Question

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find

(i) the volume of water which can completely fill the bucket;

(ii) the area of the metal sheet used to make the bucket.

Answer 1

(i). volume of frustum of cone

$=\frac{1}{3}\times \pi \times h\times (R^2+Rr+r^2)=\frac{1}{3}\times \frac{22}{7}\times 24\times ({14}^2+14\times 7+7^2)=22\times \frac{8}{7}\times (196+98+49)=22\times \frac{8}{7}\times 343=22\times 8\times 49=8624c{m}^3$

(ii) slant height of frustum,

$l=\sqrt{(R-r{)}^2+h^2}=\sqrt{(14-7{)}^2+{24}^2}=\sqrt{7^2+{24}^2}=\sqrt{49+576}=\sqrt{625}=25cm$

area of metallic sheet = CSA of frustum + area of base

$=\pi \times l\times (R+r)+\pi \times r^2=\frac{22}{7}\times 25\times (14+7)+\frac{22}{7}\times 7^2=\frac{550}{7}\times 21+22\times 7=550\times 3+154=1650+154=1804c{m}^2$