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Quantitative Aptitude practice

Mensuration Questions with Answers

Mensuration is the measurement of lengths, areas and volumes of flat and solid shapes. Mensuration questions in SSC CGL, CHSL and MTS, RRB NTPC and Group D, and bank prelims cover rectangles, triangles, circles, cubes, cuboids, cylinders, cones and spheres. Typical tasks are finding area, surface area or volume, or reshaping one solid into another. Tip: write the formula first, check that all measurements are in the same unit, and only then substitute values.

Formulas you need

Rectangle: Area = l × b; Perimeter = 2(l + b)
Triangle: Area = ½ × base × height
Circle: Area = πr²; Circumference = 2πr (π = 22/7 when radius is a multiple of 7)
Cube: Volume = a³; Total surface area = 6a²; Diagonal = a√3
Cuboid: Volume = l × b × h; Total surface area = 2(lb + bh + hl)
Cylinder: Volume = πr²h; Curved surface = 2πrh; Total surface = 2πr(r + h)
Cone: Volume = ⅓πr²h; Slant height l = √(r² + h²); Curved surface = πrl; Total surface = πr(r + l)
Sphere: Volume = (4/3)πr³; Surface area = 4πr²
When a solid is melted and recast, volume stays the same

Worked examples

Q1. Find the area of a circular field of radius 14 m. (Use π = 22/7)

Area = πr²
= 22/7 × 14 × 14
= 22 × 2 × 14 = 616

Answer: 616 m²

Q2. The total surface area of a cube is 384 cm². Find its volume.

6a² = 384, so a² = 64
a = 8 cm
Volume = a³ = 8 × 8 × 8 = 512

Answer: 512 cm³

Q3. A conical vessel of radius 12 cm and height 15 cm is full of water. The water is poured into a cylindrical vessel of radius 6 cm. Find the height of water in the cylinder.

Volume of cone = ⅓ × π × 12² × 15 = 720π cm³
Volume of water in cylinder = π × 6² × h = 36πh
Equate: 36πh = 720π
h = 720 ÷ 36 = 20

Answer: 20 cm

Shortcuts that save time

Cancel π and common factors before multiplying, as in the cone and cylinder example. Smaller numbers mean fewer arithmetic slips.
Memorise the common triples (3-4-5, 5-12-13, 7-24-25, 8-15-17). They give slant heights and diagonals instantly, with no square roots to get wrong.
Convert units first. A mix of cm and m is the most common trap, and it produces an option that looks right but is off by a factor of 100.

Where this appears

SSC CGL Tier-1	Quantitative Aptitude section
SSC CHSL Tier-1	Quantitative Aptitude section
SSC MTS	Numerical and Mathematical Ability section
RRB NTPC CBT-1	Mathematics section
Bank PO/Clerk Prelims	Quantitative Aptitude section

Mensuration

Practice Set

Attempted: 0Correct: 0Incorrect: 0

1 of 1 (adapting live)Medium

Find the area of a triangle with base 8 m and height 6 m.

Quantitative Aptitude · stress-free practice · answers reveal instantly · difficulty adapts as you go

Mensuration questions with answers

10 sample questions from this topic with the correct option and the solution.

