Mathematics Questions and Answers – Surface Area & Volumes – Frustum of a Cone

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This set of Mathematics Problems for Class 10 focuses on “Surface Area & Volumes – Frustum of a Cone”.

1. The frustum of a cone has its circular ends radius as 15 cm and 30 cm where the height is 10cm. Find the volume of the frustum.
a) 21544.4 cm2
b) 16485 cm2
c) 16570 cm2
d) 16572.5 cm2
View Answer

Answer: b
Explanation: r = 15 cm, R = 30 cm, h = 10 cm
The volume of a Frustum = \(\frac {1}{3}\)πh(r2 + rR + r2)
= \(\frac {1}{3}\) × 3.14 × 10 (225 + 450 + 900)
= 16485 cm3
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2. Find the slant height if the frustum of height 10cm and having its diameter as 20 cm and 40 cm.
a) 10.4 cm
b) 14.14 cm
c) 15.50 cm
d) 89.4 cm
View Answer

Answer: b
Explanation: h = 10 cm, R = 20 cm, r = 10 cm
The slant height of a cone s2 = h2 + (R – r)2
s2 = √(100 + (20 – 10)2)
s = 14.14 cm

3. Find the Lateral surface area of the frustum, if the slant height is 15cm and having radii as 14 cm and 21 cm.
a) 1648.5 cm2
b) 1115 cm2
c) 2105.4 cm2
d) 1237.2 cm2
View Answer

Answer: a
Explanation: s = 15 cm, R = 42 cm, r = 28 cm
Lateral Surface area of Frustum (L) = π(R + r)s
L = 3.14(21 + 14 )15
= 1648.5 cm2
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4. Find the total surface area of a frustum whose slant height is 10 cm and having radii as 10 cm and 20 cm.
a) 1923.21 m3
b) 2512 cm2
c) 2184.21 m3
d) 2842.21 m3
View Answer

Answer: b
Explanation: s = 10 cm, R = 20 cm, r = 10 cm
Total surface area of a cone = π(R + r)s + πR2 + πr2
= 3.14(20 + 10)10 + (3.14 × 202) + (3.14 × 102)
= 942 + 1256 + 314
= 2512 cm2

5. Find the difference between the radii of a Frustum, if the slant height and height are 10 cm and 8 cm.
a) 4 cm
b) 1 cm
c) 2 cm
d) 6 cm
View Answer

Answer: d
Explanation: The slant height of a cone = h2 + (R – r)2
102 = 82 + (R – r)2
(R – r)2 = 100 – 64
(R – r) = 6 cm
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6. The frustum of a cone has its circular ends radius as 5 cm and 15 cm where the height is given as 8 cm. Find the volume of the frustum.
a) 2152.76 cm3
b) 1254 cm3
c) 2721.33 cm3
d) 1421.76 cm3
View Answer

Answer: c
Explanation: r = 5 cm, R = 15 cm and h = 8 cm
The volume of a Frustum = \(\frac {1}{3}\) πh(r2 + rR + r2)
= \(\frac {1}{3}\) × 3.14 × 8(25 + 75 + 225)
= 2721.33 cm3

7. Find the Lateral surface area of the frustum, if the slant height is 12 cm and having radii as 15 cm and 35 cm.
a) 1984 cm2
b) 1804 cm2
c) 1984 cm2
d) 1884 cm2
View Answer

Answer: d
Explanation: s = 12 cm, R = 35 cm, r = 15 cm
Lateral Surface area of frustum = π(r + R)s
L = 3.14 × (15 + 35) × 12
= 1884 cm2
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8. The difference between the radii is 12 cm and slant height is given as 15 cm. Find the height of the frustum.
a) 5.4 cm
b) 1.2 cm
c) 3 cm
d) 9 cm
View Answer

Answer: d
Explanation: Given R – r = 12 cm and slant height = 15 cm
The slant height of a cone s2 = h2 + (R – r)2
225 = h2 + 144
h2 = 225 – 144
h2 = 81
h = 9 cm

9. Find the total surface area of a frustum whose slant height is 18 cm and having radii as 8 cm and 16 cm.
a) 2361.28 cm2
b) 2205.6 cm2
c) 1628.23 cm2
d) 2604 cm2
View Answer

Answer: a
Explanation: s = 18cm, R = 16 cm, r = 8 cm
Total surface area of a cone = π(R + r)s + πR2 + πr2
= 3.14(16 + 8)18 + (3.14 × 256) + (3.14 × 64)
= 1356.48 + 803.84 + 200.96
= 2361.28 cm2
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10. Find the slant height if the Frustum of height 22 cm has its diameter as 34 and 50cm.
a) 83.40 cm
b) 43.90 cm
c) 23.40 cm
d) 73.48 cm
View Answer

Answer: c
Explanation: h = 22 cm, R = 25 cm, r = 17 cm
The slant height of a cone (s2) = h2 + (R – r)2
s2 = 484 + (25 – 17)2
s2 = 548
s = 23.40 cm

Sanfoundry Global Education & Learning Series – Mathematics – Class 10.

To practice Mathematics Problems for Class 10, here is complete set of 1000+ Multiple Choice Questions and Answers.

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