SOLUTION: A rectangle is 4 times as long as it is wide. a second rectangle is 5 cm longer and 2 cm wider than the first. the are of the second rectangle is 270 square cm greater than the fir

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Question 186249: A rectangle is 4 times as long as it is wide. a second rectangle is 5 cm longer and 2 cm wider than the first. the are of the second rectangle is 270 square cm greater than the first. What are the dimensions of the original rectangle?


Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A rectangle is 4 times as long as it is wide. a second rectangle is 5 cm longer and 2 cm wider than the first. the are of the second rectangle is 270 square cm greater than the first. What are the dimensions of the original rectangle?
:
Let x = width of the 1st rectangle:
:
"A rectangle is 4 times as long as it is wide."
4x = length of the 1st rectangle
then
4x^2 sq/cm = area of 1st rectangle
:
a second rectangle is 5 cm longer and 2 cm wider than the first."
(4x + 5) = length of 2nd rectangle
(x + 2) = width of 2nd rectangle
then find the area of the 2nd rectangle
FOIL:(4x+5)*(x+2) = 4x^2 + 8x + 5x + 10 = 4x^2 + 13x + 10 sq cm
:
the area of the second rectangle is 270 square cm greater than the first.
4x^2 + 13x + 10 - 4x^2 = 270
4x^2 - 4x^2 + 13x = 270 - 10
13x = 260
x = 260%2F13
x = 20
:
What are the dimensions of the original rectangle?
4(20) = 80 cm is the length
80 by 20 is the dimensions of the original rectangle
;
;
Check solution, find the area of both rectangles
4(20) + 5 = 85 cm
20 + 2 = 22 cm
85 * 22 = 1870 sq cm area of 2nd rectangle
and
80 * 20 = 1600 sq cm area of 1st rectangle
:
1870 - 1600 = 270 confirms our solution