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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 Total surface area of cylinder is 704 cm2 and ratio of height and base radius of the cylinder is 3 : 4, then what will be the volume of that cylinder?

1. 384π cm3

2. 243π cm3

3. 518π cm3

4. 423π cm3

(Difficulty: 2, Estimated Time: 15 Seconds) This was an easy one. Did get it right?

Q:2 If the length of the side of the rhombus is 10 cm and the length of the shorter diagonal is 12 cm then what is the area of the rhombus?

1. 96 cm2

2. 108 cm2

3. 80 cm2

4. 144 cm2

(Difficulty: 3, Estimated Time: 20 Seconds) This was a simple one, don&#39;t get stuck in unnecessary calculations! 

Q:3 What is the volume of a hexagonal prism of base length 5 cm and height 4√3 cm?

1. 360 cm3

2. 420 cm3

3. 450 cm3

4. 480 cm3

(Difficulty: 2, Estimated Time: 15 Seconds) A question of seconds.....

Q:4 The area of the circular park is 5544 m2. There is a 7m wide path for running inside the park. Park owners decide to pave the running track. If the cost of paving Rs 12 per meter square then find the total cost of paving the running track of the park.

1. Rs 12500

2. Rs 24520

3. Rs 18540

4. Rs 20328

(Difficulty: 3, Estimated Time: 20 Seconds) It is not an easy one but I think now you&#39;re prepared for it.

Q:5 Amal has a toy that is in the shape of a regular tetrahedron and has a curved surface area of 800 cm2. Each face of the tetrahedron has a height of 20 cm. What is the length of the base of one of the faces?

1. 20 cm

2. 10 cm

3. 15 cm

4. 21 cm

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 Find the volume of a pyramid whose base is square in shape with an area of 225 sq.cm. and height is 11 cm.

1. 965 cu.cm

2. 741 cu.cm

3. 825 cu.cm

4. 786 cu.cm

(Difficulty: 2, Estimated Time: 15 Seconds) Should we raise the level of questions?

Q:7 Ratio of curved surface area to total surface area of cylinder is 11 : 18. Also, area of its circular end is 616 cm2. Find the volume of cylinder.

1. 11576 cm3

2. 13552 cm3

3. 12785 cm3

4. 16415 cm3

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks to save time..

Q:8 The length and breadth of a rectangle are in a ratio 2 : 1 and the perimeter of the rectangle is 48 cm. Find the area of the rectangle.

1. 128 cm2

2. 256 cm2

3. 64 cm2

4. 144 cm2

(Difficulty: 3, Estimated Time: 20 Seconds) This was a bit calculative one...

Q:9 The sides of the right-angle triangular park are in ratio 3 : 4. The sum of all sides is 144 m. Find the area of the triangular park.

1. 894 m2

2. 864 m2

3. 926 m2

4. 1024 m2

(Difficulty: 3, Estimated Time: 20 Seconds) Try decreasing time in your calculations

Q:10 What is the volume of the sphere whose radius is 14cm?

1. 22158.33

2. 11699.67

3. 11250.33

4. None of these

(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. (1)</td>
			<td>2. (1)</td>
			<td>3. (3)</td>
			<td>4. (4)</td>
			<td>5. (1)</td>
		</tr>
		<tr>
			<td>6. (3)</td>
			<td>7. (2)</td>
			<td>8. (1)</td>
			<td>9. (2)</td>
			<td>10. (4)</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 1 i.e. 384π cm3

Let height and base radius of the cylinder is 3x and 4x respectively.

Total surface area of the cylinder = 2πr(h + r)

704 = 2 × (22/7) × 4x × (3x + 4x)

⇒ 704 = 176x/7 × (3x + 4x)

⇒ 704 = 176x2

⇒ x2 = 4

⇒ x = 2

Base radius = 4x = 8 cm

Height = 3x = 6 cm

Volume of the cylinder = πr2h = π × 64 × 6 = 384π cm3 

 

Q:2 The correct answer is option 1 i.e. 96 cm2

<img loading="lazy" alt="" src="https://hranker.com/admin/assets/uploads/ckeditor/Urgent Diagram_08.04.2022_D1.png" />

By observing the given diagram and as we know that AOB is a right-angle triangle.

⇒ OB =12/2 = 6cm

⇒ AO = √(100 - 36) (Using Pythagoras theorem)

⇒ 8cm

⇒ AC = 16cm

⇒ Area of the diagonal = (1/2) × AC × BD

⇒ (1/2) × 16 × 12 = 96 cm2

 

Q:3 The correct answer is option 3 i.e. 450 cm3

Volume of prism = Area of base × Height

Area of hexagonal base = 3√3/2 × side2

⇒ 3√3/2 × 52 = 75√3/2

Volume of prism = 75√3/2 × 4√3

⇒ 450 cm3

 

Q:4 The correct answer is option 4 i.e. Rs 20328

 

Area of circular park = πr2

⇒ 5544 = πr2

⇒ r2 = 5544/π = 1764 (where π = 22/7)

⇒ r = √1764 = 42 m

Path length is 7 m (given)

Inner circle area (without including path area)

Radius of inner circle = (42 - 7) m = 35 m

Area2 = 22/7 × 35 × 35 = 3850 m2

Total area of path = Area1 - Area2

= (5544 - 3850) m2

= 1694 m2

Total cost for paving = 1694 × 12 

= Rs 20328

 

Q:5 The correct answer is option 1 i.e. 20 cm.

As we know that the regular tetrahedron has 4 triangular faces.

Area of one face = 800/4 = 200 cm2 

Area of triangle = 1/2 × base × height.

200 = 1/2 × base × 20

Base = 20 cm.

 

Q:6 The correct answer is option 3 i.e. 825cm3.

Given 

Area of square base = 225 cm2

h = 11 cm

The volume of pyramid = 1/3 × Area of square base × h = 1/3 × 225 × 11 = 825 cm3

 

Q:7 The correct answer is option 2 i.e. 13552 cm3​​​​​​.

The curved surface area of the cylinder = 2πrh

The total surface area of the cylinder = 2πr(r + h)

Ratio = 2πrh : 2πr(r + h) = h : (r + h)

Now, h : (r + h) = 11 : 18

⇒ 18h = 11r + 11h

⇒ 7h = 11r

⇒ h/r = 11/7

Also, πr2 = 616 

⇒ r2 = 616/π

⇒ r2 = 196 = 14 

⇒ h/r = 11/7 (if r is 14, then h must be 22)

Volume = π × r2 × h = (22/7) × 14 × 14 × 22 = 13552 cm3

 

Q:8 The correct answer is Option 1 i.e. 128 cm2

Area of rectangle = Length × Breadth

Perimeter of rectangle = 2(Length + Breadth)

Let, the length is 2x and the breadth is x (since the ratio is 2 : 1).

Now given,

⇒ 48 = 2(2x + x)

⇒ 48 = 6x

⇒ x = 8

So, Length = 16 cm and Breadth = 8 cm

Area = 16 × 8 = 128 cm2

 

Q:9 The correct answer is option2 i.e. 864 m2.

The park is of right angle triangle shape so, the triplet of sides is (3,4,5).

Let the side of the park be 3x, 4x, 5x

Sum of sides = 12x = 144

⇒ x = 12

Thus, sides are 36 m, 48 m, 60m

The area of triangle = (1/2) × 36 × 48 = 864 m2

 

Q:10 The correct answer is option 4 i.e. None of these

The volume of the sphere = (4/3)πr3

Where r is the radius of the sphere.

⇒ (4/3) × (22/7) × 14 × 14 × 14 

⇒ 11,498.66 cm3

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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