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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.

 Mar 27, 2024
 #1
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\($10 \sqrt{2 }\)

 

Let the sides be   a , b

2a + 2b =40

a + b =20      square both sides

a^2 + 2ab + b^2  = 400

a^2 + b^2 + 2ab  = 400

 

a^2 + b^2 = (10sqrt 2)^2

a^2 + b^2  = 200

 

200 + 2ab = 400

100 +ab  = 200

ab = 200 - 100

ab = 100  =  the area

 

 

cool cool cool

 Mar 28, 2024

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