# Find the lengths of each of the following right triangles. In triangle A, the longer leg is 4 more than the shorter leg and the hypotenuse is 4 more than the longer leg. I think the short leg=12,...

Find the lengths of each of the following right triangles.

In triangle A, the longer leg is 4 more than the shorter leg and the hypotenuse is 4 more than the longer leg.

I think the short leg=12, the long leg =16 and the hypotenuse=20

Is this right?

In triangle B, the short leg =x the long leg =1/2x+11 and the hypotenuse=2x+1

### 1 Answer | Add Yours

For triangle A:

Let l = the longer arm

Let s = the shorter arm

Let h= the hypotenuse

We are given that: `l=4+s ` and `h=4+l`

And we also know that: `h^2=l^2+s^2`

Rearrange the given equations to be a function of l:

`s=l-4 ` and `h=4+l`

Substitute into the Pythagorean equation:

`(4+l)^2=l^2+(l-4)^2`

`16+8l+l^2=l^2+l^2-8l+16`

`16l-l^2=0`

`l^2=16l`

`l=16`

`s=l-4=16-4=12`

`h=l+4=16+4=20`

Therefore, your answers for triangle A are correct :)

For triangle B:

Let x = the short leg

Let l = the long leg = `1/2x+11`

Let h = the hypotenuse = `2x+1`

We know that:

`h^2=l^2+x^2`

`(2x+1)^2=(1/2x+11)^2+x^2`

`4x^2+4x+1=1/4x^2+11x+121+x^2`

`4x^2-1/4x^2-x^2+4x-11x+1-121=0`

`11/4x^2-7x-120=0`

`11x^2-28x-480=0`

a=11; b=-28; c=-480

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(28+-sqrt((-28)^2-4(11)(-480)))/(4(11))`

`x_1=(28+148)/44=4`

`x_2=(28-148)/44=-2.73`

Since the length of a side cannot be negative,` x=x_1=4` :

Therefore:

`s=x=4`

`l=1/2x+11=1/2(4)+11=13`

`h=2x+1=2(4)+1=9`

**Sources:**