We have 2 different equations with which we need to work to find the answer to this. First off, the area formula is A = L * W. Our area is 12, so 12 = L * W. Then we have the Pythagorean's theorem we have to worry about, which says that [tex] 5^2=l^2+w^2 [/tex]. We are going to solve the area equation for L to get [tex] l=\frac{12}{w} [/tex]. What we are going to do now is sub that into the other formula in place of L to get [tex] 5^2=(\frac{12}{w})^2+w^2 [/tex] and [tex] 25=\frac{144}{w^2}+w^2 [/tex]. In order to solve for w, we need to multiply all those terms by w^2 to get rid of the denominator. That leaves us with this then: [tex] 25w^2=144+w^4 [/tex]. We need to factor that now to solve for w. We will bring everything over to the one side of the equation and set it equal to 0. [tex] w^4-25w^2+144=0 [/tex]. We are going to use a u substitution here to make the factoring easier. We are going to let [tex] u=w^2 [/tex] so [tex] u^2=w^4 [/tex]. Now we will factor [tex] u^2-25u+144=0 [/tex]. When we factor that we get (u - 16)(u - 9). By the Zero Product Property, either u - 16 = 0 or u - 9 = 0. u = 16 and u = 9. But remember that u = w^2, so what we really have, after we make the replacement, is [tex] w^2=16 [/tex] so w = 4, and [tex] w^2=9 [/tex] and w = 3. Now it just so happens that if the length is 4 and the width is 3, the product of those is 12 (the area), and if you remember, in a right triangle with a hypotenuse of 5, the side lengths will be 3 and 4. So your dimensions of the rectangle are that the length is 4 and the width is 3. There you go!
