## Britt_d 2 years ago In a circle with a 12-inch radius, find the length of a segment joining the midpoint of a 20-inch chord and the center of the circle. x =

1. jb1515g

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2. jb1515g

Use triangles to solve.

3. Britt_d

Pythagorean theorem?

4. jb1515g

That's your next step, yes. But it's also important that you understand how I set up that triangle to get to the Pythagorean theorem.

5. Britt_d

Ok this is really confusing how did you set it up?

6. jb1515g

Alright, so we know that the circle has radius 12, yes? So ANY point on the circle that connects with the center is 12. The chord is 20 inches long, and the definition of chord means that both end points touch the circle. We want to find the length of the segment connecting the MIDPOINT of the chord to the center, meaning that the chord will be split in half. That's where the 10 measurements came from. So one side of the triangle is 10, the other (from the center to the point on the chord) is the radius, or 12. That just leaves x. Clearer?

7. Britt_d

Yes so much more clear haha so $x^{2}+10^{2}=12^{2}$ ?

8. jb1515g

yes

9. Britt_d

Oh my gosh thank you so much! so it would be $2\sqrt{22}$ ?

10. jb1515g

You might want to check your math on that last part. That's not what I got.

11. Britt_d

I'm not to sure what I did wrong......? I did 100-144= 44 and then there isn't a perfect square for 44 so 44\2 = 22 so $2\sqrt{22}$

12. jb1515g

$\sqrt{44} =/= 2\sqrt{22}$

13. jb1515g

Evaluate both numbers and see. Sqrt(44) is 6.63, 2*sqrt(22) is 9.38

14. jb1515g

Remember that $\sqrt{ab} = \sqrt{a} * \sqrt{b}$ I think you just simplified the square root wrong.

15. Britt_d

Hmm ok so maybe $2\sqrt{11}$ ?

16. jb1515g

Yeah, that's what I got.

17. Britt_d

Awesome I just forgot to simplify the 4 thank you!

18. jb1515g

Any time. Good job.

19. wolf1728

Basically, you are calculating the length of the apothem. Go to the calculator here: http://1728.org/circsect.htm radius = 12 chord=20 calculator states the apothem = 6.6332 (which is 2 * sqroot(11))

20. goformit100

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