Practice Set 9.2 : Textbook Page 119

1. Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.

Solution:

Here, h = 12 cm, l = 13 cm, r = ?

Now,
l² = h² + r²
∴ 13² = 12² + r²
∴ 169 = 144 + r²
∴ 169 − 144 + r²
∴ 25 = r²
∴ \(\displaystyle \sqrt {25}\) = r
∴ 5 = r
i.e. r = 5 cm ... (i)

∴ The radius of the base of that cone is 5 cm.

2. Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm.

\(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

Solution:

Here, S_t = 7128 cm², r = 28 cm, V = ?

First, let’s find the slant height and perpendicular height of the cone.

Total surface area of a cone:
S_t = πr(l + r)

∴ 7128 = \(\displaystyle \frac {22}{7}\) × 28 (l + 28)

∴ 7128 = 22 × 4 (l + 28)
∴ 7128 = 88(l + 28)

∴ \(\displaystyle \frac {7128}{88}\) = l + 28

∴ 81 = l + 28
∴ 81 − 28 = l
∴ 53 = l
i.e. l = 53 cm ... (i)

Now,
l² = h² + r²
∴ 53² = h² + 28²
∴ 2809 = h² + 784
∴ 2809 − 784 = h²
∴ 2025 = h²
∴ \(\displaystyle \sqrt {2025}\) = h
∴ 45 = h
i.e. h = 45 cm ... (ii)

Also,Volume of a cone:

V = \(\displaystyle \frac {1}{3}\) × πr²h

∴ V = \(\displaystyle \frac {1}{3}\) × \(\displaystyle \frac {22}{7}\) × 28 × 28 × 45

∴ V = 22 × 4 × 28 × 15
∴ V = 36960 cm³ ... (iii)
∴ The volume of the cone is 36960 cm³.

3. Curved surface area of a cone is 251.2 cm² and radius of its base is 8 cm. Find its slant height and perpendicular height. (π = 3.14)

Solution:

Here, S_c = 251.2 cm², r = 8 cm, h = ?, l = ?

Curved surface area of a cone:
S_c = πrl
∴ 251.2 = 3.14 × 8 × l

∴ \(\displaystyle \frac {251.2}{3.14 \times 8}\) = l

∴ 10 = l
i.e. l = 10 cm ... (i)

Now,
l² = h² + r²
∴ 10² = h² + 8²
∴ 100 = h² + 64
∴ 100 − 64 = h²
∴ 36 = h²
∴ \(\displaystyle \sqrt {36}\) = h
∴ 6 = h
i.e. h = 6 cm ... (ii)

∴ The slant height of that cone is 10 cm and its perpendicular height is 6 cm.

4. What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is ₹ 10 per sq. m.?

Solution:

Here, r = 6 m, l = 8 m, S_t = ?

Total surface area of a cone:
S_t = πr(l + r)

∴ S_t = \(\displaystyle \frac {22}{7}\) × 6 (8 + 6)

∴ S_t = \(\displaystyle \frac {22}{7}\) × 6 × 14
∴ S_t = 22 × 6 × 2
∴ S_t = 264 m² ... (i)

Now,
Cost of making the cone
= Area × Rate
= 264 × 10
= 2640 ... (ii)
∴ The cost of making the cone is ₹ 2,640/-

5. Volume of a cone is 6280 cubic cm and base radius of the cone is 20 cm. Find its perpendicular height. (π = 3.14)

Solution:

Here, V = 6280 cm³, r = 20 cm, h = ?

Now,
Volume of a cone:

V = \(\displaystyle \frac {1}{3}\) × πr²h

∴ 6280 = \(\displaystyle \frac {1}{3}\) × 3.14 × 20 × 20 × h

∴ \(\displaystyle \frac {6280 \times 3}{3.14 \times 20 \times 20}\) = h

∴ 15 = h
i.e. h = 15 cm ... (i)

∴ The perpendicular height of that cone is 15 cm.

6. The curved surface area of a cone is 188.4 sq. cm and its slant height is 10 cm. Find its perpendicular height (π = 3.14)

Solution:

Here, S_c = 188.4 cm², l = 10 cm, h = ?
Now,
Curved surface area of a cone:
S_c = πrl
∴ 188.4 = 3.14 × r × 10

∴ \(\displaystyle \frac {188.4}{3.14 \times 10}\) = r

∴ 6 = r
i.e. r = 6 cm ... (i)

Now,
l² = h² + r²
∴ 10² = h² + 6²
∴ 100 = h² + 36
∴ 100 − 36 = h²
∴ 64 = h²
∴ \(\displaystyle \sqrt {64}\) = h
∴ 8 = h
i.e. h = 8 cm ... (ii)

∴ The perpendicular height of that cone is 8 cm.

7. Volume of a cone is 1232 cm³ and its height is 24 cm. Find the curved surface area of the cone.

\(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

Solution:

Here, V = 1232 cm³, h = 24 cm, S_c = ?

