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.
Let's let the two sides of the rectange x and y.
We can use the problem to write the equation
2x+2y=40x+y=20
Isolating y, we have y=20−x.
Subsituting this value for y into the second part of the qeustion and using the pythaogrean theorem, we have
x2+y2=(10√2)2x2+(20−x)2=200x2+x2−40x+400=2002x2−40x+200=0x2−20x+100=0
Finally, we find that we have x = 10. From the first equation, we find that y = 10 as well.
The area is just 10*10=100!
Thanks! :)