SOLUTION: A rectangle has perimeter 54 inches, and its area is 162 sq-inches. Find the dimensions of this rectangle. (For purposes of this problem only, the width is shorter than the length.

Algebra ->  Rectangles -> SOLUTION: A rectangle has perimeter 54 inches, and its area is 162 sq-inches. Find the dimensions of this rectangle. (For purposes of this problem only, the width is shorter than the length.      Log On


   

Question 1030218: A rectangle has perimeter 54 inches, and its area is 162 sq-inches. Find the dimensions of this rectangle. (For purposes of this problem only, the width is shorter than the length.)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width = +w+
The perimeter = +54+ in
The sides are:
+w+, +w+, +%28+54+-+2w+%29+%2F+2+ , +%28+54+-+2w+%29+%2F+2+
-------------------
The area = +162+ in2
+w%2A%28+54+-+2w+%29+%2F+2+=+162+
+w%2A%28+54+-+2w+%29+=+324+
+-2w%5E2+%2B+54w+=+324+
+2w%5E2+-+54w+=+-324+
+w%5E2+-+27w+=+-162+
Complete the square:
+w%5E2+-+27w+%2B+%2827%2F2%29%5E2+=+-162+%2B+%28+27%2F2+%29%5E2+
+w%5E2+-+27w+%2B+729%2F4+=+-648%2F4+%2B+729%2F4+
+w%5E2+-+27w+%2B+729%2F4+=+81%2F4+
+%28+w+-+27%2F2+%29%5E2+=+%28+9%2F2+%29%5E2+
+w+-+27%2F2+=+9%2F2+
+w+=+36%2F2+
+w+=+18+
--------------
The length is:
+%28+54+-+2w+%29%2F2+=+%28+54+-+2%2A18+%29%2F2+
+54+-+2w+=+%28+54+-+36+%29%2F2+
+54+-+2w+=+18%2F2+
+54+-+2w+=+9+
-----------------------
The dimensions are 9 x 18
-----------------------
check:
Perimeter = +2%2A9+%2B+2%2A18+
Perimeter = +18+%2B+36+
Perimeter = +54+ in
------------------
Area = +9%2A18+
Area = +162+ in2