Mensuration MCQ Quiz - Objective Question with Answer for Mensuration - Download Free PDF

Last updated on May 4, 2025

Mensuration MCQs Quiz for high school students, college students, and other candidates who wish to appear for competitive exams, interview or entrance exams. Testbook provides the complete set Mensuration Question Answers along with detailed solutions, tricks and shortcuts so that candidates can practice mensuration problems with ease. Candidates' final aim should be to solve the questions with accuracy. Start your practise today.

Latest Mensuration MCQ Objective Questions

Mensuration Question 1:

Perimeter of square is 64 m. Length of rectangle is 4 m more than the side of square. Breadth of rectangle is 12 m. Find the area of rectangle?

  1. 240
  2. 220
  3. 260
  4. 200
  5. 310

Answer (Detailed Solution Below)

Option 1 : 240

Mensuration Question 1 Detailed Solution

Calculation

First, side of square:

Side = 64/4 = 16 m

Length of rectangle = 16 + 4 = 20 

Area of rectangle = Length × Breadth = 20 × 12 = 240

Mensuration Question 2:

The volumes of two cones are in the ratio of 3 ∶ 2 and their radii are in the ratio 3 ∶ 4. The ratio of their heights is:

  1. 8 ∶ 3
  2. 9 ∶ 4
  3. 4 ∶ 9
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 8 ∶ 3

Mensuration Question 2 Detailed Solution

Given:

Ratio of volumes of two cones = 3 : 2

Ratio of radii of the two cones = 3 : 4

Formula:

The volume of a cone is given by the formula:

V = (1/3)πr2h

Solution:

Let's assume the radii of the two cones are 3x and 4x, where x is a common factor.

So, the ratio of their radii is 3x : 4x.

The volume of the first cone (V1) can be expressed as:

V1 = (1/3)π(3x)2h1

V1 = (1/3)π9x2h1

The volume of the second cone (V2) can be expressed as:

V2 = (1/3)π(4x)2h1

V2 = (1/3)π16x2h1

Given that the ratio of volumes of the two cones is 3 : 2, we have:
V1/V2 = 3/2

9x​2h1/16x​2h2 = 3/2

h1 / h2 = 48x2/18x​2

h1/h2 = 8/3

Therefore, the ratio of the heights of the two cones is 8 : 3.

Mensuration Question 3:

If a regular polygon has 10 sides, then the measure of its interior angle is greater than the measure of its exterior angle by how many degrees?

  1. 120
  2. 132
  3. 108
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 108

Mensuration Question 3 Detailed Solution

Concept used:

If the number of sides of a regular polygon be n, then

The interior angle = \( \frac{(n - 2) \times 180^\circ }{n}\)

Calculation:

Let the required difference be x.

No. of sides of a regular polygon = 10

According to the question

\(\Rightarrow \frac{{\left( {n - 2} \right) \times 180^\circ }}{n} - \frac{{360^\circ }}{n} = x\)

\(\Rightarrow \frac{{\left( {10 - 2} \right) \times 180^\circ }}{{10}} - \frac{{360^\circ }}{{10}} = x\)

\(\Rightarrow \frac{{8 \times 180^\circ }}{{10}} - 36^\circ = x\)

⇒ 144 - 36° = x

∴ x = 108°

Mensuration Question 4:

What is the radius (in m) of a circular field whose area is equal to six times the area of a triangular field whose sides are 35 m, 53 m and 66 m?. (Take π = 22/7)

  1. 42° 
  2. 14√3
  3. 14√6
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 42° 

Mensuration Question 4 Detailed Solution

Given:

Area of circular field = 6 × area of triangular field

Formula used:

Area of circular field = πr2, where r is a radius of circle.

Area of triangle = √s (s - a) (s - b) (s - c),

where s = (a + b + c)/2

a, b and c are sides of triangle respectively.

