CBSE Class 9 Mathematics Quadrilaterals MCQs Set 10

Mathematics Objective Questions and Answers: Chapter 08 Quadrilaterals

Explore reliable objective questions for Chapter 08 Quadrilaterals tailored for Class 9 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Download Chapter 08 Quadrilaterals MCQs with Answers

Access the complete set of multiple-choice questions for Chapter 08 Quadrilaterals below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: In a parallelogram ABCD, ∠D = 105°. Then ∠A and ∠B are
(a) 75°, 105°
(b) 105°, 75°
(c) 105°, 105°
(d) 75°, 75°
Show Answer & Explanation

Answer: (a) 75°, 105°

Explanation:
1. Adjacent angles of a parallelogram are supplementary: ∠A = 180° − 105° = 75°.
2. Opposite angles are equal: ∠B = ∠D = 105°.

Teacher's Note:
1. Adjacent angles add up to 180°.
2. Opposite angles are equal.

Question 2: The diagonals AC and BD of a parallelogram ABCD meet at O. If AC = 12.8 cm and BD = 7.6 cm, then OC and OD are
(a) 1.9 cm, 6.4 cm
(b) 6.4 cm, 3.8 cm
(c) 3.8 cm, 3.2 cm
(d) 6.4 cm, 7.6 cm
Show Answer & Explanation

Answer: (b) 6.4 cm, 3.8 cm

Explanation:
1. Diagonals of a parallelogram bisect each other.
2. OC = ½ × 12.8 = 6.4 cm and OD = ½ × 7.6 = 3.8 cm.

Teacher's Note:
1. Each diagonal is cut into two equal halves at O.

Question 3: Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. The value of x is
(a) 17
(b) 16
(c) 13
(d) 15
Show Answer & Explanation

Answer: (c) 13

Explanation:
1. Opposite angles are equal: 3x − 2 = 50 − x.
2. 4x = 52, so x = 13. Each angle is 37°.

Teacher's Note:
1. Equate opposite angles, not add them.
2. Check: 3 × 13 − 2 = 37 = 50 − 13.

Question 4: When the diagonals of a parallelogram are perpendicular to each other, it is a
(a) rectangle
(b) square
(c) trapezium
(d) rhombus
Show Answer & Explanation

Answer: (d) rhombus

Explanation:
1. In △AOB and △AOD: AO common, BO = DO, ∠AOB = ∠AOD = 90°, so AB = AD.
2. Adjacent sides equal makes the parallelogram a rhombus.

Teacher's Note:
1. A rectangle has equal diagonals; a rhombus has perpendicular diagonals.
2. A square satisfies both, but perpendicular diagonals alone only guarantee a rhombus.

Question 5: In a parallelogram ABCD, E is the midpoint of BC. DE and AB, when produced, meet at F. Then
(a) AF = 2AB
(b) AF = ½AB
(c) AF = 4AB
(d) AF = 3AB
Show Answer & Explanation

Answer: (a) AF = 2AB

Explanation:
1. △DCE ≅ △FBE (ASA: ∠DEC = ∠FEB, CE = BE, ∠DCE = ∠FBE), so BF = DC.
2. DC = AB, so BF = AB and AF = AB + BF = 2AB.

Teacher's Note:
1. Use alternate angles on AB ∥ DC.
2. Congruent triangles give BF = CD = AB.

Question 6: ABCD is a rhombus with ∠ABC = 56°. Then ∠ACD is
(a) 56°
(b) 62°
(c) 124°
(d) 34°
Show Answer & Explanation

Answer: (b) 62°

Explanation:
1. ∠BCD = 180° − 56° = 124°.
2. A diagonal of a rhombus bisects the angle, so ∠ACD = 124° ÷ 2 = 62°.

Teacher's Note:
1. Diagonals of a rhombus bisect the vertex angles.
2. Alternative: in isosceles △ABC, ∠ACB = 62° and ∠ACD = ∠ACB.

Question 7: In △ABC, P, Q, R are the midpoints of BC, CA and AB. If AC = 16 cm, BC = 20 cm and AB = 24 cm, the perimeter of the quadrilateral ARPQ is
(a) 60 cm
(b) 30 cm
(c) 40 cm
(d) 80 cm
Show Answer & Explanation

Answer: (c) 40 cm

Explanation:
1. PQ ∥ AB and PR ∥ AC by the midpoint theorem, so ARPQ is a parallelogram.
2. AR = ½AB = 12 cm and RP = ½AC = 8 cm, so PQ = AR = 12 cm and AQ = RP = 8 cm.
3. Perimeter = 12 + 8 + 12 + 8 = 40 cm.

