Question

One leg of a right triangle is 96 inches. How do you find the hypotenuse and the other leg if the length of the hypotenuse exceeds 2 times the other leg by 4 inches?

Answer 1

hypotenuse `180.5`, legs `96` and `88.25` approx.

Explanation:

Let the known leg be `c_0`, the hypotenuse be `h`, the excess of `h` over `2c` as `delta` and the unknown leg, `c`. We know that `c^2+c_0^2=h^2`(Pytagoras) also `h-2c = delta`. Subtituting according to `h` we get: `c^2+c_0^2=(2c+delta)^2`. Simplifiying, `c^2+4delta c+delta^2-c_0^2=0`. Solving for `c` we get.
`c = (-4delta pm sqrt(16delta^2-4(delta^2-c_0^2)))/2`
Only positive solutions are allowed
`c = (2sqrt(4delta^2-delta^2+c_0^2)-4delta)/2=sqrt(3delta^2+c_0^2)-2delta`