+677 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.
+129771 Let adjacent sides be x, y
So.....
2x + 2y = 40
x + y = 20
y = 20 - x
By the Pythagorean Theorem
x^2 + y^2 = (10 sqrt 2) ^2
x^2 + ( 20 - x)^2 = 200
x^2 + x^2 - 40x + 400 = 200
2x^2 - 40x + 200 = 0
x^2 - 20x + 100 = 0
(x -10)^2 = 0 take positive root
x - 10 = 0
x = 10
y = 20 - 10 = 10
Area = xy = 10 * 10 = 100
