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Mensuration questions are a common type of questions regularly asked in competitive exams. These questions carry a weightage of 1-2 questions(2-4 marks) in SSC exams and 1-2 questions(1-2 marks) bank exams. To perform well in competitive exams, your mensuration concepts should be clear.

Here are some tips for solving Mensuration questions: Familiarize yourself with formulas for various 2D and 3D shapes, Visualize shapes to better understand relationships between dimensions and Work on a range of problems involving different shapes to enhance skills.

So, we have attached 10 questions of Mensuration for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Mensuration</h2>

You can also download the Mensuration questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 There is a semi-circular park of a radius of 70m in the city. The mayor decided to fence the border of the park. The cost of fencing is Rs.50/m then what is the total cost in dollars? (1$ = Rs.80)

1. 137.5

2. 225

3. 150

4. 200

(Difficulty: 2, Estimated Time: 15 Seconds) This was an easy one. Did get it right?

Q:2 The circular path has an inner diameter and the outer diameter 330 m and 358 m respectively. Find the area of the circular path.

1. 644 π m2

2. 688 π m2

3. 988 π m2

4. 4816 π m2

(Difficulty: 3, Estimated Time: 20 Seconds) This was a simple one, don&#39;t get stuck in unnecessary calculations! 

Q:3 Find the number of revolutions of the wheel of a bicycle to cover a distance of 22 km if the radius of the wheel is 50 cm.

1. 3500 

2. 7000

3. 2200

4. 6500

(Difficulty: 2, Estimated Time: 15 Seconds) A question of seconds.....

Q:4 The length and breadth of a rectangular sheet are 24 m and 15 m respectively. 4 triangles are cut from all 4 corners with base 5 m and height 6 m. Find the area of remaining part of rectangle.

1. 360 m2

2. 250 m2

3. 280 m2

4. 300 m2

(Difficulty: 3, Estimated Time: 20 Seconds) It is not an easy one but I think now you&#39;re prepared for it.

Q:5 The area of the rectangle is four times the area of the square. If the length of the rectangle is 60 cm and the breadth of the rectangle is equal to the side of the square, then find the side of the square.

1. 20 cm

2. 25 cm

3. 18 cm

4. 15 cm

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 Area of the circle of diameter 28 cm is equal to the perimeter of the rectangle in magnitude whose length is 8 cm more than its breadth. Find the area of the rectangle.

1. 27300 cm2

2. 234500 cm2

3. 23700 cm2

4. 21500 cm2 

(Difficulty: 2, Estimated Time: 15 Seconds) Should we raise the level of questions?

Q:7 The cost of fencing a square lawn at Rs. 20 per metre is Rs. 1200. Find the area of the square lawn.

1. 225 m2

2. 400 m2

3. 3600 m2

4. 900 m2

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks to save time..

Q:8 A triangle is cut from inside a circle in such a way that one of its bases coincides with the diameter of length 14 cm and another base of the triangle makes 45º with the diameter of the circle. Find the remaining area of the circle.  

1. 616 cm2

2. 580 cm2

3. 105 cm2

4. 154 cm2

(Difficulty: 3, Estimated Time: 20 Seconds) This was a bit calculative one...

Q:9 A prism of square base with its sides of 12cm. Find the volume of prism if its height is 10.5 cm?

1. 1512 cm3

2. 1528 cm3

3. 3024 cm3

4. 3056 cm3

(Difficulty: 3, Estimated Time: 20 Seconds) Try decreasing time in your calculations

Q:10 In the figure given below, ABCD is a rhombus. If the perimeter of the triangle AOB is 12 cm and the product of the lengths and the diagonals of the rhombus is 48 cm2. What is the perimeter (in cm) of the rhombus?

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/SSC CGL Tier 2 FT3 Quant_04.05.2020-D11.png" />

1. 16

2. 20

3. 24

4. 28

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (2)</td>
			<td>2. (4)</td>
			<td>3. (2)</td>
			<td>4. (4)</td>
			<td>5. (4)</td>
		</tr>
		<tr>
			<td>6. (3)</td>
			<td>7. (1)</td>
			<td>8. (3)</td>
			<td>9. (1)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is Option 2 i.e. 225.

Perimeter of the semi-circular park = (2πr)/2 + 2r

⇒ πr + 2r = r(π + 2)

⇒ 70 × (22/7 + 2)

⇒ 70 × (36/7) = 360m

Cost of fencing = 360 × 50

⇒ Rs.18000

Converting Indian currency to $

⇒ 18000/80

⇒ 225$

 

Q:2 The correct answer is option 4 i.e. 4816 π m2.

