Rectangle

Question

-1

6

1

+1040

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.

LiIIiam0216 Mar 16, 2024

+29

+1

Best Answer

$\sqrt{200} = 10\sqrt2$, so $a^2 + b^2 = 200$

Assuming that a and b are both integer side lengths, $10^2+10^2=200$, so the sides are 10 and 10.

10 times 10 is 100, so the area of the rectangle is $\bf100$

jilin73 Mar 16, 2024

jilin73 · Answer 1 · 2024-03-16T02:39:54

$\sqrt{200} = 10\sqrt2$, so $a^2 + b^2 = 200$

Assuming that a and b are both integer side lengths, $10^2+10^2=200$, so the sides are 10 and 10.

10 times 10 is 100, so the area of the rectangle is $\bf100$