Rectangle

Question

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39

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+2729

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

+39

+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 $\bf100$