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.
\($10 \sqrt{2 }\)
Let the sides be a , b
2a + 2b =40
a + b =20 square both sides
a^2 + 2ab + b^2 = 400
a^2 + b^2 + 2ab = 400
a^2 + b^2 = (10sqrt 2)^2
a^2 + b^2 = 200
200 + 2ab = 400
100 +ab = 200
ab = 200 - 100
ab = 100 = the area
+359 
