# Thread: applications of quadratic equations, geometric problem

1. ## applications of quadratic equations, geometric problem

The problem: find the dimensions of a rectangle whose area is 180cm2 and whose perimeter is 54cm.

What I know: area = length*width. If the perimeter is 54cm, then would the length * width be 1/2 the perimeter?

I am also unsure about the dimensions because I do not know how to represent 180cm2 in the equation. I tried to use a metric converter to feet and inches, but I had difficulty converting CM2 to feet and inches. Not that the dimensions need to be converted, I think all the calculations can be done in centimeters, but I am confused.

2. ## Re: applications of quadratic equations, geometric problem

$A = \ell \cdot w = 180$

$P = 2(\ell + w) = 54$

$\ell + w = 27$

$\ell = 27-w$

$A = (27-w)w = 27w-w^2 = 180$

$w^2 - 27w + 180 = 0$

$(w-12)(w-15) = 0$

$w = 12 \text{ or }w=15$

$w=12 \Rightarrow \ell = 27-12 = 15$

$w=15 \Rightarrow \ell = 27-15 = 12$

so the dimensions of the rectangle are $15cm \times 12cm$ in either horizontal or vertical orientation.

3. ## Re: applications of quadratic equations, geometric problem

Originally Posted by retro
The problem: find the dimensions of a rectangle whose area is 180cm2 and whose perimeter is 54cm.

What I know: area = length*width. If the perimeter is 54cm, then would the length * width be 1/2 the perimeter?
No, why would you even think that? "Perimeter" is the total distance around the rectangle, "length plus width plus length plus width"= 2*length+ 2*width= 2(length+ width). Half the perimeter is length plus width, NOT "length times width".

I am also unsure about the dimensions because I do not know how to represent 180cm2 in the equation. I tried to use a metric converter to feet and inches, but I had difficulty converting CM2 to feet and inches. Not that the dimensions need to be converted, I think all the calculations can be done in centimeters, but I am confused.