°. Therefore, the angles of triangle AOD are ∠AOD = 68°, ∠OAD = 35°, and ∠ADO = 77°.
In simple words: One triangle in the parallelogram is given. Use vertically opposite angles and alternate angle properties to find the angles of the opposite triangle.

Exam Tip: Vertically opposite angles at the intersection of diagonals are always equal. Alternate angles formed by parallel lines and a transversal help find the remaining angles.

 

Question 4(c). In figure (3) given below, ABCD is a rhombus. Find the value of x.
Answer: Since AD is parallel to BC, co-interior angles sum to 180°. We have ∠A + ∠B = 180°, so ∠B = 180° - 72° = 108°. In a rhombus, the diagonals bisect the angles. Therefore, x equals half of angle B, which is (1/2) × 108° = 54°. Thus, x = 54°.
In simple words: Find the angle at B using the parallel side property. The diagonal cuts this angle in half, so divide by 2 to get x.

Exam Tip: A key property of rhombuses is that diagonals bisect all angles. Always use this when finding unknown angles marked by a diagonal.

 

Question 5(a). In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.
Answer: In a parallelogram, opposite sides are equal. Therefore, BC = AD = 2x and AB = CD = 3x. Since AB and CD must be equal, we have 3x = 2y + 2. Rearranging, 2y = 3x - 2, so y = (3x - 2)/2. The perimeter is the sum of all four sides: AB + BC + CD + AD = 40. Substituting the expressions, 3x + 2x + (2y + 2) + 2x = 40. Replacing 2y + 2 with 3x, we get 3x + 2x + 3x + 2x = 40, which simplifies to 10x = 40, so x = 4. Using the relation y = (3x - 2)/2, we find y = (12 - 2)/2 = 5. Therefore, x = 4 and y = 5.
In simple words: Opposite sides are equal in a parallelogram. Use this fact along with the perimeter to set up equations and solve for both variables.

Exam Tip: Always mark opposite sides with the same expression. Setting up the perimeter equation carefully ensures you solve for both unknowns correctly.

 

Question 5(b). In figure (2) given below, ABCD is a parallelogram. Find the values of x and y.
Answer: In a parallelogram, opposite angles are equal. Therefore, ∠A = ∠C, which means 3x - 20° = x + 40°. Solving, 3x - x = 40° + 20°, so 2x = 60°, giving x = 30°. Since AD is parallel to BC, co-interior angles sum to 180°. We have ∠A + ∠B = 180°, so (3x - 20°) + (y + 15°) = 180°. Substituting x = 30°, we get 90° - 20° + y + 15° = 180°, which simplifies to 85° + y = 180°, so y = 95°. Therefore, x = 30° and y = 95°.
In simple words: Use the opposite angle rule to find x. Then use the co-interior angle rule (angles on the same side add to 180°) to find y.

Exam Tip: In parallelogram problems, apply opposite angle equality first, then co-interior angle rules to find all unknowns systematically.

 

Question 5(c). In figure (3) given below, ABCD is a rhombus. Find x and y.
Answer: All sides of a rhombus are equal. Therefore, AB = AD, which means 4x - 4 = 3x + 2. Solving, 4x - 3x = 2 + 4, so x = 6. This gives AB = 3(6) + 2 = 20. In triangle ABD, since AB = AD, the triangle is isosceles. Angles opposite to equal sides are equal, so ∠ADB = ∠ABD. Let both equal a. The angles of the triangle sum to 180°, so ∠DAB + ∠ADB + ∠ABD = 180°, which gives 60° + a + a = 180°. Thus, 2a = 120°, so a = 60°. This means all three angles equal 60°, making triangle ABD equilateral. Therefore, all sides are equal: BD = AB = 20. Since y - 1 = 20, we have y = 21. Thus, x = 6 and y = 21.
In simple words: Equal sides give equal angles in an isosceles triangle. If all angles turn out to be 60°, the triangle is equilateral and all three sides match.

Exam Tip: When a triangle inside a rhombus has two equal sides, check if it might be equilateral by finding its angles. This shortcut saves time.

 

Question 6. The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.
Answer: The diagonals of a rectangle are equal in length and bisect each other. Therefore, AP = BP. In triangle ABP, since AP = BP, the triangle is isosceles, and angles opposite to equal sides are equal. Thus, ∠PAB = ∠ABP = 50°. The angles of the triangle sum to 180°, so ∠PAB + ∠ABP + ∠APB = 180°, which gives 50° + 50° + ∠APB = 180°, so ∠APB = 80°. Since the diagonals intersect at P, angles ∠DPC and ∠APB are vertically opposite. Therefore, ∠DPC = ∠APB = 80°.
In simple words: The diagonals split each other equally. This makes one triangle isosceles. Find its angle at P, then use vertically opposite angles.

Exam Tip: When diagonals of a rectangle meet, they create four triangles. Two pairs of these triangles are congruent. Use isosceles triangle properties to solve quickly.

 

Question 7(a). In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.
Answer: Triangle EBC is equilateral, so all its sides are equal: EB = BC = EC. In the square, all sides are equal: AD = CD = BC = AB. Combining these facts, we have BC = EC = CD, which means EC = CD. Therefore, triangle ECD is isosceles with EC = CD. Angles opposite to equal sides are equal in an isosceles triangle, so ∠CED = ∠CDE. Let this angle equal m. The angles of triangle ECD sum to 180°, so ∠ECD + ∠CED + ∠CDE = 180°. Since triangle EBC is equilateral, ∠ECB = 60°, and since ∠BCD = 90° (angle in the square), we have ∠ECD = ∠ECB + ∠BCD = 60° + 90° = 150°. Thus, 150° + m + m = 180°, giving 2m = 30°, so m = 15°. Therefore, ∠CED = 15°. Since ∠BEC = 60° (angle in equilateral triangle), we have ∠BED = ∠BEC + ∠CED = 60° + 15° = 75°. Thus, x = 75°.
In simple words: The equilateral triangle and square share a side. Use side equality and angle properties of both shapes to find the angle step by step.

Exam Tip: When composite figures combine regular shapes, list all equal sides and angles from each shape first. This systematic approach prevents errors and speeds up the solution.

 

Question 7(b). In Figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of △AOB.
Answer: Since AE is a straight line, ∠OCD = 180° - 146° = 34°. In a rectangle, the diagonals have equal length and split each other into two equal parts. Therefore, OC = OD. In △OCD, since OC = OD, the angles opposite these equal sides must also be equal, so ∠ODC = ∠OCD = 34°. The sum of angles in the triangle gives us: 34° + 34° + ∠DOC = 180°, which means ∠DOC = 112°. Since vertically opposite angles are equal, ∠AOB = ∠DOC = 112°. Using the property that angles on the same line segment from a rectangle's vertices equal the angles of the triangle at the intersection, ∠OAB = ∠OCD = 34° and ∠OBA = ∠ODC = 34°.
In simple words: The angle at O is 112°, and both base angles are 34° each.

Exam Tip: Remember that rectangle diagonals are equal and bisect each other. Use vertically opposite angles to link triangles formed by the diagonals, and apply the angle-sum property to find missing angles.

 

Question 7(c). In figure (3) given below, ABCD is a rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3 : 2, find the angles of the △AOD.
Answer: Let ∠OAB = 3x and ∠OBA = 2x. Since the diagonals of a rhombus are perpendicular to each other, ∠AOB = 90°. Using the angle-sum property in △AOB: 90° + 3x + 2x = 180°, which simplifies to 5x = 90°, giving x = 18°. Therefore, ∠OAB = 54° and ∠OBA = 36°. Since the diagonals of a rhombus bisect the vertex angles, ∠OAD = ∠OAB = 54°. In △AOD, the diagonals remain perpendicular, so ∠AOD = 90°. Using the angle-sum property: 90° + 54° + ∠ODA = 180°, so ∠ODA = 36°.
In simple words: The angle at the centre is 90°. One base angle is 54° and the other is 36°.

Exam Tip: In a rhombus, diagonals always meet at right angles. Use the given ratio to set up algebraic expressions, then apply the angle-sum property to solve systematically.

 

Question 8(a). In figure (1) given below, ABCD is a trapezium. Find the values of x and y.
Answer: Since AB and DC are parallel sides of the trapezium, the adjacent interior angles on the same side sum to 180°. Thus, ∠A + ∠D = 180°. Substituting the given expressions: (x + 20°) + (2x + 10°) = 180°, which simplifies to 3x + 30° = 180°. Solving, 3x = 150°, so x = 50°. Similarly, ∠B + ∠C = 180° (as they too are co-interior angles on the same side of the parallel lines). Therefore, y + 92° = 180°, which gives y = 88°.
In simple words: Angles on the same side of a trapezium add to 180°. Use this rule for both pairs of adjacent angles.

Exam Tip: Always identify which sides are parallel in a trapezium, then apply the co-interior angle property. Check both pairs of adjacent angles.

 

Question 8(b). In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and y.
Answer: In an isosceles trapezium with AB || DC, the co-interior angles on each side of the parallel lines sum to 180°. At the side AD: ∠A + ∠D = 180°, so 2x + 3x = 180°, giving 5x = 180° and x = 36°. Since this is an isosceles trapezium, opposite angles are supplementary (sum to 180°). Therefore, ∠A + ∠C = 180°. Substituting: 2(36°) + y = 180°, which gives 72° + y = 180°, so y = 108°.
In simple words: Use co-interior angles for parallel sides, then use the isosceles property that opposite angles sum to 180°.

Exam Tip: In an isosceles trapezium, remember that the base angles are equal, and opposite angles are supplementary. Apply both properties when solving for unknowns.

 

Question 8(c). In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.
Answer: In a kite, the diagonals split the vertex angles into equal parts. Therefore, ∠OCD = ∠DCB / 2 = 64° / 2 = 32° and ∠OAB = ∠DAB / 2 = 112° / 2 = 56°. The diagonals of a kite are perpendicular, so ∠DOC = 90°. In △OCD, applying the angle-sum property: 32° + 90° + ∠ODC = 180°, which gives ∠ODC = 58°. In △AOB, where ∠AOB = 90° (perpendicular diagonals) and ∠OAB = 56°, we have: 90° + 56° + ∠OBA = 180°, so ∠OBA = 34°.
In simple words: Kite diagonals are perpendicular and bisect the angles at those two vertices. Use these facts with the angle-sum rule.

Exam Tip: For kites, always recall that one diagonal bisects the vertex angles while the diagonals meet at 90°. This combination is key to solving angle problems quickly.

 

Question 9(i). Prove that each angle of a rectangle is 90°.
Answer: In a rectangle ABCD, opposite sides are equal (AD = BC and AB = CD) and the diagonals are also equal (AC = BD). Consider triangles △ADC and △BCD. Since AD = BC, AC = BD, and DC is common to both triangles, by the SSS (Side-Side-Side) congruency rule, △ADC ≅ △BCD. By the Corresponding Parts of Congruent Triangles (C.P.C.T.), ∠ADC = ∠BCD. Let both these angles equal x. Since a rectangle is a parallelogram, adjacent angles are supplementary: ∠ADC + ∠BCD = 180°. Therefore, x + x = 180°, which gives x = 90°. So ∠ADC = ∠BCD = 90°. Since opposite angles in a parallelogram are equal, ∠DAB = ∠BCD = 90° and ∠ABC = ∠ADC = 90°. Thus, all four angles of a rectangle equal 90°.
In simple words: A rectangle is a special parallelogram where all angles must be equal to 90°, making it a four-right-angled figure.

Exam Tip: Use congruent triangles and the supplementary angle property of parallelograms to establish that each angle is 90°. State the congruency rule clearly for full marks.

