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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.

 Apr 30, 2024
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Let adjacent  sides be x, y

 

So.....

 

2x + 2y  =  40

 

x + y  = 20

 

y = 20 - x

 

By the Pythagorean Theorem

 

x^2  + y^2  = (10 sqrt 2) ^2

 

x^2 + ( 20 - x)^2  =  200

 

x^2 + x^2  - 40x + 400 =  200

 

2x^2  - 40x + 200  = 0

 

x^2 - 20x + 100  = 0

 

(x -10)^2  = 0           take positive root

 

x  - 10  =  0

 

x = 10

 

y = 20 - 10  =  10

 

Area = xy =  10 * 10  =   100

 

 

cool cool cool

 Apr 30, 2024

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