Question

The length of the longer leg of a right triangle is 4 inches more than twice the length of the shorter leg. the length of the hypotenuse is 6 inches more than twice the length of the shorter leg. find the side lengths of the triangle.

Answer 1

The Pythagorean's Theorem for our situation would look like this:

$shortleg^2+longleg^2=hypotenuse^2$

So let's call the short leg s, the long leg l and the hypotenuse h. It appears that all our measurements are based on the measurement of the short leg. The long leg is 4 more than twice the short leg, so that expression is l=2s+4; the hypotenuse measure is 6 more than twice the short leg, so that expression is h=2s+6. And the short leg is just s. Now we can rewrite our formula accordingly:

$s^2+(2s+4)^2=(2s+6)^2$

And of course we have to expand. Doing that will leave us with

$s^2+4s^2+8s+8s+16=4s^2+12s+12s+36$

Combining like terms we have

$5s^2+16s+16=4s^2+24s+36$

Our job now is to get everything on one side of the equals sign and solve for s

$s^2-8s-20=0$

That is now a second degree polynomial, a quadratic to be exact, and it can be factored several different ways. The easiest is to figure what 2 numbers add to be -8 and multiply to be -20. Those numbers would be 10 and -2. Since we are figuring out the length of the sides, AND we know that the two things in math that will never EVER be negative are time and distance/length, -2 is not an option. That means that the short side, s, measures 10. The longer side, 2s+4, measures 2(10)+4 which is 24, and the hypotenuse, 2s+6, measures 2(10)+6 which is 26. So there you go!