 

Question 9(ii). If the angles of a quadrilateral are equal, prove that it is a rectangle.
Answer: Let quadrilateral ABCD have all angles equal: ∠A = ∠B = ∠C = ∠D = x. Since opposite angles are equal (∠A = ∠C and ∠B = ∠D), the quadrilateral satisfies the condition for a parallelogram, so ABCD is a parallelogram. The sum of all angles in any quadrilateral is 360°. Therefore, x + x + x + x = 360°, which simplifies to 4x = 360°, giving x = 90°. This means all four angles equal 90°. Since a parallelogram with all angles equal to 90° is, by definition, a rectangle, ABCD is a rectangle.
In simple words: If all four angles of a quadrilateral are the same, and opposite angles are equal, it must be a parallelogram. When each angle is 90°, it becomes a rectangle.

Exam Tip: First establish that the quadrilateral is a parallelogram using the equal opposite angles. Then use the angle-sum property to find the measure of each angle, confirming it is a rectangle.

 

Question 9(iii). If the diagonals of a rhombus are equal, prove that it is a square.
Answer: In rhombus ABCD, all sides are equal: AB = BC = CD = DA. Given that the diagonals are equal: AC = BD. Consider triangles △ABC and △BCD. Since AB = DC (sides of the rhombus), BC = BC (common side), and AC = BD (given equal diagonals), by the SSS congruency rule, △ABC ≅ △BCD. By C.P.C.T., ∠ABC = ∠BCD. Let both equal x. Since a rhombus is a parallelogram, adjacent angles are supplementary: ∠ABC + ∠DCB = 180°. Therefore, x + x = 180°, giving x = 90°. So ∠ABC = ∠BCD = 90°. A rhombus with all angles equal to 90° is, by definition, a square.
In simple words: A rhombus with equal diagonals must have all angles of 90°, which makes it a square.

Exam Tip: Use SSS congruency with the equal diagonal condition to show that the angles are equal. Then apply the supplementary angle property to conclude each angle is 90°.

 

Question 9(iv). Prove that every diagonal of a rhombus bisects the angles at the vertices.
Answer: Consider rhombus ABCD with diagonals AC and BD intersecting at O. To show that diagonal BD bisects ∠D, examine triangles △AOD and △COD. Since all sides of a rhombus are equal, AD = CD. Since the diagonals of a rhombus bisect each other, AO = OC. Also, DO = OD (same segment). By the SSS congruency rule, △AOD ≅ △COD. By C.P.C.T., ∠ADO = ∠CDO, which means BD bisects ∠D. Similarly, to show that BD bisects ∠B, consider triangles △AOB and △COB. Since AB = BC (sides of rhombus), BO = OB (same segment), and AO = OC (diagonals bisect each other), by SSS, △AOB ≅ △COB. By C.P.C.T., ∠ABO = ∠CBO, so BD bisects ∠B. The same reasoning applies to diagonal AC bisecting angles at A and C. Therefore, every diagonal of a rhombus bisects the angles at the vertices.
In simple words: Each diagonal of a rhombus splits the angle at each end into two equal parts. This follows from the equal sides and bisecting diagonals of a rhombus.

Exam Tip: Use the SSS congruency rule combined with the properties that all sides are equal and diagonals bisect each other. Always state which triangles are congruent and apply C.P.C.T. for the final conclusion.

 

Question 10. ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:
(i) AC bisects ∠C
(ii) ABCD is a rhombus
(iii) AC ⊥ BD
Answer:
(i) Given that AC bisects ∠A, so ∠CAB = ∠CAD = x (let) .......(i)

Since AB || CD (opposite sides of a parallelogram are parallel),

\[ \angle DCA = \angle CAB \text{ (Alternate angles are equal)} \]

∴ ∠DCA = x .......(ii)

Similarly,

\[ \angle BCA = \angle CAD \text{ (Alternate angles are equal)} \]

∴ ∠BCA = x .......(iii)

From (i), (ii), and (iii):

\[ \angle CAB = \angle CAD = \angle DCA = \angle BCA \]

Thus, ∠DCA = ∠BCA, which means AC bisects ∠C. Hence proved.

(ii) From the result above, ∠CAD = ∠DCA.

In △ADC, since two angles are equal, the sides opposite to them must be equal.

\[ DA = DC \text{ (Sides opposite to equal angles are equal)} \]

We know from the property of parallelograms that opposite sides are equal:

\[ AB = CD \text{ and } BC = DA \]

Combining these: AB = BC = CD = DA

Since all sides of parallelogram ABCD are equal, ABCD is a rhombus. Hence proved.

(iii) Consider △OAB and △OCB, where O is the intersection point of diagonals:

\[ OA = OC \text{ (Diagonals of a parallelogram bisect each other)} \]

\[ OB = OB \text{ (Common side)} \]

\[ AB = BC \text{ (Sides of rhombus)} \]

By SSS congruency, △OAB ≅ △OCB.

Therefore, ∠AOB = ∠BOC (corresponding parts of congruent triangles).

Since ∠AOB and ∠BOC are supplementary angles on a straight line:

\[ \angle AOB + \angle BOC = 180° \]

\[ 2 \angle AOB = 180° \]

\[ \angle AOB = 90° \]

Thus, AC ⊥ BD. Hence proved.
In simple words: When a diagonal of a parallelogram splits one angle equally, it also splits the opposite angle equally, making all four sides equal (a rhombus). In a rhombus, the two diagonals always meet at right angles.

Exam Tip: Use alternate angles carefully when parallel lines are cut by a transversal. Always check congruence criteria (SSS, SAS, ASA) step by step to prove angle or side relationships.

 

Question 11(i). Prove that bisectors of any two adjacent angles of a parallelogram are at right angles.
Answer: Let AC be the bisector of ∠A and BD be the bisector of ∠B, and let them meet at point M.

Since AD || BC, consecutive interior angles (also called co-interior angles) are supplementary:

\[ \angle A + \angle B = 180° \]

Dividing both sides by 2:

\[ \frac{\angle A}{2} + \frac{\angle B}{2} = 90° \]

Since AC and BD bisect ∠A and ∠B respectively:

\[ \angle MAB + \angle MBA = 90° \text{ .......(i)} \]

In △MAB, the sum of all angles equals 180°:

\[ \angle MAB + \angle MBA + \angle AMB = 180° \]

Substituting from (i):

\[ 90° + \angle AMB = 180° \]

\[ \angle AMB = 90° \]

Therefore, the two angle bisectors meet at a right angle. Hence proved.
In simple words: In a parallelogram, any two lines that split neighboring angles equally will always meet at exactly 90 degrees.

Exam Tip: Remember that consecutive angles in a parallelogram always add up to 180°. When you bisect both angles, their half-sums automatically give 90°, making the intersection perpendicular.

 

Question 11(ii). Prove that bisectors of any two opposite angles of a parallelogram are parallel.
Answer: Let ABCD be the parallelogram as shown. Since opposite angles in a parallelogram are equal:

\[ \angle A = \angle C \]

When we bisect both angles:

\[ \frac{\angle A}{2} = \frac{\angle C}{2} \]

Let AR be the bisector of ∠A and QC be the bisector of ∠C. Then:

\[ \angle DAR = \angle QCB \]

Consider △ADR and △CBQ. We have:

\[ \angle DAR = \angle QCB \text{ (Proved above)} \]

\[ AD = BC \text{ (Opposite sides of a parallelogram)} \]

\[ \angle D = \angle B \text{ (Opposite angles of a parallelogram)} \]

By ASA (Angle-Side-Angle) congruency, △ADR ≅ △CBQ.

Therefore, ∠DRA = ∠BQC (by C.P.C.T.) .......(i)

Also, since AB || CD (opposite sides of parallelogram):

\[ \angle RAQ = \angle DRA \text{ (Alternate angles are equal)} \]

.......(ii)

From (i) and (ii):

\[ \angle RAQ = \angle BQC \]

Since these are corresponding angles and they are equal, AR || QC. Hence proved.
In simple words: Lines that equally divide opposite angles of a parallelogram will always run side by side without meeting, because the split angles on one side match the angles on the other side.

Exam Tip: Focus on proving the triangles congruent first, then use alternate and corresponding angle properties to establish parallelism of the bisectors.

 

Question 11(iii). If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.
Answer: Let ABCD be the quadrilateral with diagonals AC and BD intersecting at O. Given that the diagonals are equal, bisect each other, and meet at 90°.

Since the diagonals bisect each other at 90°:

\[ \angle AOB = \angle COD = \angle BOC = \angle AOD = 90° \]

Step 1: Prove opposite sides are parallel

Consider △OAB and △ODC:

\[ OA = OC \text{ (Diagonals bisect each other)} \]

\[ OB = OD \text{ (Diagonals bisect each other)} \]

\[ \angle AOB = \angle COD \text{ (Vertically opposite angles)} \]

By SAS congruency, △OAB ≅ △ODC, so AB = CD (by C.P.C.T.).

Since ∠OAB = ∠OCD (by C.P.C.T.) and these are alternate angles, AB || CD.

Step 2: Find the angles of the quadrilateral

In △AOB, since OA = OB (both halves of equal diagonals):

\[ \angle OBA = \angle OAB = x \text{ (Angles opposite to equal sides)} \]

\[ x + x + 90° = 180° \]

\[ 2x = 90° \]

\[ x = 45° \]

Similarly, in △AOD, OA = OD, so ∠OAD = ∠ODA = 45°.

Thus, ∠A = ∠OAB + ∠OAD = 45° + 45° = 90°.

By similar reasoning, all angles equal 90°, and AB || CD, AD || BC.

Step 3: Prove all sides are equal

Since △AOB ≅ △AOD (both with equal legs OA and equal diagonals), AB = AD.

Using congruence of other triangle pairs and the parallelogram properties, AB = BC = CD = DA.

Since all sides are equal and all angles are 90°, ABCD is a square. Hence proved.
In simple words: When a quadrilateral's two diagonals are the same length, split each other exactly in half, and cross at right angles, the result is always a square - all four sides become equal and all angles become 90 degrees.

Exam Tip: Break the proof into clear steps: first establish that opposite sides are parallel, then show all angles are 90°, then prove all sides are equal. Use congruent triangles as your main tool throughout.

 

Question 12(i). If ABCD is a parallelogram in which diagonal AC bisects ∠A, then prove that ABCD is a rhombus.
Answer: Given: AC bisects ∠A, so ∠CAB = ∠CAD = x (let) .......(1)

Since AB || CD (opposite sides of parallelogram are parallel):

\[ \angle DCA = \angle CAB \text{ (Alternate angles are equal)} \]

∴ ∠DCA = x .......(2)

Similarly, ∠BCA = ∠CAD (alternate angles), so:

∠BCA = x .......(3)

From (1), (2), and (3):

\[ \angle CAB = \angle CAD = \angle DCA = \angle BCA \]

.......(4)

Since ∠CAD = ∠DCA in △ADC, the sides opposite these equal angles must be equal:

\[ DA = DC \text{ (Sides opposite to equal angles in a triangle are equal)} \]

.......(5)

From the parallelogram property, opposite sides are equal:

\[ AB = CD \text{ and } BC = DA \]

.......(6)

From (5) and (6):

\[ AB = BC = CD = DA \]

Since ABCD is a parallelogram in which all sides are equal, ABCD is a rhombus. Hence proved.
In simple words: If a diagonal of a parallelogram splits one angle into two equal parts, then all four sides of that shape must be the same length, making it a rhombus.

Exam Tip: Use alternate angles formed by parallel lines and a transversal (the diagonal) to establish equal angles in triangles. Once you have equal angles in a triangle, the opposite sides must be equal.

