Question 1

A rectangle has length 24 cm and breadth 18 cm. What is the perimeter of the rectangle?

Accepted Answer

Step 1: Recall the formula for perimeter of a rectangle: P = 2(l + b). Step 2: Substitute l = 24 cm and b = 18 cm. Step 3: P = 2(24 + 18) = 2(42) = 84 cm. Therefore, the perimeter is 84 cm.

Question 2

The area of a square is 256 cm². What is the length of its side?

Accepted Answer

Step 1: Recall that the area of a square is A = s², where s is the side length. Step 2: Given A = 256 cm², we have s² = 256. Step 3: Taking the square root: s = √256 = 16 cm. Therefore, the side length is 16 cm.

Question 3

A parallelogram has a base of 20 cm and a height of 15 cm. What is its area?

Accepted Answer

Step 1: Recall the formula for the area of a parallelogram: A = base × height. Step 2: Substitute base = 20 cm and height = 15 cm. Step 3: A = 20 × 15 = 300 cm². Therefore, the area is 300 cm².

Question 4

A rhombus has diagonals of length 16 cm and 12 cm. What is its area?

Accepted Answer

Step 1: Recall the formula for the area of a rhombus: A = (d₁ × d₂) / 2, where d₁ and d₂ are the diagonals. Step 2: Substitute d₁ = 16 cm and d₂ = 12 cm. Step 3: A = (16 × 12) / 2 = 192 / 2 = 96 cm². Therefore, the area is 96 cm².

Question 5

A trapezium has parallel sides of 10 cm and 14 cm, and a height of 8 cm. What is its area?

Accepted Answer

Step 1: Recall the formula for the area of a trapezium: A = [(a + b) / 2] × h, where a and b are the parallel sides and h is the height. Step 2: Substitute a = 10 cm, b = 14 cm, and h = 8 cm. Step 3: A = [(10 + 14) / 2] × 8 = [24 / 2] × 8 = 12 × 8 = 96 cm². Therefore, the area is 96 cm².

Question 6

The diagonals of a rectangle are 10 cm each. If one side is 6 cm, what is the length of the adjacent side?

Accepted Answer

Step 1: In a rectangle, the diagonals are equal and bisect each other. The diagonal, length, and breadth form a right triangle. Step 2: Using the Pythagorean theorem: d² = l² + b², where d = 10 cm and l = 6 cm. Step 3: 10² = 6² + b² → 100 = 36 + b² → b² = 64 → b = 8 cm. Therefore, the adjacent side is 8 cm.

Question 7

A parallelogram has a base of 24 cm and a height of 15 cm. If the base is increased by 25% and the height is decreased by 20%, what is the percentage change in the area of the parallelogram?

Accepted Answer

Step 1: Original area = base × height = 24 × 15 = 360 cm². Step 2: New base = 24 + (25% of 24) = 24 + 6 = 30 cm. Step 3: New height = 15 − (20% of 15) = 15 − 3 = 12 cm. Step 4: New area = 30 × 12 = 360 cm². Step 5: Percentage change = [(360 − 360) / 360] × 100 = 0%. Therefore, the area remains unchanged.

Question 8

The diagonals of a rhombus are 16 cm and 12 cm. What is the perimeter of the rhombus?

Accepted Answer

Step 1: In a rhombus, the diagonals bisect each other at right angles. Step 2: Half of diagonal 1 = 16/2 = 8 cm. Half of diagonal 2 = 12/2 = 6 cm. Step 3: These halves form a right triangle with the side of the rhombus as the hypotenuse. Step 4: Using Pythagoras theorem: side = √(8² + 6²) = √(64 + 36) = √100 = 10 cm. Step 5: Perimeter = 4 × side = 4 × 10 = 40 cm.

Question 9

In a rectangle, the length is 8 cm more than the width. If the perimeter is 56 cm, what is the area of the rectangle?

Accepted Answer

Step 1: Let width = w cm. Then length = (w + 8) cm. Step 2: Perimeter = 2(length + width) = 2(w + 8 + w) = 2(2w + 8) = 4w + 16 = 56. Step 3: 4w = 40, so w = 10 cm. Step 4: Length = 10 + 8 = 18 cm. Step 5: Area = length × width = 18 × 10 = 180 cm².

Question 10

A parallelogram PQRS has sides PQ = 13 cm and QR = 8 cm. The angle between them is 60°. A perpendicular is drawn from R to side PQ, meeting it at point T. What is the length of RT?

Accepted Answer

Step 1: In parallelogram PQRS, angle PQR = 60°. Step 2: RT is the perpendicular from R to line PQ (or its extension). Step 3: In triangle QRT, angle QTR = 90°, angle TQR = 60°. Step 4: RT = QR × sin(60°) = 8 × (√3/2) = 4√3 cm ≈ 6.93 cm. For SSC purposes, this is approximately 7 cm or exactly 4√3 cm.

Question 11

In a kite PQRS, the diagonals PR and QS intersect at O. If PR = 24 cm, QS = 18 cm, and the diagonals intersect such that PO = 16 cm and QO = 10 cm, what is the area of the kite?

Accepted Answer

Step 1: In a kite, the diagonals are perpendicular to each other. Step 2: Area of a kite = (1/2) × d₁ × d₂, where d₁ and d₂ are the lengths of the diagonals. Step 3: Area = (1/2) × 24 × 18 = (1/2) × 432 = 216 cm². Note: The positions of O on the diagonals (PO = 16, QO = 10) are given to verify the kite's properties but do not change the area formula, which depends only on the total diagonal lengths.

SSC MTS Quadrilaterals

Quadrilaterals— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Quadrilaterals — Practice Questions

60-Second Revision — Quadrilaterals