Chapter 9 Surface Area and Volume Practice Set 9.2 Solutions

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

ii. Now, l²= r²+ h²
∴ 53²= 28²+ h²
∴ 2809 = 784 + h²
∴ 2809 – 784 = h²
∴ h²= 2025
∴ h = …… [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.

ii. Now, l²= r²+ h²
∴ 10²= 8²+ h²
∴ 100 = 64 + h²
∴ 100 – 64 = h²
∴ h²= 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.

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.

ii. Now, l²= r²+ h²
∴ 10²= 6²+ h²
∴ 100 = 36 + h²
∴ 100 – 36 = h²
∴ h²= 64
∴ h = … [Taking square root on both sides]
= 8 cm
∴ The perpendicular height of the cone is 8 cm.

ii. Now, l²= r²+ h²
∴ l²= 7²+ 24²
= 49 + 576 = 625
∴ l = … [Taking square root on both sides]
= 25

iii. Curved surface area of cone = πrl
= x 7 x 25
= 22 x 25
= 550 sq.cm
∴The surface area of the cone is 550 sq.cm.

ii. Total surface area of cone = πr (l + r)
= x 14 x (50 + 14)
= x 14 x 64
= 22 x 2 x 64
= 2816 sq.cm
∴ The total surface area of the cone is 2816 sq.cm.

ii. Surface area of the base of the tent = πr²
∴ 100 = πr²
∴ πr²= 100

iii. Volume of the tent= πr²h
= x 100 x 18 …….[∵ πr²= 100]
= 100 x 6
= 600 cubic metre
∴ The volume of the tent is 600 cubic metre.

ii. Now, l²= r²+ h²
= (3.6)²+ (2.1)²
= 12.96 + 4.41
∴ l²=17.37
∴ l²= ...[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
= 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.

Maharashtra Board Class 9 Maths Solutions