Teacher's Note:
1. Perimeter equals AB + AC = 40.
2. Check: 2 × (12 + 8) = 40.

Question 8: LMNO is a trapezium with LM ∥ NO. P and Q are the midpoints of LO and MN. If LM = 5 cm and ON = 10 cm, PQ =
(a) 2.5 cm
(b) 5 cm
(c) 15 cm
(d) 7.5 cm
Show Answer & Explanation

Answer: (d) 7.5 cm

Explanation:
1. The segment joining the midpoints of the non-parallel sides is parallel to the bases and half their sum.
2. PQ = ½(5 + 10) = 7.5 cm.

Teacher's Note:
1. Be careful: half the sum, not half the difference.

Question 9: ABCD is an isosceles trapezium with ∠A = 45°. Then ∠C is
(a) 135°
(b) 45°
(c) 90°
(d) 115°
Show Answer & Explanation

Answer: (a) 135°

Explanation:
1. Angles on the same parallel-side base are equal in an isosceles trapezium, and co-interior angles are supplementary.
2. So the angles of the other base are 180° − 45° = 135°, giving ∠C = 135°.

Teacher's Note:
1. Base angles of an isosceles trapezium are equal.
2. Angles along a non-parallel side add to 180°.

Question 10: In a right-angled triangle ABC with ∠B = 90°, AB = 9 cm and AC = 15 cm, D and E are the midpoints of AB and AC. The area of △ADE is
(a) 67.5 cm²
(b) 13.5 cm²
(c) 27 cm²
(d) 54 cm²
Show Answer & Explanation

Answer: (b) 13.5 cm²

Explanation:
1. BC = √(15² − 9²) = 12 cm. DE ∥ BC and DE = 6 cm.
2. △ADE is right-angled at D with AD = 4.5 cm: area = ½ × 4.5 × 6 = 13.5 cm².

Teacher's Note:
1. Area of △ABC is 54 cm² and △ADE is one fourth of it.
2. Midpoint theorem gives DE ∥ BC.

Question 11: The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. The angles are
(a) 30°, 50°, 90°, 130°
(b) 36°, 60°, 100°, 164°
(c) 36°, 60°, 108°, 156°
(d) 45°, 75°, 108°, 132°
Show Answer & Explanation

Answer: (c) 36°, 60°, 108°, 156°

Explanation:
1. Sum of angles = 360° and 3 + 5 + 9 + 13 = 30, so one part = 12°.
2. Angles = 36°, 60°, 108°, 156°.

Teacher's Note:
1. Angle sum of a quadrilateral is 360°.
2. Check: 36 + 60 + 108 + 156 = 360.

Question 12: In a parallelogram, one angle is 24° less than twice the smallest angle. The angles are
(a) 68°, 12°, 68°, 12°
(b) 56°, 124°, 56°, 124°
(c) 78°, 102°, 78°, 102°
(d) 68°, 112°, 68°, 112°
Show Answer & Explanation

Answer: (d) 68°, 112°, 68°, 112°

Explanation:
1. Let the smallest angle be x. The adjacent angle 2x − 24° satisfies x + 2x − 24° = 180°.
2. 3x = 204°, x = 68°, other angle = 112°. Opposite angles are equal.

Teacher's Note:
1. Adjacent angles are supplementary.
2. Check: 2 × 68 − 24 = 112.

Question 13: In △ABC, D, E, F are the midpoints of BC, CA and AB, and AB = 17 cm, BC = 18 cm, CA = 19 cm. The perimeter of △DEF is
(a) 27 cm
(b) 54 cm
(c) 18 cm
(d) 13.5 cm
Show Answer & Explanation

Answer: (a) 27 cm

Explanation:
1. Each side of △DEF is half of the parallel side of △ABC.
2. Perimeter = ½(17 + 18 + 19) = 27 cm.

Teacher's Note:
1. The medial triangle has half the perimeter.

