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 2.5 times the other leg by 4 inches?

Answer 1

Use Pythagoras to establish `x=40` and `h =104`

Explanation:

Let `x` be the other leg
then the hypotenuse `h=5/2x +4`
And we are told the the first leg `y= 96`

We can use Pythagoras' equation `x^2 +y^2 = h^2`
`x^2 +96^2 = (5/2x+4)^2`
`x^2 + 9216 = 25x^2/4 +20x +16`

Reordering gives us
`x^2 - 25x^2/4 -20x +9200 = 0`

Multiply throughout by `-4`
`21x^2 +80x -36800 = 0`

Using the quadratic formula `x= (-b+-sqrt(b^2 - 4ac))/(2a)`

`x =( -(80) +- sqrt(6400 + 3091200))/(-42)`
`x = (-80 +-1760)/42`

so `x = 40` or `x = -1840/42`

We can ignore the negative answer as we re dealing with a real triangle, so the other leg `=40`

The hypotenuse `h = 5*40/2 +4 =104`