Question

A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm³ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of bucket and the cost of making it at the rate of Rs.10 per cm².

Answer 1

Dear Student,

Answer:
Height of Frustum of cone would be = 15 cm

Cost = 17090.20 Rupees

Solution:

Capacity of Frustum of a cone = 12308.8 cm³

R = 20 cm

r = 12 cm

Volume of Frustum of cone =( π /3)h ( r^2 + rR + R^2)

12308.8 = (π/3)h ( 12*12 + 20*12 + 20*20)

12308.8* 3 = π h( 144+240+400)

12308.8*3 = π h(784)

$h = \frac{12308.3 \times3 }{\pi \times 784} \\ \\ h = 14.99 \: cm$
Height of Frustum of cone would be = 15 cm

Cost of making Frustum= surface area of Frustum x 10

Surface area= π( R+ r) √{(R-r)^2 +h^2}

= π (20+12) √{(20-12)^2 +15^2}

= π(32)√(64+225)

= π (32)(17)

= 1709.02 cm².

Cost = 1709.02*1 0

Cost = 17090.20 Rupees

Hope it helps you.

Answer 2

SOLUTION :

Given :

Bucket is in the form of the frustum of a cone.

Capacity of Frustum of a cone = volume of the frustum of a cone = 12308.8 cm³

Bigger radius (r1) = 20 cm

Smaller radius (r2) = 12 cm

Volume of Frustum of cone =(⅓ π)h ( r1² + r2² + r1r2)

12308.8 = (π/3)h ( 20×20 + 12×12 + 20×12)

12308.8× 3 = π h( 400 + 144+240)

12308.8×3 = 22/7 h(784)

h = 12308.8×3 ×7 / (22 × 784)

h = 12308.8×3 / (22 × 112)

h = 6,154.4 × 3 / 11× 112

h= 18,463.2/1232 = 14.99

h=15 (approximately)

Height of Frustum of cone (h) = 15 cm

Slant height (l) of a frustum cone =√h² + (r1 - r2)²

l = √15² +(20 - 12)² = √225 + (8)²

l = √225 + 64 = √ 289 = 17

l = 17 cm

Surface area of a frustum cone = πl (r1 + r2)

= π × 17 (20+12)

= 22/7 × 17(32)

= 11968/7 = 1,709.7

= 1709.7 cm²

Cost of making 1 cm² = ₹10

Cost of making 1709.7 cm² = ₹(1709.7 × 10) = ₹ 17097

Cost of making 1709.7 cm² = ₹ 17097

Hence, the height of the bucket is 15 cm and the cost of making the bucket is ₹ 17097.

HOPE THIS ANSWER WILL HELP YOU..