A reservoir in the form of the frustum of a right

Solution:

Let the depth of the frustum cone like reservoir is h 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 $26145.225 \mathrm{~m}^{2}$