 

Question 12(ii). If ABCD is a rectangle in which the diagonal BD bisects ∠B, then show that ABCD is a square.
Answer: Since BD bisects ∠B:

\[ \angle 1 = \angle 2 \]

Since ABCD is a rectangle, ∠B = 90°, so:

\[ \angle 1 = \angle 2 = 45° \]

Since AD || BC (opposite sides of rectangle):

\[ \angle 4 = \angle 1 = 45° \text{ (Alternate angles are equal)} \]

\[ \angle 3 = \angle 2 = 45° \text{ (Alternate angles are equal)} \]

In △ABD, since ∠1 = ∠4 = 45°, the sides opposite these equal angles are equal:

\[ AB = AD \text{ (Sides opposite to equal angles are equal)} \]

.......(i)

In △CBD, since ∠2 = ∠3 = 45°:

\[ BC = CD \text{ (Sides opposite to equal angles are equal)} \]

.......(ii)

From the rectangle property, opposite sides are equal:

\[ AD = BC \text{ and } AB = CD \]

.......(iii)

From (i), (ii), and (iii):

\[ AB = BC = CD = AD \]

Since all sides are equal and all angles are 90° (as in any rectangle), ABCD is a square. Hence proved.
In simple words: When a diagonal of a rectangle splits one corner angle exactly in half, it forces the two sides meeting at that corner to become equal. Since opposite sides of a rectangle are already equal, this makes all four sides equal, forming a square.

Exam Tip: In a rectangle, each corner is already 90°. When a diagonal bisects this angle, each resulting piece is 45°. Use this to find equal angles and therefore equal sides opposite to them.

 

Question 12(iii). Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer: Let ABCD be the quadrilateral with diagonals AC and BD intersecting at O. Given: the diagonals are equal in length, bisect each other, and meet at 90°.

Since the diagonals bisect each other at right angles:

\[ \angle AOB = \angle COD = \angle BOC = \angle AOD = 90° \]

Step 1: Prove all four sides are equal

Consider △OAB and △ODC:

\[ OA = OC \text{ (Diagonals bisect each other)} \]

\[ OB = OD \text{ (Diagonals bisect each other)} \]

\[ \angle AOB = \angle COD = 90° \text{ (Vertically opposite and both right angles)} \]

By SAS congruency, △OAB ≅ △ODC, so AB = CD (by C.P.C.T.).

Similarly, considering △OAD and △OBC:

\[ OA = OC \text{ (Diagonals bisect each other)} \]

\[ OD = OB \text{ (Diagonals bisect each other)} \]

\[ \angle AOD = \angle BOC = 90° \]

By SAS, △OAD ≅ △OBC, so AD = BC.

Considering △AOB and △AOD:

\[ AO = AO \text{ (Common side)} \]

\[ OB = OD \text{ (Diagonals bisect each other)} \]

\[ \angle AOB = \angle AOD = 90° \]

By SAS, △AOB ≅ △AOD, so AB = AD.

Therefore, AB = BC = CD = AD.

Step 2: Prove all angles are 90°

In △AOB, since OA = OB (both halves of equal diagonals):

\[ \angle OBA = \angle OAB \text{ (Angles opposite to equal sides)} \]

\angle OBA + \angle OAB + 90° = 180°

\[ \angle OAB = 45° \]

Similarly, in △AOD, ∠OAD = 45°.

Thus, ∠A = ∠OAB + ∠OAD = 45° + 45° = 90°.

By the same reasoning and using the congruence results, all angles of ABCD equal 90°.

Since all sides are equal and all angles are 90°, ABCD is a square. Hence proved.
In simple words: If two diagonals of a four-sided shape are identical in length, meet each other halfway, and cross at a perfect right angle, that shape must be a square with four equal sides and four 90-degree corners.

Exam Tip: Systematically use SAS congruency on different pairs of triangles formed by the diagonals to establish that opposite sides are equal and adjacent sides are equal (proving all four are equal). Then show each angle is 90° by using the isosceles triangle property in the four right triangles created at the intersection point.

 

Question 12. If the diagonals of a quadrilateral are equal and bisect each other at right angles, prove that it is a square.
Answer: The quadrilateral has diagonals that cut each other in half at right angles. When the diagonals bisect each other at right angles, all four angles at the intersection point equal 90°. By comparing triangles OAB and ODC, we find that OA - OC (since diagonals bisect), OB - OD (since diagonals bisect), and angle AOB - angle COD (vertically opposite). Using the SAS axiom, triangle OAB is congruent to triangle ODC, so AB - CD. Similarly, when we compare triangles OAB and OAD, we prove AB - AD. Through these congruences and angle calculations in the isosceles triangles formed, we can show that all four sides of the quadrilateral are equal to each other. Additionally, each interior angle works out to 90°. Since all four sides are equal and all four angles measure 90°, the quadrilateral is a square.
In simple words: When a four-sided shape has diagonals that are the same length, cut each other in half, and meet at right angles, all its sides are the same length and all its corners are right angles - which makes it a square.

Exam Tip: Always show the congruent triangles step-by-step and use C.P.C.T. to prove equal sides - examiners reward systematic working more than jumping to the conclusion.

 

Question 13. P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Answer: In the parallelogram ABCD, the diagonals AC and BD intersect at point O. We look at triangles OAP and OCQ. Since AB is parallel to CD and PQ cuts across these parallel lines, the alternate angles OAP and OCQ are equal. The diagonals bisect each other, so OA - OC. The angles AOP and COQ are vertically opposite angles, so they are equal. By the ASA axiom, triangle OAP is congruent to triangle OCQ. Using C.P.C.T., we get OP - OQ, which shows that O is the midpoint of PQ. Therefore, PQ is bisected at point O.
In simple words: When a line segment passes through the point where two diagonals of a parallelogram cross, that crossing point divides the line segment into two equal parts.

Exam Tip: Identify alternate angles carefully when parallel lines are involved - this is often the key step examiners look for.

 

Question 14(i). In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that the triangles ABX and QCX are congruent.
Answer: We examine triangles ABX and QCX. Because AB is parallel to QC (from the parallel lines AB and the extension through C), angles XAB and XQC are alternate angles and therefore equal. Since X is the midpoint of BC, we have XB - XC. The angles AXB and CXQ are vertically opposite angles, so they are also equal. By the ASA axiom, triangle ABX is congruent to triangle QCX.
In simple words: The two triangles have one equal angle (from parallel lines), one equal side (because X divides BC in half), and another equal angle (from opposite angles at X).

Exam Tip: When identifying vertically opposite angles, mark them clearly in your diagram - examiners want to see that you recognize these angles form from the intersection of two straight lines.

 

Question 14(ii). In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that DC = CQ = QP.
Answer: From the congruence of triangles ABX and QCX proved in part (i), we know AB - CQ by C.P.C.T. In a parallelogram, opposite sides are equal, so AB - CD and AB - QP. Combining these three facts (AB - CQ, AB - CD, and AB - QP), we get DC - CQ - QP. Therefore, all three segments are equal to one another.
In simple words: Since the two triangles are the same shape, AB matches CQ. And because ABCD and ABPQ are both parallelograms, AB also equals both CD and QP. So all three segments have the same length.

Exam Tip: Chain together the equations logically - use the congruence result from part (i) as your starting point, then bring in the parallelogram properties to complete the chain of equalities.

 

Question 14(b). In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.
Answer: Let O be the intersection point of diagonals AC and PQ. We examine triangles AOP and COQ. Since AB is parallel to CD and PQ is a transversal, angles OAP and OCQ are alternate angles and therefore equal. We are given that AP - QC. Angles AOP and COQ are vertically opposite angles, so they are equal. By the ASA axiom, triangle AOP is congruent to triangle COQ. From C.P.C.T., we obtain AO - OC and OP - OQ. This shows that O bisects both AC and PQ, so the two segments bisect each other at point O.
In simple words: When two line segments have one pair of opposite sides on a parallelogram that are equal in length, they cut each other into two equal halves at their intersection point.

Exam Tip: State clearly which angles are alternate angles and which are vertically opposite - this structure shows the examiner your logical approach.

 

Question 15. ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ, prove that AP and DQ are perpendicular to each other.
Answer: We examine triangles ABP and ADQ. Both have a right angle at their first vertex (angles ABP and ADQ both equal 90° because ABCD is a square). We are given that AP - DQ. Since all sides of a square are equal, AB - AD. By the RHS axiom (Right angle - Hypotenuse - Side), triangle ABP is congruent to triangle ADQ. From C.P.C.T., angle BAP equals angle ADQ. Since angle BAD equals 90° (interior angle of the square), we can write angle BAP plus angle PAD equals 90°. Substituting angle ADQ for angle BAP, we get angle ADQ plus angle PAD equals 90°. From the figure, angle ADQ equals angle ADM and angle PAD equals angle MAD, where M is the intersection of AP and DQ. In triangle AMD, the sum of angles ADM, MAD, and AMD must equal 180°. Since angle ADM plus angle MAD equals 90°, we have 90° plus angle AMD equals 180°, so angle AMD equals 90°. Therefore, AP and DQ are perpendicular to each other.
In simple words: Two line segments drawn from opposite corners of a square will meet at a right angle if they have the same length.

Exam Tip: Use the RHS axiom carefully - make sure you identify the right angle, the hypotenuse, and the third side correctly before applying it.

 

Question 16. If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.
Answer: In the parallelogram, AB is parallel to CD, making BD a transversal. Therefore, angles ABP and CDQ are alternate angles and are equal. Since P and Q trisect diagonal BD, the segments BP and QD are equal in length. In a parallelogram, opposite sides are equal, so AB - CD. By the SAS axiom, triangle ABP is congruent to triangle CDQ. From C.P.C.T., angle APB equals angle CQD. Multiplying both sides by negative one gives us negative angle APB equals negative angle CQD. Adding 180° to both sides yields 180° minus angle APB equals 180° minus angle CQD, which simplifies to angle APQ equals angle CQP. These are alternate angles with respect to lines AP and CQ cut by the transversal PQ. Since the alternate angles are equal, lines AP and CQ are parallel.
In simple words: When you cut a diagonal of a parallelogram into three equal parts at two points, the line connecting one of those points to the opposite corner will be parallel to the line from the other corner to the near point on the diagonal.

Exam Tip: Always check that the angles you form actually are alternate angles - verify which transversal you are using and which pair of lines are potentially parallel.

 

Question 17. A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B; the four bisectors form a quadrilateral ACBD. Prove that (i) ACBD is a rectangle. (ii) CD is parallel to the original parallel lines.
Answer:
(i) Let the two parallel lines be LM and PQ, with the transversal cutting them at points A and B respectively. The angle bisectors of the interior angles create four half-angles at each point. At point A, the two interior angles LAB and BAM form a straight line, summing to 180°. The bisectors AC and AD divide these angles in half, so angle CAD (formed by the two bisectors) equals half of 180°, which is 90°. Similarly, at point B, the bisectors BC and BD divide angles PBA and ABQ, yielding angle CBD equal to 90°. Using the parallel line property that co-interior angles sum to 180°, we find that angles formed between the bisectors and the transversal yield 90° angles at C and D as well. By examining the triangles formed and using congruence results, all four angles of quadrilateral ACBD equal 90°. Additionally, opposite sides are equal. Therefore, ACBD is a rectangle.
(ii) The diagonals or sides of the rectangle have special angle relationships with the original parallel lines. Since ACBD is a rectangle with specific angle measures derived from the original parallel lines and their bisectors, the side CD maintains the same slope as the original parallel lines. This can be verified by showing that CD makes equal alternate angles with the transversal as the original parallel lines did, or by showing that CD makes co-interior angles with one of the original lines that sum to 180°. Therefore, CD is parallel to both LM and PQ.
In simple words: When you bisect all the interior angles formed by a transversal cutting two parallel lines, the four bisector lines form a rectangle, and the side of that rectangle connecting two opposite vertices is parallel to the original two lines.

Exam Tip: Label all eight angle pieces clearly (∠1 through ∠8) and track which angles are equal through the bisector property - this systematic labeling helps avoid confusion and shows your work clearly.

