+834 The perimeter of a rectangle is $40,$ and the length of one of its diagonals is $10 \sqrt{2}.$ Find the area of the rectangle.
+37146 Let sides be h and w
2h + 2w = 40 <=== Given so: h = 20 - w
Pythagorean theorem:
h^2 + w^2 = (10 sqrt2) ^2
(20-w)^2 + w^2 = 200
2w^2 -40w + 200 = 0 Use Quadratic Formula with a = 2 b = -40 c = 200 to find w = 10 then h = 20 -w shows h = 10 too.
area = w * h = 10 * 10 = 100 units^2
