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## anonymous one year ago Will medal and fan Circle P is tangent to the x-axis and the y-axis. If the coordinates of the center are (r, r), find the length of the chord whose endpoints are the points of tangency. 2r r² r√2

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1. anonymous

2. anonymous

Find the coordinates of the other endpoint if the midpoint is M(8, 2) and the other endpoint is P(5, 6). (6.5, 4) (21, 10) (11, -2)

3. anonymous

|dw:1439227683307:dw| Use Pythagorean Theorem to find length of chord AB

4. anonymous

i dont know how..

5. anonymous

Triangle AOB is a right triangle. The lengths of the sides are related byt the Pythagorean Theorem. If a right triangle has legs of lengths a and b, and a hypotenuse of length c, these lengths are related by$a^2 + b^2 = c^2$Have you seen this before?

6. anonymous

no...

7. anonymous

May I ask what grade you're in and what course you're working on?

8. anonymous

im in 11 and in geometry but im doing online math this summer....

9. anonymous

OK. You know what a right triangle is, right?

10. anonymous

yes.

11. anonymous

Can you identify which sides are the legs and which side is the hypotenuse?

12. anonymous

i think so

13. anonymous

OK. In the figure I drew, which side is the hypotenuse?

14. anonymous

AB?

15. anonymous

That's right.

16. anonymous

So the other two sides are the legs. Both of these legs are $$r$$ unit in length. Do you see where that comes from?

17. anonymous

ya

18. anonymous

Good.

19. anonymous

The Pythagorean Theorem says that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. In our diagram$\left( AB \right)^2 = \left( AO \right)^2 + \left( OB \right)^2$Do you understand so far?

20. anonymous

no u lost me

21. anonymous

$$AB^2$$ is the length of the hypotenuse squared. OK?

22. anonymous

ok

23. anonymous

$$AO^2$$ is the length of one of the legs squared. $$OB^2$$ is the length of the other leg squared. OK?

24. anonymous

ok

25. anonymous

Is there an easier way of learning this?

26. anonymous

Now, for any right triangle, the lengths of the sides are related, no matter the size. The relationship is, as stated earlier$\left( AB \right)^2 = \left( AO \right)^2 + \left( OB \right)^2$Now, we know that the lengths of the legs (AO & OB) are equal to $$r$$. So, substituting that value in, we get$\left( AB \right)^2 = r^2 + r^2 = 2r^2$Understand?

27. anonymous

no not really.. Im really slow with math.. ive never been any good

28. anonymous

If $\left( AO \right) = r$then $\left( AO \right)^2 = r^2$OK?

29. anonymous

ok

30. anonymous

And if $\left( OB \right) =r$then$\left( OB \right)^2 = r^2$Then by adding them together$\left( AB \right)^2 + \left( OB \right)^2 = r^2 + r^2$Still with me?

31. anonymous

um kinda

32. anonymous

Well if $AO^2 = r^2$ and $OB^2 = r^2$then $AO^2 + OB^2 = r^2 + r^2$

33. anonymous

okay

34. anonymous

Now $r^2 + r^2 = 2r^2$You OK with that?

35. anonymous

yas

36. anonymous

Alright. Putting all of that together, we have$AB^2 = AO^2 + OB^2$$AB^2 = r^2 + r^2$$AB^2 = 2r^2$Now the question requires that we solve for AB. To do that, you need to take the square root of both sides. Can you do that?

37. anonymous

i dont think so

38. anonymous

What is the square root of AB^2?

39. anonymous

umm.. No clue. Im going into all this blind

40. anonymous

What is the square root of any number squared? If 2^2 = 4, what is the square root of 4?

41. anonymous

2??

42. anonymous

Right. And if 3^2 = 9, the square root of 9 is 3. So the square root of any number squared is that number. Now just do it with a variable instead of a number. So what is the square root of AB^2?

43. anonymous

ohhh okay

44. anonymous

So$AB^2 = 2r^2$$\sqrt{AB^2} = \sqrt{2r^2}$$AB = \sqrt{2} \sqrt{r^2}$And what is the square root of r^2?

45. anonymous

umm i have no clue

46. anonymous

We just covered it. The square root of any number squared is that same number. So the square root of r^2 is r. So your answer is$AB = r \sqrt{2}$

47. anonymous

ohhh okay thank u soo much

48. anonymous

You're welcome.

49. anonymous

thats what i had too!!!!

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