 

Question 18. In parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that (i) BE bisects ∠B (ii) ∠AEB = a right angle.
Answer: (i) Since ABCD is a parallelogram, opposite sides are equal: AB = CD and AD = BC. From the diagram, since AE bisects ∠A, we have ∠1 = ∠2. When AB || DC with AE as a transversal, alternate angles give ∠1 = ∠5. Therefore ∠2 = ∠5, which means sides opposite equal angles are equal, so AD = DE. Given AB = 2AD and CD = AB (opposite sides), we get CD = 2AD. Since CD = DE + EC and CD = 2AD with DE = AD, we find EC = AD. Now BC = AD (opposite sides), so EC = BC. When EC = BC, angles opposite equal sides are equal: ∠6 = ∠4. Using alternate angles with the parallel lines, ∠6 = ∠3. This gives ∠3 = ∠4, proving BE bisects ∠B.
(ii) Let ∠1 = x and ∠3 = y. Then ∠2 = x and ∠4 = y. Since AD || BC, co-interior angles sum to 180°, so ∠A + ∠B = 180°. Substituting the angle values: x + x + y + y = 180°, which gives 2x + 2y = 180°, so x + y = 90°. In triangle AEB, the angles sum to 180°: ∠1 + ∠3 + ∠AEB = 180°. Substituting: x + y + ∠AEB = 180°. Using x + y = 90°, we get 90° + ∠AEB = 180°, therefore ∠AEB = 90°.
In simple words: The angle bisector from one corner creates equal segments along the opposite side. Using parallel line properties and angle rules, we can show that the other bisector also divides its angle equally, and the angle where they meet is a right angle.

Exam Tip: Always identify which properties of parallelograms apply here - opposite sides and angles are equal. Track angle relationships carefully when a bisector is involved, especially alternate and co-interior angles.

 

Question 19. ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2AD.
Answer: In parallelogram ABCD, AE and BE bisect ∠A and ∠B respectively, so ∠1 = ∠2 and ∠3 = ∠4. Since AB || DC, when AE acts as a transversal, alternate angles are equal: ∠1 = ∠5. Thus ∠2 = ∠5. When two angles in a triangle are equal, the sides opposite them are equal, so in triangle AED, we have DE = AD. Similarly, ∠3 = ∠6 by alternate angles, and since ∠3 = ∠4 (BE bisects angle B), we get ∠4 = ∠6. This means EC = BC. In a parallelogram, opposite sides are equal: AD = BC and AB = DC. Since EC = BC = AD and DE = AD, we can write: AB = DC = DE + EC = AD + AD = 2AD.
In simple words: When both angle bisectors from adjacent corners of a parallelogram meet on the opposite side, the longer sides turn out to be exactly twice as long as the shorter sides.

Exam Tip: Focus on the alternate angle relationships created when bisectors cross parallel sides - these lead directly to equal side segments that can be combined.

 

Question 20. ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO = AP, prove that 3∠POB = ∠AOP.
Answer: In square ABCD, the diagonals bisect the vertex angles, so ∠CAB = 45°. This means ∠OAP = 45°. In triangle AOP, we have AO = AP (given), so the triangle is isosceles. When a triangle is isosceles, base angles are equal, so ∠AOP = ∠APO. Let ∠AOP = ∠APO = x. Using the angle sum property of triangles: ∠AOP + ∠APO + ∠OAP = 180°, which becomes x + x + 45° = 180°. Solving: 2x = 135°, so x = 67.5° or \( \frac{135°}{2} \). Since the diagonals of a square meet at right angles, ∠AOB = 90°. The angles around point O give: ∠AOP + ∠POB = 90°. Substituting ∠AOP = 67.5°: 67.5° + ∠POB = 90°, so ∠POB = 22.5° or \( \frac{45°}{2} \). To verify the required relation: 3∠POB = 3 × 22.5° = 67.5° = ∠AOP. Thus 3∠POB = ∠AOP is proved.
In simple words: The diagonal of a square creates a 45-degree angle with the side. If a point P is placed so its distance from the corner equals the distance to the diagonal intersection, a special triple relationship appears between the angles formed.

Exam Tip: Use the isosceles triangle property immediately when two sides are equal - this gives you the base angles at once. Don't forget that square diagonals are perpendicular.

 

Question 21. ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.
Answer: All sides of square ABCD are equal: AB = BC = CD = AD. Since equal lengths are cut from each side (AE = BF = CG = DH), the remaining parts are also equal: EB = FC = GD = AH. Consider triangles AEH and BFE. We have AE = BF (given), AH = EB (just shown), and ∠A = ∠B = 90° (angles of square). By the SAS axiom of congruency, triangle AEH is congruent to triangle BFE. Using corresponding parts of congruent triangles, EH = EF and ∠4 = ∠2. In triangle AEH, the angles sum to 180°: ∠1 + ∠4 + 90° = 180°, giving ∠1 + ∠4 = 90°. Since ∠4 = ∠2, we have ∠1 + ∠2 = 90°. From the figure geometry, ∠1 + ∠HEF + ∠2 = 180°, so ∠HEF + 90° = 180°, giving ∠HEF = 90°. Comparing triangles DGH and CGF with DH = GC, GD = FC, and ∠D = ∠C = 90°, we get GH = FG by congruence. Comparing triangles DGH and AEH with DH = AE, GD = HA, and ∠D = ∠A = 90°, we get GH = HE. Therefore all four sides are equal: EF = FG = GH = HE. Since one angle of this rhombus equals 90° (∠HEF = 90°), the shape is a square.
In simple words: When you mark equal distances from each corner of a square and connect those points, the new shape you get is also a square. This works because the equal cuts create four identical right triangles around the edges.

Exam Tip: Prove all four sides are equal first (using congruent triangles), then show one angle is 90° - this definitively establishes a square. The congruence statements follow a pattern, so identify which triangles to compare early.

 

Question 22(a). In the figure (1) given below, ABCD and ABEF are parallelograms. Prove that (i) CDFE is a parallelogram. (ii) FD = EC (iii) △AFD ≅ △BEC.
Answer: (i) In parallelogram ABCD, opposite sides are parallel and equal: DC || AB and DC = AB. In parallelogram ABEF, opposite sides are also parallel and equal: FE || AB and FE = AB. From these two facts, DC || FE and DC = FE. When one pair of opposite sides of a quadrilateral is both parallel and equal, the quadrilateral is a parallelogram. Therefore, CDFE is a parallelogram.
(ii) Since CDEF is a parallelogram (from part i), opposite sides are equal. Thus FD = EC.
(iii) In triangles AFD and BEC, we use the properties of the parallelograms. From ABCD: AD = BC (opposite sides). From ABEF: AF = BE (opposite sides). From CDEF: FD = CE (opposite sides). All three pairs of corresponding sides are equal, so by the SSS axiom of congruency, triangle AFD is congruent to triangle BEC.
In simple words: When two parallelograms share the same side, the opposite sides of one match the opposite sides of the other. This creates equal segments and congruent triangles across the combined figure.

Exam Tip: Identify which sides belong to which parallelogram - tracking "opposite sides of a parallelogram are equal" through multiple shapes is key. The SSS congruence test works directly once you've gathered all three side pairs.

 

Question 22(b). In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC.
Answer: At point A, angles around that point sum to 360°: ∠FAG + ∠GAB + ∠BAD + ∠DAF = 360°. Since ADEF and AGHB are squares, ∠FAD = 90° and ∠GAB = 90°. Substituting: ∠FAG + 90° + ∠BAD + 90° = 360°, which simplifies to ∠FAG = 180° - ∠BAD. In parallelogram ABCD, adjacent angles sum to 180°: ∠ABC + ∠BAD = 180°, so ∠ABC = 180° - ∠BAD. Therefore ∠FAG = ∠ABC. In the parallelogram, AB = CD and AD = BC. Since ADEF is a square, all its sides are equal: AD = DE = EF = FA. Therefore FA = AD, and since BC = AD, we have BC = FA. Now consider triangles AFG and ABC. We have AF = BC (just shown), AG = AB (since AGHB is a square with AG = AB), and ∠FAG = ∠ABC (proved above). By the SAS axiom of congruency, triangle AFG is congruent to triangle ABC. Using corresponding parts of congruent triangles, FG = AC.
In simple words: Two squares built on adjacent sides of a parallelogram create a special angle relationship. This angle match, combined with equal side lengths from the parallelogram and squares, produces two congruent triangles whose matching sides are FG and AC.

Exam Tip: Calculate the angle ∠FAG carefully using the 360° angles-around-a-point rule, then match it to an angle in the parallelogram. The SAS setup is clean once you've gathered AF, AG, and the included angle.

 

Question 23. ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.
Answer: In rhombus ABCD, all sides are equal. Let each side have length a. The diagonals of a rhombus bisect each other at right angles. Let the diagonals intersect at O. Since ∠A = 60° and the diagonals bisect the vertex angles, ∠OAB = 30°. In the right triangle AOB, we have ∠AOB = 90°, ∠OAB = 30°, and ∠ABO = 60°. The side AB = a. Using trigonometry in this right triangle: \( OB = AB \sin(30°) = a \times \frac{1}{2} = \frac{a}{2} \) and \( OA = AB \cos(30°) = a \times \frac{\sqrt{3}}{2} = \frac{a\sqrt{3}}{2} \). Since the diagonals bisect each other, the full diagonal lengths are: AC = 2 × OA = 2 × \( \frac{a\sqrt{3}}{2} \) = \( a\sqrt{3} \) and BD = 2 × OB = 2 × \( \frac{a}{2} \) = a. Therefore, the ratio AC : BD = \( a\sqrt{3} \) : a = \( \sqrt{3} \) : 1.
In simple words: In a rhombus where one angle is 60 degrees, the two diagonals are not the same length. The longer diagonal is about 1.73 times the shorter diagonal.

Exam Tip: Remember that rhombus diagonals bisect the vertex angles and meet at right angles. This creates useful right triangles where trigonometry applies directly. Always label the half-diagonals clearly before finding the full ratio.

 

Exercise 12.2

 

Question 1. Using ruler and compasses only, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6cm, BC = 3.6cm, CD = 5cm. Measure ∠BCD.
Answer: Steps of construction:
(1) Draw a line segment AB = 6 cm
(2) At A, construct ∠BAX = 45°
(3) From AX, cut off AD = 6 cm
(4) With B as centre and radius 3.6 cm, draw an arc
(5) With D as centre and radius 5 cm, draw an arc to meet the previous arc at C
(6) Join BC and DC. Then, ABCD is the required quadrilateral.
On measuring ∠BCD, it is 63°.
In simple words: Follow the six steps to draw the four-sided shape. When you measure the angle at C, you will get 63 degrees.

Exam Tip: Always use a protractor to measure the final angle accurately. Check that all four sides match the given measurements before taking your final measurement.

 

Question 2. Draw a quadrilateral ABCD with AB = 6 cm, BC = 4 cm, CD = 4 cm and ∠ABC = ∠BCD = 90°.
Answer: Steps of construction:
(1) Draw a line segment BC = 4 cm
(2) At B, construct ∠CBX = 90° and cut off BA = 6 cm
(3) At C, construct ∠BCY = 90° and cut off CD = 4 cm
(4) Join AD. Then, ABCD is the required quadrilateral.
In simple words: Draw the base BC. Make two right angles at B and C. Join the remaining points to complete the shape.

Exam Tip: Use a set square or compass to construct the 90° angles accurately. Make sure both right angles are correctly drawn before joining the final side.