Question 14: In a quadrilateral ABCD, ∠B = 70° and ∠C = 105°. AO and DO are the bisectors of ∠A and ∠D. Then ∠AOD is
(a) 67.5°
(b) 87.5°
(c) 77.5°
(d) 99.75°
Show Answer & Explanation

Answer: (b) 87.5°

Explanation:
1. ∠A + ∠D = 360° − 70° − 105° = 185°.
2. In △AOD, ∠AOD = 180° − ½(185°) = 87.5°.

Teacher's Note:
1. Sum of angles of a quadrilateral is 360°.
2. The bisectors halve ∠A and ∠D.

Question 15: In a parallelogram, the bisectors of two adjacent angles meet at an angle of
(a) 30°
(b) 45°
(c) 90°
(d) 60°
Show Answer & Explanation

Answer: (c) 90°

Explanation:
1. Adjacent angles add up to 180°, so their halves add up to 90°.
2. In the triangle formed, the third angle is 180° − 90° = 90°.

Teacher's Note:
1. Half of 180° is 90°.

Question 16: In parallelogram ABCD the diagonal AC makes ∠ACD = 60° and ∠ACB = 30°. Then ∠CAB and ∠CAD are
(a) 30°, 60°
(b) 45°, 45°
(c) 90°, 90°
(d) 60°, 30°
Show Answer & Explanation

Answer: (d) 60°, 30°

Explanation:
1. AB ∥ DC gives ∠CAB = ∠ACD = 60° (alternate angles).
2. AD ∥ BC gives ∠CAD = ∠ACB = 30°.

Teacher's Note:
1. Alternate angles between parallel lines are equal.
2. Check: ∠BAD = 90°, so ABCD is in fact a rectangle.

Question 17: In parallelogram PQRS, ∠P = 70° and ∠SQR = 50°. Then ∠SQP and ∠QSP are
(a) 60°, 50°
(b) 60°, 45°
(c) 70°, 35°
(d) 35°, 70°
Show Answer & Explanation

Answer: (a) 60°, 50°

Explanation:
1. SP ∥ QR gives ∠QSP = ∠SQR = 50° (alternate angles).
2. ∠SQP = 180° − 70° − 50° = 60°.

Teacher's Note:
1. Use the triangle angle sum in △PQS.

Question 18: ABCD is a parallelogram, ∠DAO = 40°, ∠BAO = 35° and ∠COD = 65°, where O is the intersection of the diagonals. Then ∠ABO is
(a) 75°
(b) 80°
(c) 65°
(d) 40°
Show Answer & Explanation

Answer: (b) 80°

Explanation:
1. ∠AOB = ∠COD = 65° (vertically opposite).
2. In △AOB, ∠ABO = 180° − 35° − 65° = 80°.

Teacher's Note:
1. Vertically opposite angles are equal.

Question 19: In a parallelogram ABCD with ∠A = 60°, the bisectors of ∠A and ∠B meet at P on side DC. If AD = 7 cm, then DC is
(a) 7 cm
(b) 10.5 cm
(c) 14 cm
(d) 21 cm
Show Answer & Explanation

Answer: (c) 14 cm

Explanation:
1. ∠DAP = ∠PAB = ∠APD (alternate), so DP = AD = 7 cm. Similarly PC = BC = 7 cm.
2. DC = DP + PC = 14 cm.

Teacher's Note:
1. A bisector and a parallel line make an isosceles triangle.
2. DC = 2AD.

Question 20: In a parallelogram ABCD the bisector of ∠A also bisects BC at P. If AB = 6 cm, then AD is
(a) 6 cm
(b) 9 cm
(c) 3 cm
(d) 12 cm
Show Answer & Explanation

Answer: (d) 12 cm

Explanation:
1. The bisector gives ∠BAP = ∠APB (alternate angle with AD ∥ BC), so BP = AB = 6 cm.
2. P is the midpoint of BC, so BC = 12 cm = AD.

Teacher's Note:
1. BP = AB because of the isosceles triangle ABP.

Question 21: The quadrilateral formed by joining the midpoints of the sides of a square is a
(a) square
(b) rhombus that is not a square
(c) rectangle that is not a square
(d) parallelogram only
Show Answer & Explanation

Answer: (a) square

Explanation:
1. Each side of the new figure is half a diagonal of the square, so all four are equal.
2. The diagonals of the square are equal and perpendicular, so the new angles are 90°.

Teacher's Note:
1. Equal and perpendicular diagonals give a square.

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Chapter 08 Quadrilaterals Objective Questions & Solutions for Class 9 Mathematics

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