SOLUTION: Can you help me with this? The length of a rectangle is 3ft longer then its width. if the perimeter of the triangle is 34 feet then what is the area?

Algebra ->  Rectangles -> SOLUTION: Can you help me with this? The length of a rectangle is 3ft longer then its width. if the perimeter of the triangle is 34 feet then what is the area?       Log On


   



Question 866446: Can you help me with this?
The length of a rectangle is 3ft longer then its width. if the perimeter of the triangle is 34 feet then what is the area?

Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,

P = 2L + 2w = 34ft
or L + w = 17 L+=+w%2B3
(w+3)+w = 17
2w = 14
w = 7ft and L = 10ft 7%2B3%29
A = Lw = (10ft)(7ft) = 70ft^2

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
let the length of a rectangle be L
let the width of a rectangle be W
by definition:
the perimeter is P=2L%2B2W and the area is A=L%2AW
given: the length of a rectangle L is 3ft longer then its width W => L=W%2B3ft......eq.1
if the perimeter of the triangle is P=34ft then we have:
P=2L%2B2W ...substitute L and P
34ft=2%28W%2B3ft%29%2B2W ...solve for W
34ft=2W%2B6ft%2B2W
34ft-6ft=4W
28ft=4W
28ft%2F4=W
highlight%287ft=W%29
now find L

L=W%2B3ft......eq.1
L=7ft%2B3ft
highlight%28L=10ft%29
finally, you can find the area:
A=L%2AW ...............plug in highlight%28L=10ft%29 and highlight%287ft=W%29

A=10ft%2A7ft
highlight%28A=70ft%5E2%29