TIME: 3 HOURS
MAX MARKS: 100
SECTION – A (MCQs / VERY SHORT ANSWER)
1 mark × 20 = 20 marks
Q1. If ∠A = 78° in parallelogram ABCD, then ∠D equals:
(a) 102° (b) 78° (c) 90° (d) 180°
Q2. The sum of the interior angles of any quadrilateral is:
(a) 90° (b) 180° (c) 270° (d) 360°
Q3. In a rectangle, the diagonals are always:
(a) Perpendicular (b) Equal (c) Unequal (d) Parallel
Q4. A quadrilateral with one pair of parallel sides is called a:
(a) Parallelogram (b) Trapezium (c) Rhombus (d) Kite
Q5. If three angles of a quadrilateral are 75°, 110°, and 90°, then the fourth angle equals:
(a) 85° (b) 80° (c) 110° (d) 95°
Q6. Opposite angles of a cyclic quadrilateral are:
(a) Equal (b) Acute (c) Supplementary (d) Complementary
Q7. Diagonals of a rhombus always:
(a) Bisect each other (b) Are equal
(c) Are perpendicular (d) Both (a) and (c)
Q8. A parallelogram has adjacent sides 12 cm and 8 cm. Its perimeter equals:
(a) 20 cm (b) 40 cm (c) 28 cm (d) 16 cm
Q9. Diagonals of square PQRS measure 10 cm. Each side equals:
(a) 5√2 cm (b) 10 cm (c) 4√2 cm (d) 6 cm
Q10. If ∠A = 95° in cyclic quadrilateral ABCD, then ∠C equals:
(a) 95° (b) 85° (c) 45° (d) 100°
Q11. The exterior angles of any quadrilateral always add up to:
(a) 360° (b) 720° (c) 180° (d) 540°
Q12. In parallelogram PQRS, ∠P = 65°. Then ∠Q equals:
(a) 115° (b) 65° (c) 180° (d) 120°
Q13. A quadrilateral with diagonals that bisect each other is a:
(a) Kite (b) Rhombus (c) Parallelogram (d) Trapezium
Q14. Diagonals of which quadrilateral are equal and perpendicular?
(a) Square (b) Rhombus (c) Rectangle (d) Trapezium
Q15. Angles of quadrilateral ABCD are in the ratio 2 : 3 : 4 : 3. The largest angle equals:
(a) 120° (b) 135° (c) 144° (d) 150°
Q16. In parallelogram LMNO, ∠L = 70°. Then ∠M equals:
(a) 110° (b) 70° (c) 35° (d) 90°
Q17. The diagonal BD of rhombus ABCD is 24 cm and diagonal AC is 10 cm. The side equals:
(a) 13 cm (b) 12 cm (c) 10 cm (d) 14 cm
Q18. In trapezium ABCD, AB ∥ CD. If ∠A = 60°, then ∠D equals:
(a) 120° (b) 60° (c) 90° (d) None
Q19. The midpoints of the sides of any quadrilateral form a:
(a) Triangle (b) Kite (c) Parallelogram (d) Rectangle
Q20. Diagonals of a rectangle bisect each other at:
(a) 45° (b) 30° (c) 90° (d) Depends on side lengths
SECTION – B (VERY SHORT QUESTIONS)
2 marks × 5 = 10 marks
Q21. In quadrilateral ABCD, ∠A = 3x − 5°, ∠C = x + 35°. Find ∠A and ∠C.
Q22. In parallelogram PQRS, ∠P = 72°. Find ∠R and ∠Q.
Q23. In rectangle ABCD, diagonal AC makes 34° with AB. Find ∠AOC.
Q24. In parallelogram LMNO, ∠L = (5y − 10)° and ∠M = (2y + 40)°. Find y.
Q25. A trapezium has parallel sides 12 cm and 20 cm. Find the length of the mid-segment.
SECTION – C (SHORT ANSWER QUESTIONS)
3 marks × 6 = 18 marks
Q26. ABCD is a parallelogram. Prove that ∠A = ∠C.
Q27. In rhombus PQRS, diagonal PR = 18 cm and diagonal QS = 24 cm. Find the side length.
Q28. In quadrilateral ABCD, diagonals AC and BD intersect at O. AO = 6 cm, OC = 6 cm, BO = 8 cm. Find BD.
Q29. In parallelogram ABCD, ∠A = 65°. Find ∠C and ∠B.
Q30. Angles of quadrilateral ABCD are in ratio 3 : 4 : 5 : 6. Find all angles.
Q31. In triangle ABC, D and E are midpoints of AB and AC. Prove that DE ∥ BC.
SECTION – D (LONG ANSWER QUESTIONS)
5 marks × 4 = 20 marks
Q32. ABCD is a parallelogram. Prove that its diagonals divide it into two pairs of congruent triangles.
Q33. In rhombus PQRS, diagonals intersect at O. PR = 26 cm and QS = 20 cm.
(i) Prove that diagonals are perpendicular.
(ii) Find the side length.
(iii) Find ∠POQ.
Q34. In ΔXYZ, P, Q, and R are midpoints of XY, YZ, and ZX.
Show that the quadrilateral formed by joining P, Q, R, and midpoints of diagonals is a parallelogram.
SECTION – E (VERY LONG ANSWER / CASE STUDY)
6 marks × 2 = 12 marks
CASE STUDY 1 – GARDEN DESIGN
A gardener designs a garden in the shape of quadrilateral ABCD, where AB ∥ CD.
He plants grass in triangular regions formed by diagonals AC and BD.
