The hypotenuse of a right triangle is 39 inches, and the length of one leg is 6 inches longer than twice the other leg. How do you find the length of each leg?

Nov 14, 2016

The legs are of length $15$ and $36$

Explanation:

Method 1 - Familiar triangles

The first few right angled triangles with an odd length side are:

$3 , 4 , 5$

$5 , 12 , 13$

$7 , 24 , 25$

Notice that $39 = 3 \cdot 13$, so will a triangle with the following sides work:

$15 , 36 , 39$

i.e. $3$ times larger than a $5 , 12 , 13$ triangle ?

Twice $15$ is $30$, plus $6$ is $36$ - Yes.

$\textcolor{w h i t e}{}$
Method 2 - Pythagoras formula and a little algebra

If the smaller leg is of length $x$, then the larger leg is of length $2 x + 6$ and the hypotenuse is:

$39 = \sqrt{{x}^{2} + {\left(2 x + 6\right)}^{2}}$

$\textcolor{w h i t e}{39} = \sqrt{5 {x}^{2} + 24 x + 36}$

Square both ends to get:

$1521 = 5 {x}^{2} + 24 x + 36$

Subtract $1521$ from both sides to get:

$0 = 5 {x}^{2} + 24 x - 1485$

Multiply both sides by $5$ to get:

$0 = 25 {x}^{2} + 120 x - 7425$

$\textcolor{w h i t e}{0} = {\left(5 x + 12\right)}^{2} - 144 - 7425$

$\textcolor{w h i t e}{0} = {\left(5 x + 12\right)}^{2} - 7569$

$\textcolor{w h i t e}{0} = {\left(5 x + 12\right)}^{2} - {87}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(5 x + 12\right) - 87\right) \left(\left(5 x + 12\right) + 87\right)$

$\textcolor{w h i t e}{0} = \left(5 x - 75\right) \left(5 x + 99\right)$

$\textcolor{w h i t e}{0} = 5 \left(x - 15\right) \left(5 x + 99\right)$

Hence $x = 15$ or $x = - \frac{99}{5}$

Discard the negative solution since we are seeking the length of the side of a triangle.

Hence the smallest leg is of length $15$ and the other is $2 \cdot 15 + 6 = 36$