+0  
 
0
88
2
avatar+936 

 

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 Jul 15, 2020
 #1
avatar+781 
+1

Not sure about the first, but for the second:

 

We can apply the pythagorean theorem to this. The hypotenuse must be 6a+1, and the two legs a+1 and 6a.

 

\(a^2+b^2=c^2\\ (a+1)^2+(6a)^2=(6a+1)^2\\ a^2+2a+1+36a^2=36a^2+12a+1\\ a^2+2a+1=12a+1\\ a^2-10a=0\\ a=0, 10\)

 

We find that a can be 0 or 10. However, the only correct value is 10 because if we plug in the value 0 into the given sides, one of them will have a side length of 0.

 Jul 15, 2020
 #2
avatar+936 
0

Thank you!

qwertyzz  Jul 15, 2020

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