Question

The radii of the circles ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm).
If the volume of bucket is 5390 $c{m}^3$, then find the value of r.
$[Use\pi =\frac{22}{7}]$

Answer 1

Height of the bucket (which is in the shape of a frustum of a cone), h = 15 cm

Radius of one end of bucket, R = 14 cm

Radius of the other end of the bucket is r.

It is given that the volume of the bucket is 5390 $c{m}^3$.
$\Rightarrow \frac{1}{3}\pi (R^2+r^2+Rr)h=5390$
$\Rightarrow \frac{1}{3}\times \frac{22}{7}\times [{14}^2+r^2+14r]\times 5390$
$\Rightarrow 196+r^2+14r=\frac{5390\times 7}{22\times 5}=343$
$\Rightarrow r^2+14r+196-343=0$
$\Rightarrow r^2+14r-147=0$
$\Rightarrow r^2+21r-7r-147=0$
$\Rightarrow r(r+21)-7(r+21)=0$
$\Rightarrow (r+21)(r-7)=0$
$\Rightarrow r-7-0orr+21=0$
$\Rightarrow r=7orr=-21$
Since the radius cannot be a negative number, r= 7 cm
Thus, the value of r is 7 cm.