# Rectangle with perimeter 68 feet and diagonal 26 feet, then what is its width?

May 13, 2018

$w = 24$

#### Explanation:

I came to check an answer, but it's gone.

The length $l$ and width $w$ satisfy

${l}^{2} + {w}^{2} = {26}^{2}$

I've probably been doing these for too long, but a diagonal or hypotenuse of $26 = 2 \setminus \times 13$ probably means we have the right triangle ${\left(2 \setminus \cdot 5\right)}^{2} + {\left(2 \setminus \cdot 12\right)}^{2} = {\left(2 \setminus \cdot 13\right)}^{2}$

$2 l + 2 w = 68$

$l + w = 34$

We already see the solutions are 10 and 24. But let's keep going.

$w = 34 - l$

${\left(l + w\right)}^{2} = {34}^{2}$

${l}^{2} + {w}^{2} + 2 l w = {34}^{2}$

$2 l w = {34}^{2} - {26}^{2}$

$2 l \left(34 - l\right) = {34}^{2} - {26}^{2}$

$0 = 2 {l}^{2} - 68 l + \left(34 - 26\right) \left(34 + 26\right)$

$0 = 2 {l}^{2} - 68 l + 480$

$0 = {l}^{2} - 34 l + 240$

$\left(l - 10\right) \left(l - 24\right) = 0$

$l = 10$ and $w = 24$ or vice versa. We'll call the long side the width.

$w = 24$

I'm too sleepy to look at this any more . Goodnight.