SOLUTION: Please help me solve the following equation, "A rectangle is 2 centimeters longer than it is wide. If both the length and width are doubled, its area is increased by 72 square cen
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Question 737218: Please help me solve the following equation, "A rectangle is 2 centimeters longer than it is wide. If both the length and width are doubled, its area is increased by 72 square centimeters. Find the dimensions of the original rectangle."
Found 2 solutions by lwsshak3, josmiceli:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Please help me solve the following equation, "A rectangle is 2 centimeters longer than it is wide. If both the length and width are doubled, its area is increased by 72 square centimeters. Find the dimensions of the original rectangle."
***
let x=width of rectangle
x+2=length of rectangle
Area=length*width
..
area of original rectangle=x(x+2)=x^2+2x
area of enlarged rectangle=2x*2(x+2)=2x(2x+4)=4x^2+8x
area of enlarged rectangle-area of original rectangle=72
(4x^2+8x)-(x^2+2x)=72
4x^2+8x-x^2-2x)=72
3x^2+6x-72=0
x^2+2x-24=0
(x+6)(x-4)=0
x=-6 (reject)
or
x=4
x+2=6
width of original rectangle=4 cm^2
length of original rectangle=6 cm^2
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = the width in cm of the original window
= the length in cm of the original window
The area of the original window is
-----------------
The widow with increased area is:
-----------------
By substitution:
( by inspection )
( reject the other solution - can't be negative )
The original rectangle is 4 x 6 cm2
------------
check:
If both the length and width are doubled,
it is 8 x 12 cm2
cm2
-----------------
cm2
OK
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