Question

in the given right triangle, find the side labeled x. x = box × the triangle is a right triangle with hypotenuse 58, one leg labeled x, and the other leg labeled x + 2

Explanation:

Step1: Apply Pythagorean theorem

For a right triangle, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse. Here, \(a = x\), \(b = x + 2\), \(c = 58\). So, \(x^2+(x + 2)^2 = 58^2\).

Step2: Expand and simplify

Expand \((x + 2)^2\): \(x^2+x^2 + 4x+4 = 3364\). Combine like terms: \(2x^2+4x+4 - 3364 = 0\), \(2x^2+4x - 3360 = 0\). Divide by 2: \(x^2+2x - 1680 = 0\).

Step3: Solve quadratic equation

Factor or use quadratic formula. Factoring: \(x^2+2x - 1680=(x + 42)(x - 40)=0\). Solutions: \(x=-42\) (discard, length can't be negative) or \(x = 40\).

Answer:

\(40\)