**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 cmTo 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 =22/7)Given: Radius (r) = 28 cm,Total surface area of cone = 7128 sq.cmTo find: Volume of the coneSolution:**

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 = 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

= 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 areaof cone = 251.2 cm^{2}To find: Slant height (l) and the perpendicular height (h) of the coneSolution:**

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 mTo find: Total cost of making the coneSolution:**

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)

= 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 cmTo find: Perpendicular height (h) of the coneSolution:**

∴ 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.cmTo find: Perpendicular height (h) of the coneSolution:**

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

= 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 =22/7)Given: Height (h) = 24 cm,Volume of cone = 1232 cm^{3}To find: Surface area of the coneSolution:**

**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 =22/7)Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cmTo find: Total surface area of the coneSolution:**i. Curved surface area of cone = πrl

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

= 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.mTo find: Volume of the tentSolution:**

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

= 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 polytheneTo find: Volume of the heap of the fodder and polythene sheet requiredSolution:**

= 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

= 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

= 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.