#### Answer

(a) $V = 32~in^3$
(b) $altitude = 8~in$

#### Work Step by Step

(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:
$V = \frac{1}{3}~(base~area)(altitude)$
$altitude = \frac{3V}{base~area}$
$altitude = \frac{(3)(32~in^3)}{12~in^2}$
$altitude = 8~in$