+713 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.
+975 First, from the first part of the problem, we know that
\(2x + 2y = 40 \\ x + y = 20 \\ y = 20 - x\)
Now, we use the fact that the diaganol is 10 sqrt2, we have
\(x^2 + y^2 = (10 sqrt 2) ^2 \\ x^2 + ( 20 - x)^2 = 200 \\ x^2 + x^2 - 40x + 400 = 200 \\ 2x^2 - 40x + 200 = 0 \\ x^2 - 20x + 100 = 0\)
\((x -10)^2 = 0 \\ x - 10 = 0 \\ x = 10 \\ y = 20 - 10 = 10\)
The area is just 10*10 = 20.
Thanks! :)
