+0  
 
0
1
1
avatar+1137 

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.

 Jun 8, 2024
 #1
avatar+129401 
+1

Let a,b be the sides

2 ( a + b)  = 40

a + b    = 20

b = 20  - a

 

a^2 + b^2  = (10 sqrt 2)^2

 

a^2  + (20 - a)^2  = 200

 

a^2 + a^2 -40a + 400 = 200

 

2a^2 -40a + 200  = 0

 

a^2 -20a + 100 =  0

 

(a - 10*2  = 0

 

a =10

b =20 -10  =10

 

Area =  a*b =  10 * 10   =  100

 

cool cool cool

 Jun 8, 2024

4 Online Users

avatar
avatar