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

1. the volume of water which can completely fill the bucket;
2. the area of the metal sheet used to make the bucket.

Answer 1

We have,

Height, h = 24 cm,

Upper base radius, R = 14 cm and lower base radius, r = 7 cm

Also, the slant height, ${\displaystyle l=\sqrt{{(\text{R}-\text{r})}^2+{\text{h}}^2}}$

${\displaystyle =\sqrt{{(14-7)}^2+{24}^2}}$

${\displaystyle =\sqrt{7^2+{24}^2}}$

${\displaystyle =\sqrt{49+576}}$

${\displaystyle =\sqrt{625}}$ 

= 25 cm

 i . Volume of the bucket${\displaystyle =\frac{1}{3}\pi \text{h}({\text{R}}^2+{\text{r}}^2+\text{Rr})}$

${\displaystyle =\frac{1}{3}\times \frac{22}{7}\times 24\times ({14}^2+7^2+14+7)}$

${\displaystyle =\frac{22}{7}\times 8\times (196+49+98)}$

${\displaystyle =\frac{22}{7}\times 343}$

= 8624 cm³

So, the volume of water which can completely fill the bucket is 8624 cm³

ii . surface area of the bucket = π (R + r)l + πr² 

${\displaystyle =\frac{22}{7}\times (14+7)\times 25+\frac{22}{7}\times 7\times 7}$

${\displaystyle =\frac{22}{7}\times 21\times 25+22\times 7}$

= 22 × 3 × 25 + 22 × 7

= 1650 + 154

= 1804 cm²  

So, the area of the metal sheet used to make the bucket is 1804 cm² .