## Elementary Geometry for College Students (6th Edition)

(a) $V = 32~in^3$ (b) $altitude = 8~in$
(a) We can find the area of the base $\triangle BCD$: $Area = \frac{1}{2}(4~in)(8~in) = 16~in^2$ We can find the volume: $V = \frac{1}{3}~(base~area)(altitude)$ $V = \frac{1}{3}~(16~in^2)(6~in)$ $V = 32~in^3$ (b) We can find the length of the altitude: $V = \frac{1}{3}~(base~area)(altitude)$ $altitude = \frac{3V}{base~area}$ $altitude = \frac{(3)(32~in^3)}{12~in^2}$ $altitude = 8~in$