 

Question 3. Using ruler and compasses only, construct the quadrilateral ABCD given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm, ∠BAD = 90° and the diagonal AC = 5.5 cm.
Answer: Steps of construction:
(1) Draw a line segment AB = 5cm
(2) With centre A and radius 5.5 cm, draw an arc
(3) With centre B and radius 2.5 cm, draw an arc to meet the previous arc at C
(4) Join AC and BC
(5) At A, construct ∠BAX = 90°
(6) With centre C and radius 6 cm, draw an arc intersecting AX at D
(7) Join CD. Then, ABCD is the required quadrilateral.
In simple words: Draw side AB and the diagonal AC. Find point C using two arcs. Then construct a right angle at A and find point D using a third arc from C.

Exam Tip: The diagonal AC is a key measurement - draw it accurately with compasses. All arc intersections must be clear and sharp for the quadrilateral to be correctly constructed.

 

Question 4. Construct a quadrilateral ABCD in which AB = 3.3 cm, BC = 4.9 cm, CD = 5.8 cm, DA = 4 cm and BD = 5.3 cm.
Answer: Steps of construction:
(1) Draw a line segment AB = 3.3 cm
(2) With centre A and radius 4 cm, draw an arc
(3) With centre B and radius 5.3 cm, draw an arc to meet the previous arc at D
(4) Join AD and BD
(5) With centre B and radius 4.9 cm, draw an arc
(6) With centre D and radius 5.8 cm, draw an arc to meet the previous arc at C
(7) Join BC and DC. Then ABCD is the required quadrilateral.
In simple words: First, make triangle ABD using three measurements. Then find point C by drawing two more arcs from B and D to complete the quadrilateral.

Exam Tip: Use the diagonal BD as an intermediate step. First construct the triangle ABD accurately, then locate C using the remaining two sides.

 

Question 5. Construct a trapezium ABCD in which AD || BC, AB = CD = 3 cm, BC = 5.2 cm and AD = 4 cm.
Answer: Steps of construction:
(1) Draw a line segment BC = 5.2 cm
(2) From BC, cut off BE = 4 cm
(3) With centre E and radius 3 cm, draw an arc
(4) With centre C and radius 3 cm, draw an arc to meet the previous arc at D
(5) Join ED and CD
(6) With centre D and radius 4 cm, draw an arc
(7) With centre B and radius 3 cm, draw an arc to meet the previous arc at A
(8) Join BA and DA. Then ABCD is the required quadrilateral.
In simple words: Mark point E on BC. Find D so that ED and CD are both 3 cm. Then find A so that BA and DA match the required lengths.

Exam Tip: The construction uses a helper point E to ensure the parallel sides are correctly positioned. Verify that AD and BC are parallel by checking they remain equidistant.

 

Question 6. Construct a trapezium ABCD in which AD || BC, ∠B = 60°, AB = 5 cm, BC = 6.2 cm and CD = 4.8 cm.
Answer: Steps of construction:
(1) Draw a line segment BC = 6.2 cm
(2) At B, construct ∠CBX = 60°
(3) With centre B and radius 5 cm, draw an arc intersecting BX at A
(4) From A, draw a line AY parallel to BC
(5) With centre C and radius 4.8 cm, draw an arc which intersects AY at D
(6) Join CD. Then ABCD is the required trapezium.
In simple words: Draw BC and make a 60° angle at B. Mark point A on this angle line at distance 5 cm. Draw a parallel line through A and find D using an arc from C.

Exam Tip: The parallel line AY drawn from A is crucial for ensuring AD || BC. Use a set square or compass to draw this parallel line accurately.

 

Question 7. Using ruler and compasses only, construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.
Answer: Steps of construction:
(1) Draw a line segment BC = 7 cm
(2) At B, construct ∠CBX = 75°
(3) With B as center and radius 5.1 cm, draw an arc intersecting BX at A
(4) With A as center and radius 7 cm, draw an arc
(5) With C as center and radius 5.1 cm, draw an arc to meet the previous arc at D
(6) Join AD and CD. Then ABCD is the required parallelogram.
In simple words: Draw BC and the 75° angle at B. Mark A at distance 5.1 cm along this angle. Complete the parallelogram by finding D so opposite sides are equal.

Exam Tip: In a parallelogram, opposite sides are equal. So AD must equal BC and CD must equal AB. Use these equal lengths to find the final vertex D.

 

Question 8. Using ruler and compasses only, construct a parallelogram ABCD in which AB = 4.6 cm, BC = 3.2 cm and AC = 6.1 cm.
Answer: Steps of construction:
(1) Draw a line segment AB = 4.6 cm
(2) With A as center and radius 6.1 cm, draw an arc
(3) With B as center and radius 3.2 cm, draw an arc to meet the previous arc at C
(4) Join AC and BC
(5) Again, with A as center and radius 3.2 cm, draw an arc
(6) With C as center and radius 4.6 cm, draw an arc to meet the previous arc at D
(7) Join AD and CD. Then ABCD is the required parallelogram.
In simple words: Use the diagonal AC to find point C first. Then use the equal opposite sides of a parallelogram to find the final point D.

Exam Tip: The diagonal AC divides the parallelogram into two triangles. Construct triangle ABC first, then use the property that opposite sides are equal to find point D.

 

Question 9. Using ruler and compasses, construct a parallelogram ABCD given that AB = 4 cm, AC = 10 cm, BD = 6 cm. Measure BC.
Answer: Steps of construction:
(1) Draw a line segment AB = 4 cm
(2) Construct triangle OAB such that
\( OA = \frac{1}{2} \times AC = 5 \text{ cm} \)
\( OB = \frac{1}{2} \times BD = 3 \text{ cm} \)
With A as center and radius 5 cm and with centre B and radius 3 cm, draw arcs intersecting each other at O.
As diagonals of a parallelogram bisect each other
(3) Produce AO to C such that OA = OC = 5cm
(4) Produce BO to D such that OB = OD = 3cm
(5) Join AD, BC and CD. Then ABCD is the required parallelogram.
On measuring, BC = 7.2 cm
In simple words: Find the midpoint O of both diagonals by using half their lengths. Then extend both diagonals to complete the parallelogram. The fourth side BC will measure 7.2 cm.

Exam Tip: The key property used here is that diagonals of a parallelogram bisect each other. Point O is the intersection of both diagonals, found using the half-lengths.

 

Question 10. Using ruler and compasses only, construct a parallelogram ABCD such that BC = 4 cm, diagonal AC = 8.6 cm and diagonal BD = 4.4 cm. Measure the side AB.
Answer: Steps of construction:
(1) Draw a line segment BC = 4 cm
(2) Construct triangle OBC such that
\( OB = \frac{1}{2} \times BD = 2.2 \text{ cm} \)
\( OC = \frac{1}{2} \times AC = 4.3 \text{ cm} \)
(3) With B as center and radius 2.2 cm and with centre C and radius 4.3 cm, draw arcs intersecting each other at O. Since diagonals of a parallelogram bisect each other
(4) Produce BO to D such that BO = OD = 2.2 cm
(5) Produce CO to A such that CO = OA = 4.3 cm
(6) Join AB, AD and CD. Then ABCD is the required parallelogram.
On measuring AB = 5.6 cm
In simple words: Find point O using half of each diagonal length from B and C. Extend both diagonals outward to get points D and A. The side AB measures 5.6 cm.

Exam Tip: When the diagonals are given instead of the sides, use the property that they bisect each other at point O. This point is the center of the parallelogram.

 

Question 11. Use ruler and compasses to construct a parallelogram with diagonals 6 cm and 8 cm in length having given the acute angle between them is 60°. Measure one of the longer sides.
Answer: Steps of construction:
Let AC = 6 cm and BD = 8 cm
(1) Draw \( AO = \frac{1}{2} AC = \frac{1}{2} \times 6 = 3 \) cm and produce AO to C such that OC = OA
(2) At O construct ∠COP = 60°
(3) From OP cut off \( OD = \frac{1}{2} BD = \frac{1}{2} \times 8 = 4 \) cm, produce DO to B such that OB = OD
(4) Join AB, BC, CD and DA. Then ABCD is the required parallelogram.
In simple words: Draw half of diagonal AC (which is 3 cm) and complete the diagonal by extending it equally on both sides of point O. At O, make a 60° angle and repeat for the other diagonal (4 cm on each side). Join all four points to form the parallelogram.

Exam Tip: The angle between the diagonals is drawn at their intersection point O. The diagonals bisect each other, so each half-length must be marked carefully on either side of O.

 

Question 12. Using ruler and compasses only, draw a parallelogram whose diagonals are 4 cm and 6 cm long and contain an angle of 75°. Measure and write down the length of one of the shorter sides of the parallelogram.
Answer: Let AC = 4 cm and BD = 6 cm.

Steps of construction:
(1) Draw AO = \( \frac{1}{2} \) AC = \( \frac{1}{2} \times 4 = 2 \) cm and extend AO to C such that OC = OA.
(2) At O, construct \( \angle \)COP = 75°.
(3) From OP, measure off OD = \( \frac{1}{2} \) BD = \( \frac{1}{2} \times 6 = 3 \) cm, then extend DO to B so that OB = OD.
(4) Join AB, BC, CD and DA. ABCD is now the required parallelogram.

On measuring the shorter sides, we get AB = CD = 3.1 cm.

Therefore, the length of the shorter side = 3.1 cm.
In simple words: Draw both diagonals by finding their midpoints and marking the angle between them. Then connect the four endpoints to form the shape.

Exam Tip: Always verify that opposite sides are equal by measuring - this confirms your construction is correct. The diagonals must intersect at their midpoints.

 

Question 13. Using ruler and compasses only, construct a parallelogram ABCD with AB = 6 cm, altitude = 3.5 cm and side BC = 4 cm. Measure the acute angles of the parallelogram.
Answer: Steps of construction:
(1) Draw AB = 6 cm.
(2) At B, construct BP perpendicular to AB.
(3) From BP, mark off BE = 3.5 cm (the height of the parallelogram).
(4) Through E, draw QR parallel to AB.
(5) With B as centre and radius 4 cm, draw an arc that cuts QR at C. Since opposite sides of a parallelogram are equal, AD = BC = 4 cm.
(6) With A as centre and radius 4 cm, draw an arc that cuts QR at D.
(7) Join AD, BC and CD. ABCD is the required parallelogram.

On measuring the acute angle of the parallelogram, it equals 61°.

Therefore, the acute angle = 61°.
In simple words: First draw the base line, then go up at a right angle by the height amount. Draw a line across at that height, then use the radius 4 cm to find where the other two corners go.

Exam Tip: The height must be perpendicular to the base - use a set square or compass to ensure a 90° angle. Verify your answer by checking that opposite sides are equal.

 

Question 14. The perpendicular distances between the pairs of opposite sides of a parallelogram ABCD are 3 cm and 4 cm and one of its angles measures 60°. Using ruler and compasses only, construct ABCD.
Answer: Steps of construction:
(1) Draw a straight line PQ and choose a point A on it.
(2) At A, construct \( \angle \)QAF = 60°.
(3) At A, draw AE perpendicular to PQ. From AE, mark off AN = 3 cm.
(4) Through N, draw a straight line parallel to PQ to meet AF at D.
(5) At A, draw AG perpendicular to AD. From AG, mark off AM = 4 cm.
(6) Through M, draw a straight line parallel to AD to meet AQ at B and ND at C.
(7) Join AB, BC, CD and DA. ABCD is the required parallelogram.
In simple words: Start with one line and an angle. Draw perpendicular lines at the given distances, then connect the points using parallel lines to complete the shape.

Exam Tip: The perpendicular distances represent the heights of the parallelogram between opposite sides - mark these carefully with your compass. Double-check that opposite sides are parallel.

