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

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

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Answer 1 · 2016-04-22T17:10:24

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

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

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