A right triangle has a hypotenuse that is 13 inches long. If one leg is 2 inches more than twice the other leg, find the length of the longer leg of the right triangle.
You know that your legs are x and 2x + 2. You know your hypotenuse is 13. You know that the Pythagorean Theorem says that the sum of the squares of the legs is equal to the square of the hypotenuse.
So now you know that
x^2 + (2x + 2)^2 = 169
So now you need to do
(2x + 2)*(2x +2) so as to find the square of the second leg. That gets you
4x^2 + 8x + 4.
So now your equation is
x^2 + 4x^2 + 8x + 4 = 169
5x^2 + 8x - 165 = 0
Now you have a quadratic equation and can solve for x.
The Pythagorean theorem is joining th sum of the squares of the cathetus of a right angle triangle and the square of hypotenuse.
a^2 = b^2 + c^2
a = 13 inches = hypotenuse
b = one cathetus
c = the other cathetus = 2 + 2b
We'll substitute the values of the sides of the triangle in Pythagorean theorem:
13^2 = b^2 + (2+ 2b)^2
We'll raise to square and we'll get:
169 = b^2 + 4 + 8b + 4b^2
We'll combine like terms and we'll get:
5b^2 + 8b + 4 - 169 = 0
5b^2 + 8b - 165 = 0
b1 = [-8 + sqrt(64 + 3300)]/10
b1 = (-8+58)/10
b1 = 5
b2 = -66/10
Since a length of a side cannot be negative, we'll keep just the positive value of b = 5 inches long.
The length of the longer leg is:
c = 2 + 2b
c = 2 + 10
c = 12 inches long