Question

The length of a rectangle is 59 inches greater than twice the width. If the diagonal is 2 inches more than the​ length, find the dimensions of the rectangle.

Answer 1

so.. hmm notice the picture here

we know the length is 59 more than twice the width,
twice the width, 2 * w, or 2w
59 more than that
2w + 59

we also know that, whatever that is, the diagonal of it is,
2 more inches than that, or
(2w + 59) + 2

now.. use the pythagorean theorem

`\bf c^2=a^2+b^2\implies \qquad \begin{cases} a=2w+59\\ b=w\\ c=(2w+59)+2\to 2w+61 \end{cases} \\\\\\ (2w+61)^2=(2w+59)^2+(w)^2 \\ \qquad\qquad \uparrow\qquad \qquad\qquad \qquad \uparrow \\ \textit{expand using binomial theorem, or FOIL}`

you'd end up with a quadratic, after simplifying, solve for "w"