Question

How do you use the pythagorean theorem to solve a=x-5 b=x 2 c=13?

Answer 1

So acceptable value of x is 10

Explanation:

Here sides are
`a=x-5`
`b=x+2`
`c=13`
Taking c as larger side i.e. hypotenuse
we can apply Pythagorean theorem

`a^2+b^2=c^2`
`=>(x-5)^2+(x+2)^2=13^2`
`=>(x^2-10x+25+x^2+4x+4)=169`
`=>(2x^2-6x-140)=0`
`=>x^2-3x-70=0`
`=>x^2-10x+7x-70=0`
`=>x(x-10)+7(x-10)=0`
`=>(x-10)(x+7)=0`
So acceptable value of x is 10
otherwise a and b will be negative for x = -7