Question 1121823: Find the dimensions of a rectangle whose area is 192 cm2 and whose perimeter is 56 cm. (Enter your answers as a comma-separated list.)
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Half perimeter is L+W=28 cm
L*W=192 cm^2
L=28-W
(28-W)W=-W^2+28W, and that equals 192
0=W^2-28W+192
0=(W-16)(W-12)
W=16 or 12
Length longer by convention, so length is 16 cm and width is 12 cm.
(12, 16) for (W, L)
Answer by ikleyn(52782)  (Show Source):
You can put this solution on YOUR website! .
Let me show you a quick, elegant and unexpected method of solving such problems.
You are given that the perimeter of the rectangle is 56 cm; hence, the sum of the length and the width is one half of that:
x + y = 28, where x is the length, and y is the width.
Then the average of the length and the width is one half of 28, i.e. 14.
It is clear that the values of x and y are remoted at the same value/distance "u" from 14, so we can write
x = 14 + u,
y = 14 - u.
Then the area is xy = (14+u)*(14-u) =  , and it is equal to 192, according to the condition.
Hence, = 192, which gives  = 196 - 192 = 4, and then u =  = 2.
Thus the length is x = 14 + u = 14 + 2 = 16,
and the width is x = 14 - u = 14 - 2 = 12.
Answer. The dimensions of the rectangle are 12 cm and 16 cm.
Solved.
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See the lesson
- Three methods to find the dimensions of a rectangle when its perimeter and the area are given
in this site.
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