Area of the circular path = π(R + r)(R - r)

Where R is the outer radius and r is the inner radius

The inner radius of a circular path r = 330/2 = 165 m

The outer radius of a circular path R = 358/2 = 179 m 

The area of the circular path =  π (179 + 165)(179 - 165) m2 = 4816 π m2

Trick: Since the numbers are very large and we know that (14 x three-digit number) is always greater than 1000. So, just by seeing the option, we can conclude that the answer should be option 4

 

Q:3 The correct answer is Option 2 i.e. 7000.

The number of revolution = total distance/circumference of the circle

= Distance/2πr

The radius of the wheel = 50 cm = 1/2 metre

Number of revolution = (22000)/{2 × (22/7) × 1/2} = 7000

 

Q:4 The correct answer is Option 4 i.e. 300 m2.

Given,

Length of rectangle = 24 m

Breadth of rectangle = 15 m

Area of rectangle = Length × Breadth = 24 × 15 = 360 m2

Area of 1 triangle = (1/2) × base × height = (1/2) × 5 × 6 = 15 m2

Also,Base of triangle = 5m and, Height of triangle = 6 m

Since the base and height of all triangles are the same so, the area of all 4 triangles is equal

Area of 4 triangles = 15 × 4 = 60 m2

Therefore, the Area of the remaining part = 360 - 60 = 300 m2

 

Q:5 The correct answer is Option 4 i.e. 15 cm.

Suppose

The breadth of the rectangle = Side of the square = a cm

According to the question:

Length × Breadth = 4(Side)2

⇒ 60 × a = 4a2

⇒ a = 15 cm

Hence, side of the square = 15 cm

 

Q:6 The correct answer is option 3 i.e. 23700 cm2.

Area of Circle = π × r2 , Where r = Radius of the circle = 14 cm

Area = π × 14 × 14 = 22/7 × 14 × 14 = 616 cm2

But this is equal to the perimeter of the rectangle in magnitude

Hence,

Perimeter of the rectangle = 2 × (L + B) = 616 cm , Where L = Length, B = Breadth

And L = (B + 8)

So,

2 × (L + B) = 616

⇒ B + 8 + B = 308

⇒ 2B = 300 = B = 150 and L = 158

Hence,

Area = L × B = 150 × 158 = 23,700 cm2

 

Q:7 The correct answer is option 1 i.e. 225 m2

The cost of fencing

= Cost of fencing per meter × perimeter of the square.

Cost of fencing per metre = Rs 20 per metre

Total cost = Rs 1200

The perimeter of the square = 1200/20 = 60

4 × side = 60

Side = 15

Area of the square lawn = (15)2 = 225 m2

 

 

Q:8 The correct answer is option 3 i.e. 105 cm2.

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_13.07.2021_D2.png" />

Area of circle = πr2

Area of triangle = (1/2)×base×height

Diameter = 14 cm

Radius = 7 cm

∠ABC = 45º

∠OCB = 45

∠BOC = 90

So, triangle OBC is a right isosceles triangle.

OC is perpendicular to AB and OC = OB

Area of triangle = (1/2) × 14 × 7 = 49 cm2

Area of circle = (22/7) × 7 × 7 = 154 cm2

Area of remaining part = 154 - 49 = 105 cm2

 

Q:9 The correct answer is Option 1 i.e. 1512 cm3.

Area of base of prism = base area ×  height

Area of square = side2 = 122 = 144 cm2

Height = 10.5 cm = 21/2 cm

Volume = 144 × 21/2 = 1512 cm3

 

Q:10 The correct answer is Option 2 i.e. 20.

Let the lengths of AC and BD be 2x and 2y respectively

AO = AC/2 = x

BO = BD/2 = y [∵ Diagonals bisect each other at right angles]

AB = √(x2 + y2) [∵ Pythagoras theorem]

Perimeter of AOB = AB + AO + BO

⇒ 12 = √(x2 + y2) + x + y

Product of diagonals = 2x × 2y = 48

⇒ xy = 12

⇒ 12 = √{(x + y)2 – 2xy} + x + y

Let x + y = t

⇒ 12 = √(t2 – 24) + t

⇒ 12 – t = √(t2 – 24)

Squaring on both sides

⇒ 144 + t2 – 24t = t2 – 24

⇒ 168 = 24t

⇒ t = 7

⇒ x + y = 7

⇒ xy = 12

⇒ x (7 – x) = 12

⇒ 7x – x2 = 12

⇒ x2 – 7x + 12 = 0

⇒ (x – 4) (x – 3) = 0

⇒ x = 4 or 3

⇒ y = 3 or 4

∴ AB = √(x2 + y2) = 5 cm

Perimeter = 4 × Side = 4 × 5 = 20 cm

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Mensuration.

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Practice Questions on Mensuration!

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