Question

The length of a rectangle is 3131 inches greatergreater than twice the width. if the diagonal is 2 inches more than the​ length, find the dimensions of the rectangle.

Answer 1

We have a rectangle with a diagonal included. That means we have to use Pythagorean's Theorem to find the missing dimensions. We start with the width as w. The length is 31 more than twice the width, so the length is 2w+31. The hypotenuse (diagonal) is 2 more than the length, so the diagonal measures 2w+33. In a right triangle, the hypotenuse is always the longest side, so we set up Pythagorean's Theorem using the bolded values as such: $(2w+33)^2=w^2+(2w+31)^2$. Distributing by multiplying we simplify to $4w^2+132w+1089=w^2+4w^2+124w+961$. We need to solve for w. Once we do that we can use that value to find the length. What's nice is that the 4w^2 terms cancel each other out. So when we combine like terms and get them all on one side of the equals sign to factor we have $w^2-8w-128=0$. The 2 numbers that add up to -8 and multiply to -128 are 16 and -8. So the width is either 16 inches or the width is -8 inches. The 2 things in math that will never EVER be negative are time and distance/length. So -8 is out. The width is 16 inches. That means that the length is 2(16)+31, which is 63 inches. There you go!