SOLUTION: The perimeter of a rectangle is 76 ft , and the area of the rectangle is 72 ft^2 . Find the dimensions of the rectangle.

Algebra ->  Rectangles -> SOLUTION: The perimeter of a rectangle is 76 ft , and the area of the rectangle is 72 ft^2 . Find the dimensions of the rectangle.       Log On


   

Question 752841: The perimeter of a rectangle is 76 ft , and the area of the rectangle is 72 ft^2 . Find the dimensions of the rectangle.
Answer by Okeke_Christian(26) About Me  (Show Source):
You can put this solution on YOUR website!
Perimeter of Rectangle = 2l + 2b
Area of Rectangle = l * b
Perimeter of Rectangle
=76ft=2l%2B2b ------------- eqn 1


Area of Rectangle
=72ft%5E2=l%2Ab
We can remove units and solve.
2l%2B2b=76 --------------- eqn. 1
l%2Ab=72 ----------------- eqn. 2
From eqn. 2,
l=%2872%2Fb%29
Substitute 72/b for l in eqn. 1
2l%2B2b=76
2b=76-2l
2b=76-2%2872%2Fb%29
2b=76-%28144%2Fb%29
2b=%2876b-144%29%2Fb
2b%5E2=76b-144
2b%5E2-76b%2B144=0
2b%5E2-72b-4b%2B144=0
2b%28b-36%29-4%28b-36%29=0
%282b-4%29%28b-36%29=0
Either
2b-4=0
OR
b-36=0
Either
2b=4 => b=2
OR
b=36




When you substitute
b=2 into eqn. 2, you get that l=36
AND
When you substitute
b=36 into eqn. 2, you get that l=2
Therefore, dimensions of the rectangle are
l=2 and b=36 OR
l=36 and b=2
Hope I helped!