Question

1. A frustum of a regular hexagonal pyramid has an upper base edge of 5m and a lower base edge of 8.5m. Its lateral area is 160m2. Determine the slant height of the frustum.

Answer 1

The lateral area of frustum of regular pyramid is equal to one-half the sum of the perimeters of the bases multiplied by the slant height.

$$A_L=\dfrac{1}{2}n(a+b)L$$

$$L=\dfrac{2A_L}{n(a+b)}$$

$$L=\dfrac{2(160\ m^2)}{6(5\ m+8.5\ m)}=\dfrac{320}{81}\ m$$