The radii of two bases of a frustum of height 10.5 cm is 5 cm and 3 cm. What is its volume in cm$$^3$$($$\pi = \frac{22}{7}$$)?

It is given that,

radii of two bases of a frustum are $$R=5cm$$ and $$r=3cm$$

Height $$(h)=10.5$$cm

$$\pi = \frac{22}{7}$$

Volume of the frustum $$(V)=\dfrac{\pi \times h \times(R^2+r^2+R\times r)}{3}$$

$$(V) =\dfrac{22 \times 10.5 \times(5^2+3^2+5\times 3)}{7\times 3}$$

$$(V) =\dfrac{22 \times 10.5 \times(25+9+15)}{7\times 3}$$

$$(V) =\dfrac{22 \times 10.5 \times 49}{7\times 3}$$

$$(V) =22 \times 3.5 \times 7$$

$$(V)=539 cm^3$$