Calculation:

a = 35 m, b = 53 m and c = 66 m

s = (35 + 53 + 66)/2

⇒ s = 77

Area of triangle = √s(s - a) (s - b) (s - c)

⇒ √77 (77 - 35) (77 - 53) (77 - 66)

⇒ √77 × 42 × 24 × 11

⇒ √7 × 11 × 2 × 3 × 7 × 2 × 3 × 2 × 2 × 11

⇒ 11 × 7 × 2 × 2 × 3

⇒ 924 m2

Now, 6 × Area of triangle = Area of circle

⇒ 924 × 6 = πr2

⇒ (924 × 6 × 7)/22 = r2

⇒ 1764 = r2

⇒ 42 = r

∴ Radius of circle is 42 m.

Mensuration Question 5:

A bycle wheel makes 5000 revolutions in moving 11 km. How much cms will be the diameter of the wheel?

  1. 70
  2. 80
  3. 50
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 70

Mensuration Question 5 Detailed Solution

Given:

Number of revolutions = 5000

Distance covered = 11 km = 1100000 cm

Formula used:

Circumference of circle = 2πr

Calculation:

Circumference of circle = 1100000/5000

⇒ 220 cm

According to the question,

⇒ 2πr = 220 cm

⇒ 2 × (22/7) × r = 220 cm

⇒ r/7 = 5 cm

⇒ r = 35 cm

Diameter = 2r

⇒ 2 × 35

⇒ 70 cm

∴ The diameter of circle is 70 cm

Top Mensuration MCQ Objective Questions

Six chords of equal lengths are drawn inside a semicircle of diameter 14√2 cm. Find the area of the shaded region?

F4 Aashish S 21-12-2020 Swati D7

  1. 7
  2. 5
  3. 9
  4. 8

Answer (Detailed Solution Below)

Option 1 : 7

Mensuration Question 6 Detailed Solution

Download Solution PDF

Given:

Diameter of semicircle = 14√2 cm

Radius = 14√2/2 = 7√2 cm

Total no. of chords = 6

Concept:

Since the chords are equal in length, they will subtend equal angles at the centre. Calculate the area of one sector and subtract the area of the isosceles triangle formed by a chord and radius, then multiply the result by 6 to get the desired result.

Formula used:

Area of sector = (θ/360°) × πr2

Area of triangle = 1/2 × a × b × Sin θ

Calculation:

F4 Aashish S 21-12-2020 Swati D8

The angle subtended by each chord = 180°/no. of chord

⇒ 180°/6

⇒ 30°

Area of sector AOB = (30°/360°) × (22/7) × 7√2 × 7√2

⇒ (1/12) × 22 × 7 × 2

⇒ (77/3) cm2

Area of triangle AOB = 1/2 × a × b × Sin θ

⇒ 1/2 × 7√2 × 7√2 × Sin 30°

⇒ 1/2 × 7√2 × 7√2 × 1/2

⇒ 49/2 cm2

∴ Area of shaded region = 6 × (Area of sector AOB - Area of triangle AOB)

⇒ 6 × [(77/3) – (49/2)]

⇒ 6 × [(154 – 147)/6]

⇒ 7 cm2

Area of shaded region is 7 cm2

There is a rectangular garden of 220 metres × 70 metres. A path of width 4 metres is built around the garden. What is the area of the path?

  1. 2472 metre2
  2. 2162 metre2
  3. 1836 metre2
  4. 2384 metre2

Answer (Detailed Solution Below)

Option 4 : 2384 metre2

Mensuration Question 7 Detailed Solution

Download Solution PDF

Formula used

Area = length × breath

Calculation

8-July-2012 Morning 1 1 Hindi Images Q7

The garden EFGH is shown in the figure. Where EF = 220 meters & EH = 70 meters.

The width of the path is 4 meters.

Now the area of the path leaving the four colored corners

= [2 × (220 × 4)] + [2 × (70 × 4)]

= (1760 + 560) square meter

= 2320 square meters

Now, the area of 4 square colored corners:

4 × (4 × 4)

{∵ Side of each square = 4 meter}

= 64 square meter

The total area of the path = the area of the path leaving the four colored corners + square colored corners

⇒ Total area of the path = 2320 + 64 = 2384 square meter

∴ Option 4 is the correct answer.

The width of the path around a square field is 4.5 m and its area is 105.75 m2. Find the cost of fencing the field at the rate of Rs. 100 per meter.

