
$\bar{y} = 2.5 \, \text{ in. above the base}$
$A_1 = 1(f) = f \, \text{ in.}^2$
$y_1 = 0.5 \, \text{ in.}$
$A_2 = 8(1) = 8 \, \text{ in.}^2$
$y_2 = 1 + \frac{1}{2}(8) = 5 \, \text{ in.}$
$A = A_1 + A_2$
$A = f + 8$
$A\bar{y} = \Sigma ay$
$(f + 8)(2.5) = 0.5f + 8(5)$
$2.5f + 20 = 0.5f + 40$
$2f = 20$
$f = 10 \, \text{ in.}$ answer