What length is the shorter leg of a right triangle that has hypotenuse 50 in long and its perimeter is 112 inches?

2 answers:

8 0

The length of the shorter leg is 14 inches.

We will set up two equations for this; first, for the perimeter. We will call the legs a and b:

a + b + 50 = 112

Subtract 50 from both sides:
a + b + 50 - 50 = 112 - 50
a + b = 62

We will isolate a, so we can use substitution. Subtract b from both sides:
a + b - b = 62 - b
a = 62 - b

Now we will set up our Pythagorean theorem equation:
a² + b² = 50²
a² + b² = 2500

We will substitute our value for a from the first equation into this one:
(62-b)² + b² = 2500
(62 - b)(62 - b) + b² = 2500

Multiplying the binomials, we have:
62*62 - b*62 - b*62 -b(-b) + b² = 2500
3844 - 62b - 62b + b² + b² = 2500

Combining like terms:
3844 - 124b + 2b² = 2500

To set it equal to 0 and solve, we subtract 2500 from both sides:
3844 - 124b + 2b² - 2500 = 2500 - 2500
1344 - 124b + 2b² = 0

In standard form, we have
2b² - 124b + 1344 = 0

Using the quadratic formula,
$\frac{--124\pm \sqrt{(-124)^2-4(2)(1344)}}{2(2)}
\\
\\=\frac{124\pm \sqrt{15376-10752}}{4}
\\
\\=\frac{124\pm \sqrt{4624}}{4}=\frac{124\pm 68}{4}=\frac{124+68}{4}\text{ or }\frac{124-68}{4}
\\
\\=\frac{192}{4}\text{ or }\frac{56}{4}=48\text{ or }14$

The shorter leg is 14 and the longer one is 48.