Question

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

Each lateral face is a trapezoid with base lengths 5 and 8.5 m. The 6 of them will have a total area of 160 m², so one of them has an area of $\frac{160}{6}$ = 26.66 m². From the equation of the area of a trapezoid, we know

$\frac{1}{2}(b_1 + b_2)h = area$

$h = \frac{2 \times area}{b_1+b_2} = \frac{2 \times 26.66}{5 + 8.5} = \frac{53.33}{13.5} = 3.95 \;m$

This is the slant height of the face of the frustum.

Answer: 3.95 m