# A reservoir in the form of the frustum of a right

Question:

A reservoir in the form of the frustum of a right circular cone contains 44 × 107 litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take: π = 22/7)

Solution:

Let the depth of the frustum cone like reservoir is m. The radii of the top and bottom circles of the frustum cone like reservoir are r1 =100m and r2 =50m respectively.

The volume of the reservoir is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(100^{2}+100 \times 50+50^{2}\right) \times h$

$=\frac{1}{3} \times \frac{22}{7} \times 17500 \times h$

$=\frac{1}{3} \times 22 \times 2500 \times h \mathrm{~cm}^{3}$

$=\frac{1}{3} \times 22 \times 2500 \times h \times 10^{6} \mathrm{~m}^{3}$

$=\frac{1}{3} \times 22 \times 2500 \times h \times 10^{3}$ litres

Given that the volume of the reservoir is $44 \times 10^{7}$ litres. Thus, we have

$\frac{1}{3} \times 22 \times 2500 \times h \times 10^{3}=44 \times 10^{7}$

$\Rightarrow h=\frac{3 \times 44 \times 10^{7}}{22 \times 2500 \times 10^{3}}$

$\Rightarrow h=24$

Hence, the depth of water in the reservoir is $24 \mathrm{~m}$

The slant height of the reservoir is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(100-50)^{2}+24^{2}}$

$=\sqrt{3076}$

$=55.46169$ meter

The lateral surface area of the reservoir is

$S_{1}=\pi\left(r_{1}+r_{2}\right) \times l$

$=\pi \times(100+50) \times 55.46169$

$=\pi \times 150 \times 55.46169$

$=26145.225 \mathrm{~m}^{2}$

Hence, the lateral surface area is $26145.225 \mathrm{~m}^{2}$