Question

Given a frustum of a pyramid with areas of upper and lower bases equal to 20 sq.m and 30 sq. m, and the distance between 2 bases equal to 6 m, find the volume of the frustum.

Answer 1

1. Lower Base - a base of a frustum of a regular pyramid with a larger area

2. Upper Base - a base of a frustum of a regular pyramid with a smaller area

3. Altitude - the perpendicular distance between the bases of a frustum of a regular pyramid

The volume of a frustum of a regular pyramid is equal to one-third of the altitude multiplied by the sum of its bases and the geometric mean between them.

$$V=\dfrac{1}{3}h(B_1+B_2+\sqrt{B_1B_2})$$

Given $B_1=30\ m^2, B_2=20\ m^2, h=6\ m.$

$$V=\dfrac{1}{3}(6)(30+20+\sqrt{30(20)})$$

$$=20(5+\sqrt{6})\ (m^3)$$

The volume of the frustum is $20(5+\sqrt{6})$ m³.