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

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, 


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