Question 1

A rectangle has a perimeter of 56 cm and a length of 18 cm. What is its breadth?

Accepted Answer

Step 1: Recall the perimeter formula for a rectangle: P = 2(l + b). Step 2: Substitute P = 56 cm and l = 18 cm. Step 3: 56 = 2(18 + b). Step 4: 28 = 18 + b. Step 5: b = 28 - 18 = 10 cm. Therefore, the breadth is 10 cm.

Question 2

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

Accepted Answer

Step 1: Recall the formula for area of a trapezium: A = (1/2) × (sum of parallel sides) × height. Step 2: Substitute the parallel sides as 10 cm and 14 cm, and height = 8 cm. Step 3: A = (1/2) × (10 + 14) × 8 = (1/2) × 24 × 8 = (1/2) × 192 = 96 cm². Therefore, the area is 96 cm².

Question 3

A rectangle has a length of 24 cm and a breadth of 16 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 = 16 cm. Step 3: P = 2(24 + 16) = 2(40) = 80 cm. Therefore, the perimeter is 80 cm.

Question 4

The diagonals of a rhombus are 12 cm and 16 cm. Find the area of the rhombus.

Accepted Answer

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

Question 5

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 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 6

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 7

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, diagonals bisect each other at right angles. Step 2: Half-diagonals are 8 cm and 6 cm. Step 3: Each side of the rhombus is the hypotenuse of a right triangle with legs 8 cm and 6 cm. Step 4: Side = √(8² + 6²) = √(64 + 36) = √100 = 10 cm. Step 5: Perimeter = 4 × 10 = 40 cm.

Question 8

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. Step 3: 4w + 16 = 56, so 4w = 40, thus w = 10 cm. Step 4: Length = 10 + 8 = 18 cm. Step 5: Area = length × width = 18 × 10 = 180 cm².

Question 9

A square and a rectangle have the same perimeter of 48 cm. The rectangle's length is 14 cm. What is the difference between the area of the square and the area of the rectangle?

Accepted Answer

Step 1: Square perimeter = 48 cm, so each side = 48 ÷ 4 = 12 cm. Step 2: Area of square = 12² = 144 cm². Step 3: Rectangle perimeter = 48 cm, length = 14 cm. Using 2(l + w) = 48: 2(14 + w) = 48, so 14 + w = 24, thus w = 10 cm. Step 4: Area of rectangle = 14 × 10 = 140 cm². Step 5: Difference = 144 − 140 = 4 cm².

Question 10

A quadrilateral ABCD is a parallelogram with sides AB = 20 cm and BC = 15 cm. The angle between AB and BC is 60°. What is the area of the parallelogram?

Accepted Answer

Step 1: For a parallelogram, area = base × height, or equivalently, area = side₁ × side₂ × sin(angle between them). Step 2: Area = AB × BC × sin(60°). Step 3: sin(60°) = √3/2 ≈ 0.866. Step 4: Area = 20 × 15 × (√3/2) = 300 × (√3/2) = 150√3 cm². Step 5: 150√3 ≈ 150 × 1.732 = 259.8 ≈ 260 cm² (or express as 150√3 cm²).

Question 11

A rectangle PQRS has length 20 cm and width 12 cm. A diagonal PR divides the rectangle into two triangles. If a point T is chosen on side PQ such that PT = 5 cm, then the perpendicular distance from T to the diagonal PR is:

Accepted Answer

Step 1: Rectangle PQRS with PQ = 20 cm (length) and PS = 12 cm (width). Step 2: Diagonal PR = √(20² + 12²) = √(400 + 144) = √544 = 4√34 cm. Step 3: Set up coordinates: P = (0, 0), Q = (20, 0), R = (20, 12), S = (0, 12). Step 4: Point T is on PQ with PT = 5, so T = (5, 0). Step 5: The line PR goes from (0, 0) to (20, 12). Equation of PR: 12x - 20y = 0, or 3x - 5y = 0. Step 6: Perpendicular distance from T(5, 0) to line 3x - 5y = 0 is: d = |3(5) - 5(0)| / √(3² + 5²) = |15| / √34 = 15/√34 = 15√34/34 cm. Step 7: 15√34/34 ≈ 15 × 5.831 / 34 ≈ 87.46 / 34 ≈ 2.57 cm.

Question 12

In a rhombus PQRS, the diagonals PR and QS intersect at O. If PR = 30 cm and QS = 16 cm, and a point M is chosen on side PQ such that PM = 10 cm, then the perpendicular distance from M to diagonal PR is:

Accepted Answer

Step 1: In a rhombus, diagonals bisect each other at right angles. So PO = 15 cm, QO = 8 cm. Step 2: Side length of rhombus = √(15² + 8²) = √(225 + 64) = √289 = 17 cm. Step 3: Point M lies on PQ with PM = 10 cm. We need the perpendicular distance from M to PR. Step 4: Set up coordinates with O at origin, PR along x-axis, QS along y-axis. Then P = (-15, 0), Q = (0, 8), R = (15, 0), S = (0, -8). Step 5: Point M divides PQ in ratio PM:MQ = 10:7. Using section formula: M = (P + (10/17)(Q - P)) = (-15, 0) + (10/17)(15, 8) = (-15 + 150/17, 80/17) = ((-255 + 150)/17, 80/17) = (-105/17, 80/17). Step 6: Perpendicular distance from M to PR (the x-axis) is the y-coordinate of M = 80/17 ≈ 4.71 cm.

IBPS Clerk Quadrilaterals

Quadrilaterals— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Quadrilaterals — Practice Questions

60-Second Revision — Quadrilaterals