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?

Question

Eli Assouline · Accepted Answer

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

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#

Savannah Anderson · Answer

undefined To find the hypotenuse and the other leg of a right triangle, given that one leg is 96 inches and the length of the hypotenuse exceeds 2 times the other leg by 4 inches, we can use the Pythagorean theorem and algebraic equations.

Let's denote the other leg as x and the hypotenuse as y.

According to the Pythagorean theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Therefore, we have the equation:

x^2 + 96^2 = y^2

Given that the length of the hypotenuse exceeds 2 times the other leg by 4 inches, we can write another equation:

y = 2x + 4

Now, we can solve these two equations simultaneously to find the values of x and y.

By substituting the second equation into the first equation, we get:

x^2 + 96^2 = (2x + 4)^2

Expanding and simplifying this equation, we have:

x^2 + 9216 = 4x^2 + 16x + 16

Rearranging and simplifying further, we get:

3x^2 + 16x - 9200 = 0

Now, we can solve this quadratic equation for x using factoring, completing the square, or the quadratic formula. Once we find the value of x, we can substitute it back into the second equation to find the value of y.

Please note that the quadratic equation may have two solutions, but we can discard any negative values since lengths cannot be negative in this context.