SOLUTION: A rectangle is twice as long as it is wide. If the length and width are each decreased by 4cm, the area is reduced by224cm squared. Find the dimensions of the original rectangle.

Question 58533: A rectangle is twice as long as it is wide. If the length and width are each decreased by 4cm, the area is reduced by224cm squared. Find the dimensions of the original rectangle.
Found 2 solutions by checkley71, ankor@dixie-net.com:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
L=2W
(2W-4)*(W-4)=2W*W-224
THUS
2W^2-12W+16=2W^2-224
2W^2-2W^2-12W+16=224=0
-12W+240=0
-W+20=0
-W=-20
W=20 IS THE WIDTH
L=2*20
L=40 IS THE LENGTH
PROOF
(40-4)(20-4)=40*20-224
36*16=800-224
576=576

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A rectangle is twice as long as it is wide.
x = width; 2x = length
:
If the length and width are each decreased by 4cm,the area is reduced by 224 cm squared.
Original area = 2x^2
:
New dimensions: (x-4; (2x-4); new area given as 2x^2 - 224
:
2x^2 - 224 = (x-4)*(2x-4)
:
2x^2 - 224 = 2x^2 - 12x + 16
:
2x^2 - 2x^2 + 12x = 16 + 224
:
12x = 240
x = 240/12
x = 20
:
Find the dimensions of the original rectangle.
x = 20cm is the width; 40 cm is the length
:
:
Check:
20*40 = 800 sq cm
16*36 = 576 sq cm
---------------------subtract
>>>>>>>>>224 sq cm

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