For right triangles, where a = shorter leg, b = longer leg, and c = hypotenuse:
[tex] {c}^{2} = {a}^{2} + {b}^{2} \\ c = \sqrt{( {a}^{2} + {b}^{2}) } [/tex]
This is the Pythagorean Theorem.
So they tell us that b is 1in longer than a:
b = a + 1
and hypotenuse (c) is 9in longer than a:
c = a + 9
Now we plug in the Pythagorean Theorem to solve for a:
[tex] {(a + 9)}^{2} = {a}^{2} + {(a + 1)}^{2} \\ {a}^{2} + 18a + 81 = {a}^{2} + {a}^{2} + 2a + 1 \\ - {a}^{2} + 16a + 80 = 0 \\ {a}^{2} - 16a - 80 = 0 \\ [/tex]
From here, since not favorable with normal methods, use the quadratic equation to find a:
[tex]x \: (our \: a) = \\ x= (- b + - \sqrt{( {b}^{2} - 4ac)} ) \div 2a[/tex]
where b = -16, and c = -80
[tex]a = 16 + - \sqrt{ {16}^{2} - 4(1)( - 80)} \\ \div \: 2(1) \\ a = 16 + - \sqrt{256 + 320} \div 2 \\ a = 16 + - \sqrt{576} \div 2[/tex]
[tex]a = (16 + 24) \div 2 \\ and \: a = (16 - 24) \div 2 \\ so \: a = 40 \div 2 = 20 \\ and \: a = - 8 \div 2 = - 4[/tex]
We can only have positive lengths in real-life, so our a = 20
Remember our original a was the shorter leg,
then b = a + 1 = 20 + 1 = 21 (our longer leg),
and our hypotenuse (c) = a + 9
c = 20 + 9 = 29
Now our triangle's sides are:
20 in, 21 in, and 29 in