 

Question 15. Using ruler and compasses, construct a rectangle ABCD with AB = 5 cm and AD = 3 cm.
Answer: Steps of construction:
(1) Draw a straight line AB = 5 cm.
(2) At A and B, construct \( \angle \)XAB = \( \angle \)YBA = 90°.
(3) From XA and YB, mark off AD and BC = 3 cm each.
(4) Join CD. ABCD is the required rectangle.
In simple words: Draw the base line, then make two right angles at each end, measure up 3 cm on each side, and connect the top two points.

Exam Tip: Ensure both right angles are exactly 90° using a compass. All four angles in a rectangle must be right angles - verify by measuring the diagonals; they should be equal.

 

Question 16. Using ruler and compasses only, construct a rectangle each of whose diagonals measures 6 cm and the diagonals intersect at an angle of 45°.
Answer: Steps of construction:
(1) Draw a line segment AC = 6 cm.
(2) Bisect AC at O.
(3) At O, draw a ray XY making an angle of 45°.
(4) From XY, mark off OB = OD = \( \frac{6}{2} \) = 3 cm each.
(5) Join AB, BC, CD and DA. ABCD is the required rectangle.
In simple words: Draw one diagonal, find its middle point, draw another line through that point at 45°, mark equal distances on both sides, then connect the four corners.

Exam Tip: In a rectangle, diagonals are equal and bisect each other - use these properties to check your work. The angle between diagonals helps determine the rectangle's shape.

 

Question 17. Using ruler and compasses only, construct a square having a diagonal of length 5 cm. Measure its sides correct to the nearest millimeter.
Answer: Steps of construction:
(1) Draw a line segment AC = 5 cm.
(2) Draw its perpendicular bisector XY that bisects it at O.
(3) From XY, mark off OB = OD = \( \frac{5}{2} \) = 2.5 cm.
(4) Join AB, BC, CD and DA. ABCD is the required square.

On measuring, each side = 3.5 cm.

Therefore, the length of the side of the square = 3.5 cm.
In simple words: Draw the diagonal, find the middle, draw a line through the middle at right angles, measure equal distances, then join the corners.

Exam Tip: The diagonals of a square are equal, bisect each other, and meet at right angles. Verify all four sides measure the same length to confirm it is a true square.

 

Question 18. Using ruler and compasses only construct a rhombus ABCD, given that AB = 5 cm, AC = 6 cm. Measure \( \angle \)BAD.
Answer: Steps of construction:
(1) Draw a line segment AB = 5 cm.
(2) With centre A and radius 6 cm, and with centre B and radius 5 cm, draw arcs that intersect each other at C.
(3) Join AC and BC.
(4) With centre A and C and radius 5 cm, draw arcs that intersect each other at D.
(5) Join AD and CD. ABCD is the required rhombus.

On measuring, \( \angle \)BAD = 106°.

Therefore, \( \angle \)BAD = 106°.
In simple words: Use the compass to mark distances - all sides of a rhombus are equal. Draw arcs from two points to find where the other corners must be.

Exam Tip: All four sides of a rhombus are equal in length - use this to find the fourth vertex. Opposite angles in a rhombus are equal.

 

Question 19. Using ruler and compasses only, construct rhombus ABCD with sides of length 4 cm and diagonal AC of length 5 cm. Measure \( \angle \)ABC.
Answer: Steps of construction:
(1) Draw a line segment AC = 5 cm.
(2) With centre A and C and radius 4 cm, draw arcs that intersect each other above and below AC at D and B.
(3) Join AB, BC, CD, DA and BD. ABCD is the required rhombus.

On measuring, \( \angle \)ABC = 78°.

Therefore, \( \angle \)ABC = 78°.
In simple words: Draw the diagonal first. Then use the same radius (4 cm) from both ends to find the other two corners. The rhombus is complete when you join all four corners.

Exam Tip: When drawing arcs from two centres with equal radius, they will intersect at two points - these form the other two vertices. Check that all sides measure 4 cm.

 

Question 20. Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively.
Answer: Steps of construction:
(1) Draw a line segment PR = 8 cm.
(2) Draw its perpendicular bisector XY that intersects it at O.
(3) From XY, mark off OQ = OS = \( \frac{6}{2} \) = 3 cm each.
(4) Join PQ, QR, RS and SP. PQRS is the required rhombus.
In simple words: Draw one diagonal, find the middle point, draw a second line through that point at right angles, mark equal distances on both sides, then connect the four corners.

Exam Tip: The diagonals of a rhombus bisect each other at right angles - use this property to construct the shape quickly. Verify by checking that all four sides are equal.

 

Question 21. Construct a rhombus ABCD of side 4.6 cm and \( \angle \)BCD = 135°, by using ruler and compasses only.
Answer: Steps of construction:
(1) Draw a line segment BC = 4.6 cm.
(2) At C, draw a ray CX making an angle of 135° and mark off CD = 4.6 cm.
(3) With centres B and D, and radius 4.6 cm, draw arcs that intersect each other at A.
(4) Join BA and DA. ABCD is the required rhombus.
In simple words: Draw the first side, then from one end draw another side at the given angle. Use the compass to find the fourth corner, which is the same distance from both the second and third corners.

Exam Tip: All sides of a rhombus are equal - use the same compass radius throughout. Adjacent angles in a rhombus are supplementary (add up to 180°), so check your work.

 

Question 22. Construct a trapezium in which AB || CD, AB = 4.6 cm, \( \angle \)ABC = 90°, \( \angle \)DAB = 120° and the distance between parallel sides is 2.9 cm.
Answer: Steps of construction:
(1) Draw a line segment AB = 4.6 cm.
(2) At B, draw a ray BZ making an angle of 90° and mark off BC = 2.9 cm (the distance between AB and CD).
(3) At C, draw a parallel line XY to AB.
(4) At A, draw a ray making an angle of 120° that meets XY at D.
(5) Join AD and DC. ABCD is the required trapezium.
In simple words: Start with the base line, go up at a right angle, then draw a line across. From the other end, draw a line at 120° to meet that line and complete the shape.

Exam Tip: In a trapezium, one pair of opposite sides must be parallel. Use a set square or compass to ensure angles are correct and the parallel line is truly parallel to the base.

 

Question 23. Construct a trapezium ABCD when one of parallel sides AB = 4.8 cm, height = 2.6 cm, BC = 3.1 cm and AD = 3.6 cm.
Answer: Steps of construction:
(1) Draw a line segment AB = 4.8 cm.
(2) At A, draw a ray AZ making an angle of 90° and mark off AL = 2.6 cm.
(3) At L, draw a ray LX parallel to AB.
(4) With centre A and radius 3.6 cm, and with centre B and radius 3.1 cm, draw arcs that cut LX at D and C respectively.
(5) Join AD and BC. ABCD is the required trapezium.
In simple words: Draw the base, go up at a right angle by the height amount, draw a line across. Use the compass to mark off the two side lengths to find where the other corners go.

Exam Tip: The height is the perpendicular distance between the parallel sides. Mark this carefully and ensure the line you draw at that height is parallel to the base.

 

Question 24. Construct a regular hexagon of side 2.5 cm.
Answer: Steps of construction:
(1) With O as centre and radius 2.5 cm, draw a circle.
(2) Take any point A on the circumference of the circle.
(3) With A as centre and radius 2.5 cm, draw an arc that cuts the circumference at B.
(4) With B as centre and radius 2.5 cm, draw an arc that cuts the circumference at C.
(5) With C as centre and radius 2.5 cm, draw an arc that cuts the circumference at D.
(6) Continue this process to mark points E and F on the circle.
(7) Join AB, BC, CD, DE, EF and FA. The resulting shape ABCDEF is the required regular hexagon.
In simple words: Draw a circle with radius 2.5 cm. Starting from any point on the circle, use the same radius to mark six points equally spaced around the circle. Connect these points in order.

Exam Tip: In a regular hexagon, the radius equals the side length - this is why the arc radius matches the compass setting. All six sides should be equal and all angles should be 120°. Verify by measuring.

 

Question 1. Three angles of a quadrilateral are 75°, 90° and 75°. The fourth angle is
(a) 90°
(b) 95°
(c) 105°
(d) 120°
Answer: (d) 120°
In simple words: All four angles of any quadrilateral add up to 360°. Add the three given angles: 75° + 90° + 75° = 240°. The fourth angle is 360° - 240° = 120°.

Exam Tip: Always remember that the sum of interior angles in any quadrilateral equals 360°. Use this formula to find missing angles quickly.

 

Question 2. A quadrilateral ABCD is a trapezium if
(a) AB = DC
(b) AD = BC
(c) ∠A + ∠C = 180°
(d) ∠B + ∠C = 180°
Answer: (d) ∠B + ∠C = 180°
In simple words: In a trapezium, the angles on the same side of a leg that connects the two parallel sides must add up to 180°. Angles B and C are next to each other along one leg, so their sum equals 180°.

Exam Tip: Recognise that co-interior angles (also called allied angles) formed between two parallel lines and a transversal always sum to 180°. This is the defining feature tested in trapezium problems.

 

Question 3. If PQRS is a parallelogram, then ∠Q - ∠S is equal to
(a) 90°
(b) 120°
(c) 0°
(d) 180°
Answer: (c) 0°
In simple words: In a parallelogram, angles that sit opposite to each other are always the same size. Since Q and S are opposite angles, ∠Q = ∠S, which means ∠Q - ∠S = 0°.

Exam Tip: Always apply the property that opposite angles in a parallelogram are equal. This immediately tells you their difference is zero.

 

Question 4. A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is
(a) 55°
(b) 50°
(c) 40°
(d) 25°
Answer: (b) 50°
In simple words: In a rectangle, the two diagonals meet at the centre and split each other in half. Because the diagonal makes a 25° angle with one side, the triangle formed at the centre has two angles of 25° each. The angle where the diagonals cross is 180° - 25° - 25° = 130° on one side. The acute angle on the other side is 180° - 130° = 50°.

Exam Tip: Draw the triangle formed by the two half-diagonals and the rectangle side. Use the angle sum property of triangles (sum = 180°) to find the angle where diagonals meet, then find its supplement.

 

Question 5. ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is
(a) 40°
(b) 45°
(c) 50°
(d) 60°
Answer: (c) 50°
In simple words: In a rhombus, the diagonals cross at right angles. Use alternate angles - since AC and BD are the diagonals, angle DAO equals angle OCB, which is 40°. In triangle ADO, the angle at O is 90° (because diagonals cross at right angles). So angle ODA = 180° - 40° - 90° = 50°. This is the angle ADB.

Exam Tip: Remember that diagonals of a rhombus are perpendicular. Use the right angle at the intersection point to set up your triangle equation.

 

Question 6. The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32° and ∠AOB = 70°, then ∠DBC is equal to
(a) 24°
(b) 86°
(c) 38°
(d) 32°
Answer: (c) 38°
In simple words: Because AC is a diagonal crossing through two parallel sides, angle ACB equals angle DAC (alternate angles) = 32°. The diagonal AC is a straight line, so ∠AOB + ∠BOC = 180°, giving ∠BOC = 110°. In triangle OBC, the three angles are ∠BOC = 110°, ∠OCB = 32°, and ∠OBC = 180° - 110° - 32° = 38°. Since ∠OBC is the same as ∠DBC, the answer is 38°.

Exam Tip: Use alternate angles formed by the diagonal cutting parallel sides, and the straight-line angle property to find the required angle step by step.

 

Question 7. If the diagonals of a square ABCD intersect each other at O, then △OAB is
(a) an equilateral triangle
(b) a right angled but not an isosceles triangle
(c) an isosceles but not right angles triangle
(d) an isosceles right angled triangle
Answer: (d) an isosceles right angled triangle
In simple words: In a square, the diagonals meet at the middle point at 90°. The two half-diagonals OA and OB are equal in length because the diagonals of a square are equal and bisect each other. So triangle OAB has OA = OB (isosceles) and ∠AOB = 90° (right angle).

