SOLUTION: I understand the basics of Pythagorean theorem. Right now, I am solving for the unknown side. lets just say were solving for the hypotoneuse. 7^2+9^2 =c^2 49+81=c^2 c= sqre r

Question 806831: I understand the basics of Pythagorean theorem. Right now, I am solving for the unknown side. lets just say were solving for the hypotoneuse.
7^2+9^2 =c^2
49+81=c^2
c= sqre rt of 130. I just don't understand how get the answer 11.402.
2 and 65 goes into 130 and then the square root of 65 is 5 and 13... I just don't undersant how we get 11.402.

THANK YOU! I am going to be taking a test where I cant use a calculator so I need to solve this on paper or in my head.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I understand the basics of Pythagorean theorem. Right now, I am solving for the unknown side. lets just say were solving for the hypotoneuse.
7^2+9^2 =c^2
49+81=c^2
c= sqre rt of 130. I just don't understand how get the answer 11.402.
2 and 65 goes into 130 and then the square root of 65 is 5 and 13
------
The sqrt(65) is not 5 & 13, those are the factors.
---
The sqrt is the number multiplied by itself that = 130
11*11 = 121 and 12*12 = 144, so it's between 11 & 12.
11.410 is close, 11.402*11.402 = 130.005604
11.40175 is closer
11.40175425 is closer still.
I used a calculator to get those.
... I just don't undersant how we get 11.402.

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