Now,
Volume of a cone:

V = \(\displaystyle \frac {1}{3}\) πr²h

∴ 1232 = \(\displaystyle \frac {1}{3}\) × \(\displaystyle \frac {22}{7}\) × r² ×24

∴ \(\displaystyle \frac {1232 \times 3 \times 7}{22 \times 24}\) = r²

∴ 49 = r²
∴ \(\displaystyle \sqrt {49}\) = r
∴ 7 = r
i.e. r = 7 cm ... (i)

Now,
l² = h² + r²
∴ l² = 24² + 7²
∴ l² = 576 + 49
∴ l² = 625
∴ l = \(\displaystyle \sqrt {625}\)
∴ l = 25 cm ... (ii)

Also,
Curved surface area of a cone:
S_c = πrl

∴ S_c = \(\displaystyle \frac {22}{7}\) × 7 × 25

∴ S_c = 550 cm² ... (iii)
∴ The curved surface area of the cone is 550 cm².

8. The curved surface area of a cone is 2200 sq. cm and its slant height is 50 cm. Find the total surface area of cone.

\(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

Solution:

Here, S_c = 2200 cm², l = 50 cm, S_t = ?

Curved surface area of a cone:
S_c = πrl

∴ 2200 = \(\displaystyle \frac {22}{7}\) × r × 50

∴ \(\displaystyle \frac {2200 \times 7}{22 \times 50}\) = r

∴ 14 = r
i.e. r = 14 cm ... (i)

Now,
Total surface area of a cone:
S_t = πr(l + r)

∴ S_t = \(\displaystyle \frac {22}{7}\) × 14 (50 + 14)

∴ S_t = \(\displaystyle \frac {22}{7}\) × 14 × 64
∴ S_t = 22 × 2 × 64
∴ S_t = 2816 cm² ... (ii)

Now,
l² = h² + r²
∴ l² = h&sup/em>24² = h² + 7²
∴ l² = 24² + 7²
∴ l² = 576 + 49
∴ l² = 625
∴ l = \(\displaystyle \sqrt {625}\)
∴ l = 25 cm ... (ii)

9. There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq. m. of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.

Solution:

Area required on the ground
= 25 × 4
= 100 m²
∴ Base Area = 100 m²
∴ πr² = 100 m² ... (i)
Also, h = 18 m, V = ?

Now,
Volume of a cone:

V = \(\displaystyle \frac {1}{3}\) × πr²h

∴ V = \(\displaystyle \frac {1}{3}\) × 100 × 18 ... [From(i)]

∴ V = 600 m³ ... (ii)

∴ The volume of the tent is 600 m³

10. In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m. and diameter of base is 7.2 m. Find the volume of the fodder. If it is to be covered by polythene in rainy season then how much minimum polythene sheet is needed?

\(\displaystyle (\pi = \displaystyle \frac {22}{7}, \sqrt {17.37} = 4.17) \)

Solution:

Here, h = 2.1 m,
d = 7.2 m,
∴ r = 3.6 m,
V = ?, S_c = ?

Volume of a cone:

V = \(\displaystyle \frac {1}{3}\) πr²h

∴ V = \(\displaystyle \frac {1}{3}\) × \(\displaystyle \frac {22}{7}\) × 3.6 × 3.6 × 2.1

∴ V = 28.512 ≈ 28.51 m³ ... (i)

Now,
l² = h² + r²
∴ l² = 2.1² + 3.6²
∴ l² = 4.41 + 12.96
∴ l² = 17.37
∴ l = \(\displaystyle \sqrt {17.37}\)
∴ l = 4.17 m ...(Given) ... (ii)

Also,
The minimum polythene required to cover the fodder
= Curved surface area of the cone
Curved surface are of a cone

S_c = πrl

∴ S_c = \(\displaystyle \frac {22}{7}\) × 3.6 × 4.17

∴ S_c = 47.1805 ≈ 47.18 m² ... (iii)

∴ The volume of the fodder is 28.51 m³ and the minimum polythene required to cover the fodder is 47.18 m².

1. Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.

Solution:

2. Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. \(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

Solution:

3. Curved surface area of a cone is 251.2 cm² and radius of its base is 8 cm. Find its slant height and perpendicular height. (π = 3.14)

Solution:

4. What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is ₹ 10 per sq. m.?

Solution:

5. Volume of a cone is 6280 cubic cm and base radius of the cone is 20 cm. Find its perpendicular height. (π = 3.14)

Solution:

6. The curved surface area of a cone is 188.4 sq. cm and its slant height is 10 cm. Find its perpendicular height (π = 3.14)

Solution:

7. Volume of a cone is 1232 cm³ and its height is 24 cm. Find the curved surface area of the cone. \(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

Solution:

8. The curved surface area of a cone is 2200 sq. cm and its slant height is 50 cm. Find the total surface area of cone. \(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

Solution:

9. There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq. m. of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.

Solution:

Solution:

2. Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm.

\(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

7. Volume of a cone is 1232 cm³ and its height is 24 cm. Find the curved surface area of the cone.

\(\displaystyle (\pi = \displaystyle \frac {22}{7})\)

8. The curved surface area of a cone is 2200 sq. cm and its slant height is 50 cm. Find the total surface area of cone.

\(\displaystyle (\pi = \displaystyle \frac {22}{7})\)