+1694 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.
+1786 Let's let the two sides of the rectange be a and b.
From the problem, we have
\(2a + 2b =40 \\ a + b =20 \\ \)
Now, we just sqaure both sides to get
\(a^2 + 2ab + b^2 = 400 \\ a^2 + b^2 + 2ab = 400\)
From here, we have
\(a^2 + 2ab + b^2 = 400 \\ a^2 + b^2 + 2ab = 400 \\ a^2 + b^2 = (10\sqrt{ 2})^2 \\ a^2 + b^2 = 200\\ 200 + 2ab = 400 \\ 100 +ab = 200 \\ ab = 200 - 100\)
So our answer is 100.
Thanks! :)
