Question

Find the volume of a frustum cone, whose base and upper area of a circle is $40c{m}^2$ and $90c{m}^2$. (Insert answer in nearest integer).

Answer 1

Volume of a frustum cone = $\frac{1}{3}(A_1+A_2+\sqrt{A_1{A}_2})h$
= $\frac{1}{3}(40+90+\sqrt{40\times 90})10$
= $\frac{1}{3}(130+60)10$
= $633.33c{m}^3$