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A rectangle has a perimeter of 32 and an area of 56.  Find the length of its diagonal.

 Jan 8, 2020
 #1
avatar+21707 
+3

2 l + 2w = 32

l * w = 56      l = 56/w

 

Sub second into first equation

 

112/w + 2w = 32      multiply by w

2w^2 - 32w + 112 = 0      Quadratic formula shows  w = 8+ 2 sqrt 2        then  l = 8-2sqrt2

 

d^2 = (8+ 2 sqrt2)^2  + (8-2 sqrt2)^2

d^2 = 144      d = 12

 Jan 8, 2020
 #2
avatar+1068 
0

Another similar way as EP

Width = x, Height = y

2x+2y=32

xy=56

So x+y=16

Square it

x^2+y^2+2xy=256

Plug in xy=56

x^2+y^2=144

Diagonal is sqrt of that, so 12.

 

You are very welcome!

:P

 Jan 8, 2020

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