SOLUTION: Help!
A rectangle has an area of 192 and a perimeter of 56. Find the dimensions for length and width. (A = Lw & P = 2L + 2W)
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A rectangle has an area of 192 and a perimeter of 56. Find the dimensions for length and width. (A = Lw & P = 2L + 2W)
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Question 105926: Help!
A rectangle has an area of 192 and a perimeter of 56. Find the dimensions for length and width. (A = Lw & P = 2L + 2W) Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! A) L*w=192
B) 2L+2w=56
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B) 2L=56-2w subtract 2w from each side.
L=28-w divide each side by 2.
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A) w(28-w)=192 replace L with 28-w
28w-w^2=192
-w^2+28w-192=0
w^2-28w+192=0 multiply each side by -1
(w-12)(w-16)=0
w=12, w=16
Either number can be L and the other can be w.
Check:
12*16=192
24+32=56
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Ed
