Question

One leg of a right triangle is 8 inches more than four times the length of the other leg. the hypotenuse is 9 in more than the shorter leg. find the lengthsof all the sides?

Answer 1

`x=(1+sqrt5)/2` units
`y=10+2sqrt5` units
`z= (9+9sqrt5)/2` units

Explanation:

Let `x` be the shorter leg
Let `y` be the longer leg
Let `z` be the hypotenuse

So:
`y=4x+8`
`z=9x`

Just use the Pythagorean theorem:
`x^2+y^2= z^2`

Substitute for `y` and `z`:
`x^2+(4x+8)^2=(9x)^2`

Simplify:
`x^2+16x^2+64x+64= 81x^2`

Set the expression equal to 0:
`64x^2-64x-64=0`

Factor with GCF:
`64(x^2-x-1)=0`

Use the quadratic formula to solve for x:
`x=(1+-sqrt(1-4(1)(-1)))/(2(1))`
`x=(1+-sqrt5)/2`

Since sides lengths can only be positive:
`x=(1+sqrt5)/2`

`y= 4((1+sqrt5)/2)+8=`
`y= 2+2sqrt5+8=`
`y=10+2sqrt5`

`z= 9((1+sqrt5)/2)=`
`z= (9+9sqrt5)/2`