  1. Rs. 275
  2. Rs. 550
  3. Rs. 600
  4. Rs. 400

Answer (Detailed Solution Below)

Option 2 : Rs. 550

Mensuration Question 8 Detailed Solution

Download Solution PDF

Given:

The width of the path around a square field = 4.5 m

The area of the path = 105.75 m2

Formula used:

The perimeter of a square = 4 × Side

The area of a square = (Side)2

Calculation:

F2 SSC Pranali 13-6-22 Vikash kumar D6

Let, each side of the field = x

Then, each side with the path = x + 4.5 + 4.5 = x + 9

So, (x + 9)2 - x2 = 105.75

⇒ x2 + 18x + 81 - x2 = 105.75

⇒ 18x + 81 = 105.75

⇒ 18x = 105.75 - 81 = 24.75

⇒ x = 24.75/18 = 11/8

∴ Each side of the square field = 11/8 m

The perimterer = 4 × (11/8) = 11/2 m

So, the total cost of fencing = (11/2) × 100 = Rs. 550

∴ The cost of fencing of the field is Rs. 550

Shortcut TrickIn such types of questions, 

Area of path outside the Square is, 

⇒ (2a + 2w)2w = 105.75

here, a is a side of a square and w is width of a square

⇒ (2a + 9)9 = 105.75

⇒ 2a + 9 = 11.75

⇒ 2a = 2.75

Perimeter of a square = 4a

⇒ 2 × 2a = 2 × 2.75 = 5.50

costing of fencing = 5.50 × 100 = 550

∴ The cost of fencing of the field is Rs. 550

If the side of an equilateral triangle is increased by 34%, then by what percentage will its area increase?

  1. 70.65%
  2. 79.56%
  3. 68.25%
  4. 75.15%

Answer (Detailed Solution Below)

Option 2 : 79.56%

Mensuration Question 9 Detailed Solution

Download Solution PDF

Given:

The sides of an equilateral triangle are increased by 34%.

Formula used:

Effective increment % = Inc.% + Inc.% + (Inc.2/100) 

Calculation:

Effective increment = 34 + 34 + {(34 × 34)/100}

⇒ 68 + 11.56 = 79.56%

∴ The correct answer is 79.56%.

The length of an arc of a circle is 4.5π cm and the area of the sector circumscribed by it is 27π cm2. What will be the diameter (in cm) of the circle?

  1. 12
  2. 24
  3. 9
  4. 18

Answer (Detailed Solution Below)

Option 2 : 24

Mensuration Question 10 Detailed Solution

Download Solution PDF

Given : 

Length of an arc of a circle is 4.5π.

Area of ​​the sector circumscribed by it is 27π cm2.

Formula Used : 

Area of sector = θ/360 × πr2

Length of arc = θ/360 × 2πr

Calculation : 

F1 Railways Savita 31-5-24 D1

According to question,

⇒ 4.5π = θ/360 × 2πr 

⇒ 4.5 = θ/360 × 2r   -----------------(1)

⇒ 27π = θ/360 × πr2 

⇒ 27 = θ/360 × r2       ---------------(2)

Doing equation (1) ÷ (2)

⇒ 4.5/27 = 2r/πr2

⇒ 4.5/27 = 2/r

⇒ r = (27 × 2)/4.5

⇒ Diameter = 2r = 24

∴ The correct answer is 24.

A wire is bent to form a square of side 22 cm. If the wire is rebent to form a circle, then its radius will be: 

  1. 22 cm
  2. 14 cm
  3. 11 cm
  4. 7 cm

Answer (Detailed Solution Below)

Option 2 : 14 cm

Mensuration Question 11 Detailed Solution

Download Solution PDF

Given:

The side of the square = 22 cm

Formula used:

The perimeter of the square = 4 × a    (Where a = Side of the square)

The circumference of the circle = 2 × π × r     (Where r = The radius of the circle)

Calculation:

Let us assume the radius of the circle be r

⇒ The perimeter of the square = 4 × 22 = 88 cm

⇒ The circumference of the circle = 2 × π ×  r

⇒ 88 = 2 × (22/7) × r

⇒ \(r = {{88\ \times\ 7 }\over {22\ \times \ 2}}\)

⇒ r = 14 cm

∴ The required result will be 14 cm.

