## anonymous 5 years ago The perimeter of a rectangle is 110m , and the area of the rectangle is 54m^2 . Find the dimensions of the rectangle.

1. anonymous

Any ideas?

2. anonymous

You have two equations - one for area and one for perimeter. Then, if your sides are: a = length b = width $P=2a+2b$and$A=ab$From your numbers, we get$110=2a+2b$and$54=ab$Choosing any one of the last two equations, we can solve for one of the variables and sub. it into the remaining equation, so$54=ab \rightarrow b=\frac{54}{a}$Substituting this into the perimeter equation,$110=2a+2\frac{54}{a} \rightarrow 55=a+\frac{54}{a} \rightarrow 55a=a^2+54$This is quadratic in a.$a^2-55a+54=0\rightarrow a=\frac{-(-55){\pm}\sqrt{(-55)^2-4(1)(54)}}{2(1)}$i.e.$a=1,54$ then $b=54,1$ respectively.

3. anonymous

So your length/width combinations are: (a,b) = (1,54) or (54,1).