Exercise 10.3

1. The base area of a cone is 38.5cm2 Its volume is 77cm3 Find its height.

Solution:

Base Area: cm2

Volume: cm3

Volume of cone = (1/3) × Base Area × Height

= (1/3) × × Height

Height = (77 × 3)77 /

Height = / 38.5

Height = cm

2. The volume of a cone is 462 m3 Its base radius is 7 m. Find its height.

Solution:

Volume: m3

Radius: m

Volume of cone = (1/3)πr2h

= (1/3) × (22/7) × 7282 × h

462 = (1/3) × 22 × × h

462 = (/3) × h

h = (462 × 3) / 154(462 × 3) × 154

h = / 154

h = m

3. Curved surface area of a cone is 308cm2 and its slant height is 14 cm Find.

(i) radius of the base (ii) Total surface area of the cone.

Solution:

Given

Curved Surface Area (CSA): cm2

Slant Height (l): cm

(i) CSA = πrl

= () × ×

=

r = 308 / 44308 × 44

r = cm

(ii) Total Surface Area (TSA) = πr(l + r)πrl

TSA = (22/7) × 7 × (14 + 7)

TSA = ×

TSA = cm2

4. The cost of painting the total surface area of a cone at 25 paise per cm2 is ₹176. Find the volume of the cone, if its slant height is 25 cm.

Solution:

Cost of painting: ₹176 = paise

Rate: paise/cm2

TSA = 17600 / 25 = cm2

Slant Height (l): 25 cm

TSA = πr(l + r)πrl

704 = (22/7) × r × (25 + r)25

704 - 227 = r(25 + r)

= r(25 + r)

r22r + - 224 = 0

(r + 32)(r - 32)|(r - 7)(r + 7) = 0

r = cm (since radius cannot be negative)

Height (h) = √(l2 - r2)(l2 + r2) = √(252 - 72) = √( - ) = √ = cm

Volume = (1/3)πr2h = (1/3) × (22/7) × 7^2 × 24 = cm3

5. From a circle of radius 15 cm., a sector with angle 216° is cut out and its bounding radii are bent so as to form a cone. Find its volume.

Solution:

Radius of circle: cm

Sector angle: 216°

Slant height of cone (l) = Radius of circle = 15 cm

Circumference of base of cone = (216/360) × 2π × 15 = 18π19π

2πr30π = 18π

r = cm

Height (h) = √(l2 - r2)|(l2 + r2) = √(152 - 92) = √( - ) = √ = cm

{.reveal(when="blank-8")}Volume = (1/3)πr2hπr2h = (1/3) × π × 92 × 12 = 324π322π cm3

Volume = 324 × 3.14159 ≈ 1018.8761015.876 cm3

6. The height of a tent is 9 m. Its base diameter is 24 m. What is its slant height? Find the cost of canvas cloth required if it costs 14 per sq.m.

Solution:

Height (h): m

Diameter: m, Radius (r): m

Slant height (l) = √(h^2 + r2)√(h^2 - r2) = √(9^2 + 12^2) = √( + ) = √ = m

CSA of tent = πrl = π × × = 180π150π m^2

Cost of canvas = 180π × 14 = ₹2520π2720π

Cost = 2520 × 3.14159 ≈ ₹7916.817616.81

7. The curved surface area of a cone is 1159 5/7 cm2. Area of its base is 254 4/7 cm2. Find its volume.

Solution:

Curved Surface Area (CSA) = 1159 5/7 cm2 = cm2

Base Area = 254 4/7 cm2 = cm2

Base Area = πr2 =

r2 = (1782/7) / π. Assuming π = 22/7, then r2 = (1782/7) × =

r = cm

CSA = πrl =

(22/7) × × l = 8118/7

l = (8118/7) / ((22/7) × 9) = (8118/7) × (7/198) = cm

Height (h) = √(l^2 - r2) = √(41^2 - 9^2) = √(1681 - 81) = √ = cm

Volume = (1/3)πr2h = (1/3) × (22/7) × ^2 ×

Volume = (1/3) × (22/7) × 3240 = cm3 (approximately)

8. A tent is cylindrical to a height of 4.8 m and conical above it. The radius of the base is 4.5 m and the total height of the tent is 10.8 m. Find the canvas required for the tent in square meters.

Solution:

Radius of base (r) = m

Height of cylinder (h_cyl) = m

Total height of tent = m

Height of cone (h_cone) = 10.8 - 4.8 = m

Slant height of cone (l) = √(r2 + h_cone^2) = √(^2 + ^2) = √(20.25 + 36) = √ = m

{}.m-tealCanvas required = Curved surface area of cylinder + Curved surface area of cone

Curved surface area of cylinder = 2πrh_cyl = 2 × π × × =

Curved surface area of cone = πrl = π × × =

Total canvas required = 43.2π + 33.75π =

Assuming π = 3.14, total canvas ≈ 76.95 × = m^2

9. What length of tarpaulin 3 m wide will be required to make a conical tent of height 8m and base radius 6m? Assume that extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (use π = 3.14).

Solution:

Height of cone (h) = m

Base radius of cone (r) = m

Slant height of cone (l) = √(r2 + h^2) = √(^2 + ^2) = √(36 + 64) = √ = m

Curved surface area of tent = πrl = π × × = m^2

Assuming π = 3.14, Curved surface area ≈ 60 × = m^2

Width of tarpaulin = m

Length of tarpaulin required = Area / Width = / = m

Extra length for wastage = 20 cm = m

Total length of tarpaulin required = 62.8 + 0.2 = m

10. A Joker's cap is in the form of a right circular cone of base radius 7 cm and height 27 cm. Find the area of the sheet required to make 10 such caps.

Solution:

Radius (r) = cm

Height (h) = cm

Slant height (l) = √(r2 + h^2) = √(^2 + ^2) = √(49 + 729) = √ = cm (approximately)

Area of sheet for one cap = πrl = π × × = cm2

Assuming π = 3.14, area of sheet for one cap ≈ 195.23 × = cm2

Area of sheet for 10 caps = 10 × = cm2

11. Water is pouring into a conical vessel of diameter 5.2 m and slant height 6.8 m (as shown in the adjoining figure), at the rate of 1.8 m3/minute. How long will it take to fill the vessel?

Solution:

Diameter = 5.2 m, so radius (r) = 5.2 / 2 = m

Slant height (l) = m

Height (h) = √(l^2 - r2) = √(^2 - ^2) = √(46.24 - 6.76) = √ ≈ m (approximately)

Volume of cone = (1/3)πr2h = (1/3) × π × ^2 × = (1/3) × π × 6.76 × 6.28 ≈ m3

Assuming π = 3.14, volume of cone ≈ 14.14 × ≈ m3 (approximately)

Rate of pouring = m3/minute

Time to fill the vessel = Volume / Rate ≈ / ≈ minutes

12. Two similar cones have volumes 12π cu. units and 96π cu. units. If the curved surface area of the smaller cone is 15π sq. units, what is the curved surface area of the larger one?

Solution:

V1 =

V2 =

A1 =

A2 = A1*(V2/V1)^(23) = (96π12π)^(23)

A2 = 15π(8)^(23)

A2 = 15π*4 =