Exam Tip: Know the key properties of square diagonals: they are equal, bisect each other, and meet at right angles. Use these to identify the triangle type.

 

Question 8. If the diagonals of a quadrilateral PQRS bisect each other, then the quadrilateral PQRS must be a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
Answer: (a) parallelogram
In simple words: When the diagonals of a quadrilateral cut each other in half, that shape must be a parallelogram. All other shapes - rhombus, rectangle, and square - are special types of parallelograms with extra properties, but the minimum requirement is that it is a parallelogram.

Exam Tip: Remember the hierarchy: a parallelogram is the broadest category. Rectangles, rhombuses, and squares are all parallelograms with additional angle or side properties, not vice versa.

 

Question 9. If the diagonals of a quadrilateral PQRS bisect each other at right angles, then the quadrilateral PQRS must be a
(a) parallelogram
(b) rectangle
(c) rhombus
(d) square
Answer: (c) rhombus
In simple words: Both rhombuses and squares have diagonals that bisect each other at 90°. However, every square is a rhombus, but not every rhombus is a square. So the broadest answer - the shape that must have this property - is a rhombus.

Exam Tip: When a property is shared by multiple shapes, identify the most general shape that always has that property. A square is too specific; a rhombus covers all cases.

 

Question 10. Which of the following statement is true for a parallelogram?
(a) Its diagonals are equal.
(b) Its diagonals are perpendicular to each other.
(c) The diagonals divide the parallelogram into four congruent triangles.
(d) The diagonals bisect each other.
Answer: (d) The diagonals bisect each other.
In simple words: A defining feature of every parallelogram is that its two diagonals cross each other at their midpoints. This property does not need the shape to be a rectangle, rhombus, or square - it holds for all parallelograms.

Exam Tip: This is a core property that distinguishes a parallelogram from other quadrilaterals. Learn and apply it directly without needing to check other conditions.

 

Question 11. Which of the following is not true for a parallelogram?
(a) opposite sides are equal
(b) opposite angles are equal
(c) opposite angles are bisected by the diagonals
(d) diagonals bisect each other
Answer: (c) opposite angles are bisected by the diagonals
In simple words: Opposite sides being equal, opposite angles being equal, and diagonals bisecting each other are all true for every parallelogram. However, the diagonals do not bisect the opposite angles unless the parallelogram is a special type like a rhombus.

Exam Tip: Know which properties apply to all parallelograms versus which apply only to special parallelograms. Angle bisection by diagonals is a rhombus property, not a general parallelogram property.

 

Question 12. A quadrilateral in which the diagonals are equal and bisect each other at right angles is a
(a) rectangle which is not a square
(b) rhombus which is not a square
(c) kite which is not a square
(d) square
Answer: (d) square
In simple words: A square is the only quadrilateral where both diagonals are equal in length, cut each other in half, and meet at 90°. A rectangle has equal diagonals that bisect each other, but not at 90°. A rhombus has diagonals that bisect each other at 90°, but they are not equal. Only a square has all three properties together.

Exam Tip: Match the three diagonal conditions to the correct shape. When all three are combined, it can only be a square.

 

Question 13. Consider the following two statements:
Statement 1: Sum of interior angles of any polygon in 360°.
Statement 2: Sum of interior angles of a rhombus in 360°.
Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (d) Statement 1 is false, and Statement 2 is true.
In simple words: The sum of interior angles for a polygon depends on the number of sides. Use the formula: sum = (n - 2) × 180°, where n is the number of sides. For a triangle (n = 3): sum = 180°. For a rhombus or any quadrilateral (n = 4): sum = 360°. So Statement 1 is wrong because not all polygons have angle sums of 360°. Statement 2 is correct because a rhombus has four sides and its angle sum is 360°.

Exam Tip: Always use the polygon angle formula (n - 2) × 180° rather than memorising fixed values. This approach works for all shapes.

 

Question. Assertion (A): All the interior angles of a square are right angles.
Reason (R): Diagonals of a square intersect at right angles.
Answer: A square is defined as a four-sided figure with four equal sides and each angle measuring 90 degrees. This means Assertion (A) is true. In a square, the diagonals cut each other at 90 degrees, making Reason (R) also true. However, the reason why all interior angles of a square are right angles is because of how a square is defined - not because of its diagonal properties. Even though both statements are correct, the diagonal property is not the actual explanation for why the angles are right angles. Therefore, option 4 is the correct answer.
In simple words: A square has four 90-degree angles because that is what makes a square a square. The diagonals crossing at 90 degrees is a separate fact about squares, but it is not the reason the angles are right angles.

Exam Tip: Always check whether the reason given actually explains the assertion - both statements can be true, but if one does not cause the other, option 4 is correct.

 

Question. Assertion (A): If one angle of a rhombus is 70°, then its largest angle is 110°.
Reason (R): Every rhombus is a square.
Answer: A rhombus is a four-sided shape where all four sides have equal length. In a rhombus, opposite angles are equal to each other. If one angle is 70 degrees, then the angle across from it is also 70 degrees. In a rhombus, angles next to each other always add up to 180 degrees. So the next angle would be 180 degrees minus 70 degrees, which equals 110 degrees. The four angles of the rhombus are therefore 70 degrees, 110 degrees, 70 degrees, and 110 degrees. The largest among these angles is 110 degrees, so Assertion (A) is true. A rhombus is a four-sided figure where all four sides are the same length, but its angles do not have to be 90 degrees. A square is a special type of rhombus that also has all four angles equal to 90 degrees. So every square is a rhombus, but not every rhombus is a square, which means Reason (R) is false. Therefore, option 1 is the correct answer.
In simple words: A rhombus with one 70-degree angle will have angles of 70, 110, 70, and 110 degrees. But a rhombus does not have to be a square - squares are a special kind of rhombus, not the other way around.

Exam Tip: Remember the properties: opposite angles in a rhombus are equal, consecutive angles sum to 180 degrees, and a square is just a rhombus with all 90-degree angles.

 

Question. Assertion (A): Diagonals of adjoining quadrilateral bisect each other. Then ∠B = 135°.
Reason (R): If diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Answer: It is a known fact that when the diagonals of a parallelogram intersect, they cut each other in half. This makes Reason (R) true. Given that the diagonals of quadrilateral ABCD cut each other in half, ABCD must be a parallelogram. In a parallelogram, any two angles that sit next to each other always sum to 180 degrees. Using this property, if angle A is 45 degrees and angle B is next to it, then angle A plus angle B equals 180 degrees, which gives us 45 degrees plus angle B equals 180 degrees. Solving this, angle B equals 180 minus 45, which is 135 degrees. This proves Assertion (A) is true. Both statements are true, and the reason correctly explains why the assertion is true. Therefore, option 3 is the correct answer.
In simple words: If the diagonals of a shape cut each other in half, that shape is definitely a parallelogram. And in a parallelogram, angles next to each other always add up to 180 degrees.

Exam Tip: A key property to remember: if a quadrilateral has diagonals that bisect each other, it must be a parallelogram - use this to identify the figure, then apply parallelogram angle properties.

 

Chapter Test

 

Question 1. In the adjoining figure, ABCD is a parallelogram. CB is produced to E such that BE = BC. Prove that AEBD is a parallelogram.
Answer: Consider triangles AEB and BDC. We are told that EB equals BC. Since ABCD is a parallelogram, angles ABE and DCB are corresponding angles and therefore equal. Also, since ABCD is a parallelogram, opposite sides AB and DC are equal. By the Side-Angle-Side rule, triangle AEB is congruent to triangle BDC. Using the Corresponding Parts of Congruent Triangles rule, we get BD equals AE. In the parallelogram ABCD, opposite sides are equal, so BC equals AD. We are given that BC equals BE, therefore AD equals BE. Since both pairs of opposite sides of quadrilateral AEBD are equal (BD = AE and AD = BE), the quadrilateral AEBD must be a parallelogram. Hence, this completes the proof.
In simple words: Two triangles made by the construction are congruent, so their matching sides are equal. This makes opposite sides of the new quadrilateral equal, which proves it is a parallelogram.

Exam Tip: Always show congruence of triangles first using a standard axiom (SAS, SSS, etc.), then use C.P.C.T to equate the sides needed for the parallelogram proof - this structure earns full marks.

 

Question 2. In the adjoining figure, ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || BA. Show that (i) ∠DAC = ∠BCA (ii) ABCD is a parallelogram.
Answer:
(i) In triangle ABC, we know that AB equals AC. When a triangle has two sides equal, the angles opposite those sides must also be equal. Therefore, angle C equals angle B. An exterior angle of a triangle equals the sum of the two non-adjacent interior angles. So the exterior angle PAC equals angle B plus angle C. Since angle B equals angle C, we can write this as angle C plus angle C, which simplifies to 2 times angle C, or 2 times angle BCA. Since AD bisects the exterior angle PAC, angle PAC equals 2 times angle DAC. Setting these equal, we get 2 times angle DAC equals 2 times angle BCA. Dividing both sides by 2 gives us angle DAC equals angle BCA. This completes the first part of the proof.
(ii) Since angle DAC and angle BCA are equal and they are alternate angles with respect to line AC as a transversal, line AD must be parallel to line BC. We are also given that CD is parallel to BA. Since AD is parallel to BC and CD is parallel to BA, quadrilateral ABCD has both pairs of opposite sides parallel. Therefore, ABCD is a parallelogram. This completes the second part of the proof.
In simple words: An isosceles triangle has equal base angles. The exterior angle equals twice the base angle. If AD bisects the exterior angle, it creates an angle equal to the base angle, making AD parallel to BC. With both pairs of opposite sides parallel, ABCD is a parallelogram.

Exam Tip: First establish the angle equality using exterior angle and angle bisector properties, then use alternate angles to prove parallelism - presenting these steps in order is essential for full credit.

 

Question 3. Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.
Answer: Let P, Q, R, and S be the mid-points of sides AB, BC, CD, and DA respectively of isosceles trapezium ABCD. Draw both diagonals AC and BD. Since ABCD is an isosceles trapezium, its two diagonals are equal in length. Let both diagonals equal x. In triangle ABC, since P and Q are the mid-points of sides AB and BC, by the midpoint theorem, PQ is parallel to AC and PQ equals half of AC, which is x/2. Similarly, in triangle ADC, since S and R are the mid-points of sides AD and CD, SR is parallel to AC and SR equals half of AC, which is x/2. From these two results, PQ is parallel to SR and both equal x/2. In triangle CBD, since R and Q are the mid-points of sides CD and BC, QR is parallel to BD and QR equals half of BD, which is x/2. Similarly, in triangle ABD, since S and P are the mid-points of sides AD and AB, SP is parallel to BD and SP equals half of BD, which is x/2. From these results, QR is parallel to SP and both equal x/2. Combining all results, we find that PQ = QR = SR = SP = x/2, and opposite sides are parallel. When all four sides of a quadrilateral are equal and opposite sides are parallel, the quadrilateral is a rhombus. Therefore, PQRS is a rhombus. This completes the proof.
In simple words: The midpoint theorem tells us that lines connecting midpoints are half the length of parallel diagonals. Since both diagonals of an isosceles trapezium are equal, all four midpoint-connecting lines have the same length. Equal sides with parallel opposite sides make a rhombus.

Exam Tip: Apply the midpoint theorem systematically to all four triangles, show that all four sides of PQRS are equal, and verify that opposite sides are parallel - all three facts together prove it is a rhombus.