Q1. The curved surface area of a right circular cylinder is 1056 cm² and its height is 16 cm. Find the total surface area of the cylinder.
1. 1974 cm²
2. 1749 cm²
3. 1794 cm²
4. 1479 cm²
Answer: (B) 1749 cm²
Solution
Curved surface area $= 2\pi rh = 1056$, with $h = 16$ cm.
So $2\pi r = \frac{1056}{16} = 66$, which gives $r = \frac{66 \times 7}{2 \times 22} = 10.5$ cm.
Total surface area $= 2\pi r(h + r) = 66 \times (16 + 10.5) = 66 \times 26.5 = 1749$ cm².
Answer: 1749 cm² (option B).
Q2. Find the area of a circle with radius 7 m. (Use pi = 22/7)
1. 62
2. 131
3. 246
4. 154
Answer: (D) 154
Solution
Area = pi x r² = (22/7) x 7² = 154.0 sq. m.
Q3. The ratio of the radii of two right circular cylinders is 2: 3. If their heights are in the ratio 5: 3, find the ratio of their volumes.
1. 16: 19
2. 3: 4
3. 3: 1
4. 20: 27
Answer: (D) 20: 27
Solution
The volume of a cylinder is $V=\pi r^2 h$.
So $V_1:V_2 = r_1^2 h_1 : r_2^2 h_2$.
Substituting the ratios: $2^2\times5 : 3^2\times3 = 20:27$.
Answer: 20 : 27.
Q4. The curved surface area of a right circular cone of radius 28 cm is 4664 cm². What is the slant height of the cone? (use )
1. 53 cm
2. 56 cm
3. 64 cm
4. 48 cm
Answer: (A) 53 cm
Solution
Curved surface area of a cone $= \pi r l$.
Taking $\pi = \frac{22}{7}$: $\frac{22}{7} \times 28 \times l = 4664$.
This gives $88\,l = 4664$.
So $l = \frac{4664}{88} = 53$ cm.
The slant height is 53 cm (option A).
Q5. Find the area of a rectangle with length 15 m and breadth 7 m.
1. 153
2. 41
3. 105
4. 169
Answer: (C) 105
Solution
Area = length x breadth = 15 x 7 = 105 sq. m.
Q6. The length of one of the two parallel sides of a trapezium is 8 cm, while the length of the other side of this pair of parallel sides is equal to the distance between this pair of parallel sides. If the area of the trapezium is 120 cm², then what is the distance (in cm) between the said pair of parallel sides?
1. 20
2. 10
3. 15
4. 12
Answer: (D) 12
Solution
Let the other parallel side = height = $h$.
Area $= \frac{1}{2}(8+h)h = 120 \Rightarrow h^2 + 8h - 240 = 0$.
$(h+20)(h-12)=0 \Rightarrow h = 12$ (the negative root is rejected).
So the distance between the parallel sides is 12 cm.
Q7. The curved surface area of a right circular cone is 4335π cm², and the diameter of its base is 102 cm. Find the height (in cm) of the cone.
1. 68
2. 64
3. 63
4. 69
Answer: (A) 68
Solution
Radius $r = 102/2 = 51$ cm.
Curved surface area $= \pi r l = 4335\pi \Rightarrow l = 4335/51 = 85$ cm.
Height $h = \sqrt{l^2 - r^2} = \sqrt{85^2 - 51^2} = \sqrt{7225 - 2601} = \sqrt{4624} = 68$ cm.
Answer: 68 cm.
Q8. Find the area of a square with side 5 m.
1. 41
2. 13
3. 25
4. 33
Answer: (C) 25
Solution
Area = side² = 5² = 25 sq. m.
Q9. The curved surface area of a right circular cone is 20π cm². If its radius is 4 cm, then find the volume of cone.
1. 18 π cm³
2. 21 π cm³
3. 20 π cm³
4. 16 π cm³
Answer: (D) 16 π cm³
Solution
Curved surface area $= \pi r l = 20\pi \Rightarrow 4l = 20 \Rightarrow l = 5$ cm.
Height $h = \sqrt{l^2 - r^2} = \sqrt{25-16} = 3$ cm.
Volume $= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 16 \times 3 = 16\pi$ cm³.
Answer: D.
Q10. The curved surface area of a right circular cone is 2684π cm², and the diameter of its base is 44 cm. Find the height (in cm) of the cone.
1. 123
2. 121
3. 124
4. 120
Answer: (D) 120
Solution
Radius $r = 44/2 = 22$ cm.
CSA $= \pi r l = 2684\pi \Rightarrow l = 2684/22 = 122$ cm.
Height $h = \sqrt{l^2 - r^2} = \sqrt{122^2 - 22^2} = \sqrt{14884 - 484} = \sqrt{14400} = 120$ cm.
Answer: 120 cm.