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?

1 Answer
May 13, 2016

Answer:

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#