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For what value of a is there a right triangle with sides a+1 ,6a , and 6a+1?

Atroshus  Aug 29, 2017
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+1

 

We have that

 

(a + 1)^2  + (6a)^2  = (6a + 1)^2    simplify

 

a^2 + 2a+ 1 + 36a^2   =  36a^2 + 12a + 1

 

a^2  + 2a  = 12a 

 

a^2  - 10a  = 0

 

a ( a - 10)  = 0     

 

So....a  = 0  (reject )  or      a  = 10  which is correct

 

 

cool cool cool

CPhill  Aug 29, 2017

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