A milk container is made of metal sheet in the shape of frustum of a c

1.	A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 \(\frac{3}7\) cm3. The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm2
Answer» Given: volume of the frustum = 10459 \(\frac{3}7\) cm³ Radii of lower and upper ends resp.: r’ = 8 cm and r’’ = 20 cm Volume = \(\frac{1}3π(r'^2 + r"^2 + r'r")h\) ⇒ \(\frac{1}3π(8^2 + 20^2 + 8\times20)\times{h}\) = 10459 \(\frac{3}7\) ⇒ h = 16 cm Let slant height be l Slant height, l = \(\sqrt{(r' - r")^2 + h^2}\) ⇒ l = \(\sqrt{(20 - 8)^2 + 16^2}\) ⇒ l = 20 cm Total surface area of the frustum = π(r’ + r’’)l + πr’²+ πr’’² = π(20 + 8)20 + π (20)² + π (8)² = 3218.29 cm² cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm² = 3218.28 × 1.40 = Rs 4506

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 \(\frac{3}7\) cm3. The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm2

Answer»

Given:

volume of the frustum = 10459 \(\frac{3}7\) cm³

Radii of lower and upper ends resp.: r’ = 8 cm and r’’ = 20 cm

Volume = \(\frac{1}3π(r'^2 + r"^2 + r'r")h\)

⇒ \(\frac{1}3π(8^2 + 20^2 + 8\times20)\times{h}\) = 10459 \(\frac{3}7\)

⇒ h = 16 cm

Let slant height be l

Slant height, l = \(\sqrt{(r' - r")^2 + h^2}\)

⇒ l = \(\sqrt{(20 - 8)^2 + 16^2}\)

⇒ l = 20 cm

Total surface area of the frustum = π(r’ + r’’)l + πr’²+ πr’’²

= π(20 + 8)20 + π (20)² + π (8)²

= 3218.29 cm²

cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm²

= 3218.28 × 1.40

= Rs 4506