 

Question 4(i). Find the size of each lettered angle in the following figure.
Answer: Since CDE forms a straight line, angles ADE and ADC must sum to 180 degrees. We are given that angle ADE is 122 degrees, so angle ADC equals 180 minus 122, which is 58 degrees. At vertex B, the marked exterior angle is 140 degrees. The interior angle ABC is therefore 360 minus 140, which is 220 degrees. In any quadrilateral, all four interior angles sum to 360 degrees. Using the angle sum property: angle ADC plus angle BCD plus angle BAD plus angle ABC equals 360 degrees. Substituting our known values: 58 plus 53 plus x plus 220 equals 360. This simplifies to 331 plus x equals 360. Solving for x: x equals 360 minus 331, giving x equals 29 degrees. Therefore, x = 29°.
In simple words: When you have a straight line, angles on one side add to 180 degrees. The angles inside any four-sided shape add up to 360 degrees. Use these rules to find the missing angle.

Exam Tip: Always check if angles lie on a straight line (sum to 180°) or form interior angles of a quadrilateral (sum to 360°) - identify which property applies before calculating.

 

Question 4(ii). Find the size of each lettered angle in the following figure.
Answer: Since CD is parallel to BA, we can write that ED is also parallel to BA. When two parallel lines are cut by a transversal, alternate angles are equal. Therefore, angle ECB equals angle CBA. We are given that angle ECB is 75 degrees, so angle CBA equals 75 degrees. Since ABCD is a parallelogram, consecutive angles (angles on the same side) sum to 180 degrees. In particular, angle DAB plus angle CBA equals 180 degrees. Note that angle DAB can be written as x plus 66 degrees, since it is composed of x and the marked 66-degree angle. Substituting: (x + 66) + 75 equals 180. This simplifies to x plus 141 equals 180, so x equals 180 minus 141, giving x equals 39 degrees. Now, in triangle AMB, the three angles must sum to 180 degrees. We have angle MAB equals x, which is 39 degrees; angle MBA equals 30 degrees; and angle AMB is unknown. Therefore: 39 plus 30 plus angle AMB equals 180. This gives 69 plus angle AMB equals 180, so angle AMB equals 111 degrees. Since angles AMB and y are vertically opposite angles (formed where two lines cross), they are equal. Therefore, y equals 111 degrees. Hence, x = 39° and y = 111°.
In simple words: Use parallel line properties to find angle CBA, then use the parallelogram angle sum rule to find x. Next, use the triangle angle sum to find the angle at M, and recognize that vertically opposite angles are equal to find y.

Exam Tip: Identify all parallel lines first and use alternate angles; then apply parallelogram properties for consecutive angles; finally use triangle angle sums and vertically opposite angles - work through these systematically.

 

Question 4(iii). Find the size of each lettered angle in the following figure.
Answer: From the figure, we observe that ABCD appears to be a kite (a quadrilateral with two pairs of adjacent equal sides). The dashed lines indicate that AD equals AB and CD equals CB. Given that angle DBC equals 26 degrees and angle DCB equals 42 degrees, we can find angle BDC in triangle BCD. The sum of angles in triangle BCD is 180 degrees: angle DBC plus angle DCB plus angle BDC equals 180. Substituting: 26 plus 42 plus angle BDC equals 180, giving angle BDC equals 180 minus 68, which is 112 degrees. By symmetry of a kite, the diagonal BD bisects both angles ABD and CBD. Similarly, the diagonal AC creates symmetric angle relationships. Since AD equals AB, triangle ABD is isosceles, so angle ADB equals angle ABD. Let angle ABD equal angle ADB equal z. In triangle ABD: angle BAD plus z plus z equals 180, so angle BAD equals 180 minus 2z. By the kite's symmetry and the given angle at C, we can determine that angle DAC equals angle BAC due to the axis of symmetry. From the positioning and the marked angle 42 degrees at C, angle DAC equals 42 degrees. Therefore, y equals 42 degrees. For angle x at A, since ABCD is a kite with the symmetry shown: angle DAB can be divided by the diagonal. Given the constraints and the symmetry of the figure with the marked 26-degree angle at B, and noting that angle BCD is in the lower portion near the right marked angle, we determine x equals 26 degrees.
In simple words: In a kite, one diagonal acts as a line of symmetry. The angles at opposite ends of the other diagonal are related by this symmetry. Use the triangle angle sum and properties of the kite to find each angle.

Exam Tip: Recognise kite properties: two pairs of consecutive equal sides and one axis of symmetry. Use the triangle angle sum for each of the two triangles created by a diagonal, then apply symmetry to find missing angles.

 

Question 5(i). Find the size of each lettered angle in the following figure:
Answer: Since AB is parallel to CD and BC is parallel to AD, ABCD forms a parallelogram. The angle y matches angle ABC because opposite angles in a parallelogram are equal. Since the diagonals of a parallelogram bisect the vertex angles, BD splits angle ABC in half, so y equals 2 times angle ABD. This gives us y = 2 × 53° = 106°. Using the fact that DC is parallel to AB, the angles y and DAB are supplementary (they add to 180°), so angle DAB = 180° - 106° = 74°. Since AC bisects angle DAB, the angle x is half of angle DAB, giving x = 74° ÷ 2 = 37°. For angle z, since AB and CD are parallel lines cut by the transversal AC, angles z and DAC are alternate angles, which means they are equal. Therefore z = 37°.
In simple words: In a parallelogram, opposite angles are equal, and diagonals split the corner angles in half. Use these ideas to find that x = 37°, y = 106°, and z = 37°.

Exam Tip: Identify which lines are parallel and which are the transversals cutting them - this will guide you to use alternate angles or co-interior angles correctly.

 

Question 5(ii). Find the size of each lettered angle in the following figure:
Answer: Since ED forms a straight line, the angles on one side of point E must add up to 180° (linear pair). So angle AED = 180° - 60° = 120°. Similarly, because CD is a straight line, angle BCD = 180° - 50° = 130°. Now we apply the angle sum property for a pentagon. The five interior angles of pentagon ABCDE must total 540°. The angles at A and B are both 90° (right angles shown in the figure). Substituting all known values: 90° + 90° + 120° + 130° + x = 540°. This simplifies to 430° + x = 540°, so x = 110°.
In simple words: A straight line creates angles that add to 180°. A pentagon's angles add to 540°. Use these facts to work out that x = 110°.

Exam Tip: Always check that the angles you calculate form a straight line (180°) or match the polygon sum formula - this catches errors quickly.

 

Question 5(iii). Find the size of each lettered angle in the following figure:
Answer: Since AD is parallel to FE, angles on the same side add to 180° (co-interior angles in a trapezium). This gives us 60° + y = 180° and x + 110° = 180°, so y = 120° and x = 70°. Since AB is parallel to DE, we can find angle BAD using alternate angles: angle BAD = angle ADE = 70°. In quadrilateral ABCD, all four angles must add to 360°. We have angle BAD = 70°, angle ABC at B = 75°, angle BCD at C is unknown as z, and angle CDA at D is unknown. From the figure, these angles are positioned such that 70° + 75° + z + 130° = 360°, where 130° is angle ABC. Solving: 275° + z = 360°, so z = 85°.
In simple words: Parallel lines create angle relationships that help us find unknown angles. A quadrilateral's angles always add to 360°. This gives us x = 70°, y = 120°, and z = 85°.

Exam Tip: Mark parallel lines with arrows, then use co-interior and alternate angle rules - these rules often unlock the solution quickly.

 

Question 6. In the adjoining figure, ABCD is a rhombus and DCFE is a square. If ∠ABC = 56°, find
(i) ∠DAG
(ii) ∠FEG
(iii) ∠GAC
(iv) ∠AGC
Answer:
(i) In a square, all angles equal 90°. Since ABCD is a rhombus, all sides are equal: AB = BC = DC = AD. Since DCFE is a square, all its sides are equal: CD = ED = FC = EF. Combining these, all seven segments are equal: AB = BC = DC = AD = EF = FC = ED. Given angle ABC = 56°, the opposite angle in the rhombus, angle ADC, also equals 56°. From the diagram, angle EDA = angle EDC + angle ADC = 90° + 56° = 146°. In triangle ADE, we know DE = AD (from our equal segments). When two sides of a triangle are equal, the angles opposite those sides are equal, so angle DEA = angle DAE. Since these two angles are equal and must sum with angle EDA to give 180°, we have 2 × angle DAE + 146° = 180°. Solving: angle DAE = 17°. From the figure, angle DAG = angle DAE = 17°.
(ii) Since angle DAG = angle DAE = 17° and angle AED also equals 17°, we can find angle FEG. Angle FEG = angle FED - angle GED. Since G lies on ED and angle DEG = 17°, we get angle FEG = 90° - 17° = 73°.
(iii) In the rhombus, angles A and B are supplementary (they add to 180°). So angle A = 180° - 56° = 124°. The diagonal AC bisects angle A, so angle DAC = 124° ÷ 2 = 62°. Therefore angle GAC = angle DAC - angle DAG = 62° - 17° = 45°.
(iv) In triangle EDG, angle D = 90° (corner of the square), angle DEG = 17°, so angle DGE = 180° - 90° - 17° = 73°. Vertically opposite angles are equal, so angle AGC = angle DGE = 73°.
In simple words: A rhombus has all equal sides and opposite angles are equal. A square has all angles at 90°. Use these properties to find angle relationships: angle DAG = 17°, angle FEG = 73°, angle GAC = 45°, and angle AGC = 73°.

Exam Tip: When a diagonal bisects an angle or when two sides are equal, use angle bisector and isosceles triangle properties - these are the keys to unlocking multi-part geometry problems.

 

Question 7. If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.
Answer: All sides of the rhombus ABCD are 8 cm: AB = BC = CD = DA = 8 cm. Let angle A = 60°. Consider triangle ABD. Since AB = AD (sides of the rhombus), the angles opposite these equal sides must be equal. If we call these base angles x, then angle ADB = angle ABD = x. The angle sum in the triangle gives x + x + 60° = 180°, so 2x = 120°, meaning x = 60°. Since all angles equal 60°, triangle ABD is equilateral, so BD = 8 cm. The diagonals of a rhombus bisect each other at right angles. This means AO = OC, BO = OD = 4 cm, and angle AOB = 90°. In right triangle AOB, apply the Pythagorean theorem: AB² = AO² + OB². Substituting 8² = AO² + 4², we get 64 = AO² + 16, so AO² = 48. Taking the square root, AO = √48 = 4√3 cm. Since AC = 2 × AO, we have AC = 8√3 cm.
In simple words: When one angle of a rhombus is 60° and we find that triangle ABD has all angles equal to 60°, it becomes an equilateral triangle. Use the Pythagorean theorem on the right triangle formed by the half-diagonals to find the full lengths: one diagonal is 8 cm and the other is 8√3 cm.

Exam Tip: Recognize when a triangle is equilateral (all angles 60°) - this saves calculation time. Always apply the Pythagorean theorem in the right-angled triangles formed by the intersecting diagonals.

 

Question 8. Using ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and ∠BAD = 45°. If the bisector of ∠BAD meets DC at E, prove that ∠AEB is a right angle.
Answer: Steps of construction:
1. Draw AB = 5.0 cm.
2. At A, construct angle BAP = 45°.
3. With A as centre and radius 2.5 cm, mark point D on line AP.
4. With D as centre and radius 5.0 cm, draw an arc.
5. With B as centre and radius 2.5 cm, draw another arc to meet the first arc at C.
6. Join BC and CD to complete the parallelogram ABCD.
Next, draw the angle bisector of angle BAD (this line splits the 45° angle into two 22.5° angles). This bisector meets side DC at point E. Join EB to see the angle formed.
On measuring angle AEB with a protractor, it equals 90°.
In simple words: To draw a parallelogram, fix one side, then use the given angle and side length to place the other vertices. When you draw the angle bisector from A and meet it with side DC at E, the angle AEB turns out to be exactly 90°.

Exam Tip: Construction questions often ask you to prove something after drawing. Measure carefully with a protractor - a 90° angle will be exact when the geometry is correct. For the proof, note that in parallelogram ABCD with the angle bisector of the 45° angle, the triangle properties guarantee this right angle.

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