Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

## Practice Set 9.2 Geometry 9th Std Maths Part 2 Answers Chapter 9 Surface Area and Volume

Question 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.

Given: Height (h) = 12 cm, length (l) = 13 cm

To find: Radius of the base of the cone (r)

Solution:

l^{2} = r^{2} + h^{2}

∴ 13^{2} = r^{2} + 12^{2}

∴ 169 = r^{2} + 144

∴169 – 144 = r^{2}

∴ r^{2} = 25

∴ r = √25 … [Taking square root on both sides]

= 5 cm

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

Question 2.

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

Given: Radius (r) = 28 cm,

Total surface area of cone = 7128 sq.cm

To find: Volume of the cone

Solution:

i. Total surface area of cone = πr (l + r)

∴ 7128= y x 28 x (l + 28)

∴ 7128 = 22 x 4 x(l +28)

∴ l + 28 = \(\frac { 7128 }{ 22\times 4 }\)

∴ l + 28 = 81

∴ l = 81 – 28

∴ l = 53cm

ii. Now, l^{2} = r^{2} + h^{2}

∴ 53^{2} = 28^{2}+ h^{2}

∴ 2809 = 784 + h^{2}

∴ 2809 – 784 = h^{2}

∴ h^{2} = 2025

∴ h = \(\sqrt { 2025 }\) …… [Taking square root on both sides]

= 45 cm

= 22 x 4 x 28 x 15

= 36960 cubic.cm

∴ The volume of the cone is 36960 cubic.cm.

Question 3.

Curved surface area of a cone is 251.2 cm^{2} and radius of its base is 8 cm. Find its slant height and perpendicular height, (π = 3.14)

Given: Radius (r) = 8 cm, curved surface area

of cone = 251.2 cm^{2}

To find: Slant height (l) and the perpendicular height (h) of the cone

Solution:

i. Curved surface area of cone = πrl

∴ 251.2 = 3.14 x 8 x l

∴ l= 10 cm

ii. Now, l^{2} = r^{2} + h^{2}

∴ 10^{2} = 8^{2} + h^{2}

∴ 100 = 64 + h^{2}

∴ 100 – 64 = h^{2}

∴ h^{2} = 36

∴ h = √36 … [Taking square root on both sides]

= 6 cm

∴ The slant height and the perpendicular height of the cone are 10 cm and 6 cm respectively.

Question 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?

Given: Radius (r) = 6 m, length (l) = 8 m

To find: Total cost of making the cone

Solution:

i. To find the total cost of making the cone of tin sheet, first we need to find the total surface area of the cone.

Total surface area of the cone = πr (l + r)

= \(\frac { 22 }{ 7 }\) x 6 x (8 + 6)

= \(\frac { 22 }{ 7 }\) x 6 x 14

= 22 x 6 x 2 = 264 sq.m

ii. Rate of making the cone = ₹ 10 per sq.m

∴ Total cost = Total surface area x Rate of making the cone

= 264 x 10

= ₹ 2640

∴ A The total cost of making the cone of tin sheet is ₹ 2640.

Question 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)

Given: Radius (r) = 20 cm,

Volume of cone = 6280 cubic cm

To find: Perpendicular height (h) of the cone

Solution:

∴ The perpendicular height of the cone is 15 cm.

Question 6.

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

Given: Length (l) =10 cm, curved surface area of the cone = 188.4 sq.cm

To find: Perpendicular height (h) of the cone

Solution:

i. Curved surface area of the cone = πrl

∴ 188.4 = 3.14 x r x 10

ii. Now, l^{2} = r^{2} + h^{2}

∴ 10^{2} = 6^{2} + h^{2}

∴ 100 = 36 + h^{2}

∴ 100 – 36 = h^{2}

∴ h^{2} = 64

∴ h = \(\sqrt { 64 }\) … [Taking square root on both sides]

= 8 cm

∴ The perpendicular height of the cone is 8 cm.

Question 7.

Volume of a cone is 1232 cm^{3} and its height is 24 cm. Find the surface area of the cone. (π = \(\frac { 22 }{ 7 }\))

Given: Height (h) = 24 cm,

Volume of cone = 1232 cm^{3}

To find: Surface area of the cone

Solution:

∴ r^{2} = 49

∴ r = \(\sqrt { 49 }\) … [Taking square root on both sides]

= 7 cm

ii. Now, l^{2} = r^{2} + h^{2}

∴ l^{2} = 7^{2} + 24^{2}

= 49 + 576 = 625

∴ l = \(\sqrt { 625 }\) … [Taking square root on both sides]

= 25

iii. Curved surface area of cone = πrl

= \(\frac { 22 }{ 7 }\) x 7 x 25

= 22 x 25

= 550 sq.cm

∴The surface area of the cone is 550 sq.cm.

Question 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. (π = \(\frac { 22 }{ 7 }\))

Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cm

To find: Total surface area of the cone

Solution:

i. Curved surface area of cone = πrl

ii. Total surface area of cone = πr (l + r)

= \(\frac { 22 }{ 7 }\) x 14 x (50 + 14)

= \(\frac { 22 }{ 7 }\) x 14 x 64

= 22 x 2 x 64

= 2816 sq.cm

∴ The total surface area of the cone is 2816 sq.cm.

Question 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.

Given: For the tent,

height (h) = 18m,

number of people in the tent = 25,

area required for each person = 4 sq.m

To find: Volume of the tent

Solution:

i. Every person needs an area of 4 sq.m, of the ground inside the tent.

Surface area of the base of the tent = number of people in the tent × area required for each person

= 25 × 4

= 100 sq.m

ii. Surface area of the base of the tent = πr^{2}

∴ 100 = πr^{2}

∴ πr^{2} = 100

iii. Volume of the tent= \(\frac { 1 }{ 3 }\) πr^{2}h

= \(\frac { 1 }{ 3 }\) x 100 x 18 …….[∵ πr^{2} = 100]

= 100 x 6

= 600 cubic metre

∴ The volume of the tent is 600 cubic metre.

Question 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 heap of the fodder. If it is to be covered by polythene in rainy se&son then how much minimum polythene

sheet is needed? (π = \(\frac { 22 }{ 7 }\) and \(\sqrt { 17.37 }\) = 4.17 ]

Given: Height of the heap (h) = 2.1 m.

diameter of the base (d) = 7.2 m

∴Radius of the base (r) = \(\frac { d }{ 2 }\) = \(\frac { 7.2 }{ 2 }\) = 3.6 m

To find: Volume of the heap of the fodder and polythene sheet required

Solution:

i. Volume of the heap of fodder = \(\frac { 1 }{ 3 }\)πr^{2}h

= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x (3.6)^{2} x 2.1

= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 3.6 x 3.6 x 2.1

= 1 x 22 x 1.2 x 3.6 x 0.3

= 28.51 cubic metre

ii. Now, l^{2} = r^{2} + h^{2}

= (3.6)^{2} + (2.1)^{2}

= 12.96 + 4.41

∴ l^{2} =17.37

∴ l^{2} = \(\sqrt { 17.37 }\) .. .[Taking square root on both sides]

= 4.17 m

iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap

= πrl

= \(\frac { 22 }{ 7 }\) x 3.6 x 4.17

= 47.18 sq.m

∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.