Solved Problem 05 | Cube

$d = \sqrt{6^2 + 6^2}$

$d = 6\sqrt{2}$
 

$A_1 = 6(6)$

$A_1 = 36 \, \text{ ft}^2$   →   side and front
 

$A_2 = 6d = 6(6\sqrt{2})$

$A_2 = 36\sqrt{2} \, \text{ ft}^2$   →   diagonal
 

Required lateral area
$A = 2A_1 + A_2$

$A = 2(36) + 36\sqrt{2}$

$A = 122.91 \, \text{ ft}^2$           answer