Question

A container is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20cm, respectively. Find the capacity and surface area of the container. Also, find the cost of milk which can completely fill the container at the rate of rs. 25 per litre.

Answer 1

$volume \: of \: frustum = \frac{1}{3}\pi \: h( {r}^{2}+{r1}^{2} +r \times r1)$

$\frac{1}{3} \times \frac{22}{7} \times 30( {10}^{2} + {20}^{2} + 10 \times 20)$

$\frac{660}{21} (100 + 400 + 200)$

$\frac{660}{21} (700)$

$\frac{660}{3} (100)$

$220 \times 100$

$22000 {cm}^{3}$

Volume of container = 22 litres

Cost of milk = 22×25

=550Rs

$sa \: of \: frustum = \pi( {r}^{2} + {r1}^{2} + (r + r1)l)$

By using slant height formula of frustum

$l = \sqrt{ {h}^{2} + {(r1 - r)}^{2} }$

$l = \sqrt{ {30}^{2} + {(20 - 10)}^{2} }$

$l = \sqrt{900 + 100}$

$l = \sqrt{1000}$

$l = 10 \sqrt{10}$

$\frac{22}{7}(100 + 400 + (10 + 20)10 \sqrt{10} )$

$\frac{22}{7} (500 + (30)10 \sqrt{10} )$

$\frac{22}{7} (500 + 300 \sqrt{10)}$

$\frac{22}{7} (500 + 300 \times 3.1)$

$\frac{22}{7} (500 + 930)$

$\frac{22}{7} \times 1430$

$22 \times 24.28$

$534.16 {cm}^{2}$

surface area of frustum = 534.16 cm^2