Question

38. A container made of a metal sheet is in the form of a frustum of a cone of height 12 cm with radii of
lower and upper ends 3 cm and 12 cm, respectively. Find the cost of the metal sheet used, if it costs
₹7 per 100 cm? Use = 22
And
A bucket is in the form of a frustum of a cone of height 12 cm with radii of its lower and upper end
as 8 cm and 18 cm, respectively. Find its capacity. Also, find the cost of milk which can completely fill
the container, at the rate of ₹25 per litre.​

Answer 1

The cost of the metal sheet used, if it costs  ₹7 per 100 cm² is Rs. 83.16.

The cost of milk which can completely fill  the container, at the rate of ₹25 per litre is Rs. 152.

Step-by-step explanation:

Problem 1:

Height, h = 12 cm

Upper radius, R = 12 cm

Lower radius, r = 3 cm

∴ Slant Height, l = $\sqrt{h^2 + (R - r)^2} = \sqrt{12^2 + (12-3)^2} = \sqrt{144 + 81} = 15 cm$

∴ The Total Surface Area of the frustum,

= π(r + R)l + πR² + πr²

= π [(12+3) + 12² + 3²]

= $\frac{22}{7}$ × [225 + 144 + 9]

= $\frac{22}{7}$ × 378

= 22 × 54

= 1188 cm²

Thus,

The cost of the metal sheet used at the rate of Rs. 7 per 100 cm² is,

= 1188 × $\frac{7}{100}$

= Rs. 83.16

Problem 2:

Height, h = 12 cm

Upper radius, R = 18 cm

Lower radius, r = 8 cm

∴ The capacity of the container is given by,

= Volume of the frustum

= $\frac{1}{3}$ π h [r² + rh + R²]

= $\frac{1}{3}$ × $\frac{22}{7}$ × 12 × [8² + (8×12) + 18²]

= $\frac{1}{3}$ × $\frac{22}{7}$ × 12 × [64 + 96 + 324]

= $\frac{1}{3}$ × $\frac{22}{7}$ × 12 × 484

= 6084.57 cm³

∵ 1 litre = 1000 cm³

= 6.08 litre

Thus,

The cost of milk which can fill the container completely, at the rate of Rs. 25 per litre = Rs. 25 * 6.08 litre = Rs. 152

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