Question

A bucket open at the top is of the form of a frustum of a cone. The diameter of its upper and lower circular ends are 40 cm and 20 cm respectively. If total $17600c{m}^3$ of water can be filled in the bucket, find its total surface area. $(\pi =\frac{22}{7})$

Answer 1

Given
Diameter of upper end (R)=40 cm
$Radius=\frac{diameter}{2}=\frac{40}{2}=20cm$
Diameter of lower end(r) = 20 cm
$Radius=\frac{diameter}{2}=\frac{20}{2}=10cm$
Volume of the frustum = $17600c{m}^3$
Let h be the height of the frustum
We know that
Volume of the frustum = $\frac{\pi h}{3}(R^2+r^2+Rr)$
$\Rightarrow 17600=\frac{22}{7}\times \frac{1}{3}\times h({20}^2+{10}^2+20\times 10)$
$\Rightarrow 17600=\frac{22}{7}\times \frac{h}{3}(400+100+200)$
$\Rightarrow 17600=\frac{22}{7}\times \frac{h}{3}\left(700\right)$
$\Rightarrow h=\frac{176\times 3}{22}$
$\Rightarrow h=24cm$
We know that
slant height (I) = $\sqrt{h^2+(R-r{)}^2}$
$\Rightarrow l=\sqrt{{24}^2+(20-10{)}^2}$
$\Rightarrow \sqrt{576+(10{)}^2}$
$\Rightarrow l=\sqrt{576+100}$
$\Rightarrow l=\sqrt{676}$
$\Rightarrow l=26cm$
we know that
Total surface area of frustum of cone =
(area of the base) + (area of the top) + (lateral surface area)
= $\pi (R^2+r^2+l(R+r\left)\right]$
= $\frac{22}{7}\left[\right(20{)}^2+(10{)}^2+26(20+10\left)\right]$
= $\frac{22}{7}[400+100+26(30\left)\right]$
= $\frac{22}{7}[500+780]$
= $\frac{22}{7}\left(1280\right)$
= $4022.8c{m}^2$