A solid hemisphere has radius 21 cm. It is melted to form a cylinder such that the ratio of its curved surface area to total surface area is 2 ∶ 5. What is the radius (in cm) of its base (take  π = \(\frac{{22}}{7}\))?

  1. 23
  2. 21
  3. 17
  4. 19

Answer (Detailed Solution Below)

Option 2 : 21

Mensuration Question 12 Detailed Solution

Download Solution PDF

Given:

The radius of a solid hemisphere is 21 cm.

The ratio of the cylinder's curved surface area to its Total surface area is 2/5.

Formula used:

The curved surface area of the cylinder = 2πRh

The total surface area of cylinder = 2πR(R + h)

The volume of the cylinder = πR2h

The volume of the solid hemisphere = 2/3πr³ 

(where r is the radius of a solid hemisphere and R is the radius of a cylinder)

Calculations:

According to the question,

CSA/TSA = 2/5

⇒ [2πRh]/[2πR(R + h)] = 2/5

⇒ h/(R + h) = 2/5

⇒ 5h = 2R + 2h

⇒ h = (2/3)R .......(1)

The cylinder's volume and the volume of a solid hemisphere are equal.

⇒ πR2h = (2/3)πr3

⇒ R2 × (2/3)R = (2/3) × (21)3

⇒ R3 = (21)3

⇒ R = 21 cm

∴ The radius (in cm) of its base is 21 cm.

The surface area of three faces of a cuboid sharing a vertex are 20 m2, 32 m2 and 40 m2. What is the volume of the cuboid?

  1. 92 m3
  2. √3024 m3
  3. 160 m3
  4. 184 m3

Answer (Detailed Solution Below)

Option 3 : 160 m3

Mensuration Question 13 Detailed Solution

Download Solution PDF

The surface area of three faces of a cuboid sharing a vertex are 20 m2, 32 m2 and 40 m2,

⇒ L × B = 20 sq. Mt

⇒ B × H = 32 sq. Mt

⇒ L × H = 40 sq. Mt

⇒ L × B × B × H × L × H = 20 × 32 × 40

⇒ L2B2H2 = 25600

⇒ LBH = 160

∴ Volume = LBH = 160 m3

A solid cube of side 8 cm is dropped into a rectangular container of length 16 cm, breadth 8 cm and height 15 cm which is partly filled with water. If the cube is completely submerged, then the rise of water level (in cm) is:

  1. 6
  2. 4
  3. 2
  4. 5

Answer (Detailed Solution Below)

Option 2 : 4

Mensuration Question 14 Detailed Solution

Download Solution PDF

Given:

Each side of the cube = 8 cm

The rectangular container has a length of 16 cm, breadth of 8 cm, and height of 15 cm

Formula used:

The volume of cube = (Edge)3

The volume of a cuboid = Length × Breadth × Height

Calculation:

The volume of cube = The volume of the rectangular container with length of 16 cm, breadth of 8 cm, and height of the water level rise

Let, the height of the water level will rise = x cm

So, 83 = 16 × 8 × x

⇒ 512 = 128 × x

⇒ x = 512/128 = 4

∴ The rise of water level (in cm) is 4 cm

The sum of length, breadth and height of a cuboid is 21 cm and the length of its diagonal is 13 cm. Then the total surface area of the cuboid is 

  1. 272 cm2
  2. 240 cm2
  3. 314 cm2
  4. 366 cm2

Answer (Detailed Solution Below)

Option 1 : 272 cm2

Mensuration Question 15 Detailed Solution

Download Solution PDF

Given:

Sum of length,, breadth and height of a cuboid = 21 cm

Length of the diagonal(d) = 13 cm

Formula used:

d2 = l2 + b2 + h2

T.S.A of cuboid = 2(lb + hb +lh)

Calculation:

⇒ l2 + b2 + h2 = 132 = 169

According to question,

⇒ (l + b + h)2 = 441

⇒ l2 + b2 + h2 + 2(lb + hb +lh) = 441

⇒ 2(lb + hb +lh) = 441 - 169 = 272

∴ The answer is 272 cm2 .

Get Free Access Now
Hot Links: teen patti master 2025 teen patti cash teen patti real cash game teen patti wink teen patti master 2024