# Mensuration – Quiz 1

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# Mensuration – Quiz 1

### Introduction

Mensuration is one of important topic in Quantitative Aptitude Section. In Mensuration – Quiz 1 article candidates can find a question with an answer. By solving this question candidates can improve and maintain, speed, and accuracy in the exams. Mensuration - Quiz 1 questions a very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

What is the area of an equilateral triangle of side 16 cm?
A. 48 $\sqrt{3}$ ${cm}^{2}$ B. 128 $\sqrt{3}$ ${cm}^{2}$ C. 9.6 $\sqrt{3}$ ${cm}^{2}$ D. 64 $\sqrt{3}$ ${cm}^{2}$

D
Area of an equilateral triangle = $\frac{\sqrt{3}}{4} {cm}^{2}$
If S = 16, Area of triangle = $\frac{\sqrt{3}}{4} \times 16 \times 16 = 64 \sqrt{3} {cm}^{2}$

### Q2

If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area?
A. 120 ${cm}^{2}$ B. 130 ${cm}^{2}$ C. 312 ${cm}^{2}$ D. 315 ${cm}^{2}$

A
The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle = $\frac{1}{2} \times 24 \times 10 = 120 {cm}^{2}$

### Q3

Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
A. 225 ${cm}^{2}$ B. 275 ${cm}^{2}$ C. 285 ${cm}^{2}$ D. 315 ${cm}^{2}$

C
Area of a trapezium = $\frac{1}{2} (sum of parallel sides) \times (perpendicular distance between them)$
= $\frac{1}{2} (20 + 18) \times (15) = 285 {cm}^{2}$

### Q4

Find the area of a parallelogram with base 24 cm and height 16 cm.
A. 262 ${cm}^{2}$ B. 384 ${cm}^{2}$ C. 192 ${cm}^{2}$ D. 131 ${cm}^{2}$

B
Area of a parallelogram = base $\times$ height = 24 $\times$ 16 = 384 ${cm}^{2}$

### Q5

The perimeter of a triangle is 28 cm and the in radius of the triangle is 2.5 cm. What is the area of the triangle?
A. 25 ${cm}^{2}$ B. 42 ${cm}^{2}$ C. 49 ${cm}^{2}$ D. 35 ${cm}^{2}$

B
Area of a triangle = r * s
Where r is the in radius and s is the semi perimeter of the triangle.
Area of triangle = $2.5 \times \frac{28}{2} = 35 {cm}^{2}$