$\bar{y} = 2.5 \, \text{ in. above the base}$

$A_1 = 1(f) = f \, \text{ in.}^2$

$y_1 = 0.5 \, \text{ in.}$

$A_2 = 8(1) = 8 \, \text{ in.}^2$

$y_2 = 1 + \frac{1}{2}(8) = 5 \, \text{ in.}$

$A = A_1 + A_2$

$A = f + 8$

$A\bar{y} = \Sigma ay$

$(f + 8)(2.5) = 0.5f + 8(5)$

$2.5f + 20 = 0.5f + 40$

$2f = 20$

$f = 10 \, \text{ in.}$ *answer*