Strange surface area

Relevant wiki: Surface Area of a Cuboid

The surface consists of 6 rectangular faces, 2 each with area $lw$ , $wh$ , and $hl$ , so the surface area is $2lw + 2wh + 2hl$ . We expand the square of the sum of $l$ , $w$ , and $h$ to get $(l + w + h)^2 = l^2 + lw +lh + wl + w^2 + wh + hl + hw + h^2 = l^2 + w^2 + h^2 + 2lw + 2wh + 2hl$ , or $2lw + 2wh + 2hl = (l + w + h)^2 - (l^2 + w^2 + h^2)$ . We can subsitite the given values to get $2lw + 2wh + 2hl = 11^2 - 59$ , so the surface area is $121 - 59 = \boxed{62}$ .

It is unnecessary for solving the problem but might be interesting to note that the values of $l$ , $w$ , and $h$ chosen when writing this problem are 7, 3, and 1 in some order.

Surface area: $S=2(lw+wh+hl)$

Square of the sum of the dimensions:

$(l+h+w)^2=l^2+h^2+w^2+2lw+2wh+2hl=(l^2+h^2+w^2)+2(lw+wh+hl)=59+S=11^2=121$

$S=121-59=62$