Given:
AB = 30 m, CD = 18 m, AC and BD intersect at right angles.
Answer the following:
(a) Prove that ABCD is a trapezium.
(b) Find the area of ΔAOD if AO = 12 m and DO = 9 m.
(c) Find the total area covered by grass.
CASE STUDY 2 – SQUARE TILES
A company designs square tiles of side 8 cm. They join four such tiles to make a larger quadrilateral PQRS.
Given:
Tiles touch at vertices, and diagonals intersect at right angles.
Answer the following:
(a) Find diagonal length of one tile.
(b) Prove that PQRS forms a square.
(c) Find area of PQRS.
SECTION – F (4 marks × 5 = 20 marks)
Q35. In parallelogram ABCD, AC = BD iff ABCD is a rectangle. Prove the statement.
Q36. In a quadrilateral, bisectors of all interior angles form a cyclic quadrilateral. Prove the statement.
Q37. Prove that opposite sides of a parallelogram are equal using coordinate geometry.
Q38. In quadrilateral ABCD, AC ⟂ BD. AC = 20 cm and BD = 16 cm.
Find the area of quadrilateral ABCD.
Answer Key:
| Q | Answer |
|---|---|
| 1 | (a) 102° |
| 2 | (d) 360° |
| 3 | (b) Equal |
| 4 | (b) Trapezium |
| 5 | (b) 80° |
| 6 | (c) Supplementary |
| 7 | (d) Both (a) and (c) |
| 8 | (b) 40 cm |
| 9 | (a) 5√2 cm |
| 10 | (b) 85° |
| 11 | (a) 360° |
| 12 | (a) 115° |
| 13 | (c) Parallelogram |
| 14 | (a) Square |
| 15 | (c) 144° |
| 16 | (a) 110° |
| 17 | (a) 13 cm |
| 18 | (a) 120° |
| 19 | (c) Parallelogram |
| 20 | (d) Depends on side lengths |
Q21.
∠A = 3x − 5
∠C = x + 35
∠A + ∠C = 180°
⇒ 3x − 5 + x + 35 = 180
⇒ 4x + 30 = 180
⇒ x = 37.5°
∠A = 3(37.5) − 5 = 107.5°
∠C = 37.5 + 35 = 72.5°
Q22.
In parallelogram: ∠P + ∠Q = 180°
∠Q = 180 – 72 = 108°
Opposite angles equal: ∠R = 72°
Q23.
In rectangle, diagonals bisect and ∠AOC = 180° – 2×34°
= 112°
Q24.
∠L + ∠M = 180°
(5y − 10) + (2y + 40) = 180
7y + 30 = 180
7y = 150
y = 150 ÷ 7 = 21.43°
Q25.
Mid-segment of trapezium:
= ½ (sum of parallel sides)
= ½ (12 + 20)
= 16 cm
Q26.
Opposite angles of parallelogram equal → ∠A = ∠C.
(Proof of alternate interior angles and parallel lines.) ✔
Q27.
Rhombus side = ½ √(PR² + QS²)
= ½ √(18² + 24²)
= ½ √(324 + 576)
= ½ √900
= ½ × 30
= 15 cm
Q28.
BD = 2 × BO = 2 × 8 = 16 cm
Q29.
Parallelogram opposite angles equal:
∠C = 65°
Adjacent angles supplementary:
∠B = 180 − 65 = 115°
Q30.
Sum = 360
Let angles = 3x, 4x, 5x, 6x
3x + 4x + 5x + 6x = 360
18x = 360
x = 20°
Angles =
60°, 80°, 100°, 120° ✔
Q31.
Midpoint Theorem → line joining midpoints of two sides ∥ third side.
So DE ∥ BC. ✔
Q32.
Diagonals split parallelogram into triangles with:
• One pair having equal base and height
• Opposite sides parallel → alternate interior angles equal
Thus triangles are congruent. ✔
Q33.
(i) Rhombus diagonals are perpendicular (property).
(ii) Side = ½ √(PR² + QS²)
= ½ √(26² + 20²)
= ½ √(676 + 400)
= ½ √1076
= ½ × 32.8
= 16.4 cm
(iii) ∠POQ = 90° (diagonals perpendicular)
Q34.
Connecting midpoints creates two mid-segment parallelograms.
Opposite sides become parallel → quadrilateral is a parallelogram. ✔
CASE STUDY 1
(a) AB ∥ CD → definition of trapezium ✔
(b) ΔAOD is right triangle:
Area = ½ × AO × DO
= ½ × 12 × 9
= 54 m²
(c) Grass covers 4 triangular regions.
Total area = 4 × 54 = 216 m²
CASE STUDY 2
(a) Diagonal of square tile = √(8² + 8²)
= √128 = 8√2 cm
(b) Four equal squares → all angles 90°, sides equal → PQRS is square ✔
(c) Side of PQRS = 16 cm
Area = 16² = 256 cm²
Q35.
If AC = BD in parallelogram → diagonals equal → all angles become 90° → rectangle ✔
Q36.
Angle bisectors of quadrilateral → ∠A/2 + ∠B/2 + ∠C/2 + ∠D/2 = 180° → form cyclic quadrilateral ✔
Q37.
Using coordinate geometry:
Take parallelogram ABCD with A(0,0), B(a,0), D(0,b), C(a,b).
Opposite sides AB = CD and AD = BC. ✔
Q38.
Area of quadrilateral with perpendicular diagonals:
Area = ½ × AC × BD
= ½ × 20 × 16
= 160 cm²
