A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Will medal and fan
Circle P is tangent to the xaxis and the yaxis. 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
anonymous
 one year ago
Will medal and fan Circle P is tangent to the xaxis and the yaxis. 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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439227683307:dw Use Pythagorean Theorem to find length of chord AB

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Triangle 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0May I ask what grade you're in and what course you're working on?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im in 11 and in geometry but im doing online math this summer....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK. You know what a right triangle is, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you identify which sides are the legs and which side is the hypotenuse?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK. In the figure I drew, which side is the hypotenuse?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the other two sides are the legs. Both of these legs are \(r\) unit in length. Do you see where that comes from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(AB^2\) is the length of the hypotenuse squared. OK?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(AO^2\) is the length of one of the legs squared. \(OB^2\) is the length of the other leg squared. OK?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there an easier way of learning this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no not really.. Im really slow with math.. ive never been any good

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If \[\left( AO \right) = r\]then \[\left( AO \right)^2 = r^2\]OK?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well if \[AO^2 = r^2\] and \[OB^2 = r^2\]then \[AO^2 + OB^2 = r^2 + r^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now \[r^2 + r^2 = 2r^2\]You OK with that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright. 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the square root of AB^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0umm.. No clue. Im going into all this blind

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the square root of any number squared? If 2^2 = 4, what is the square root of 4?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right. 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So\[AB^2 = 2r^2\]\[\sqrt{AB^2} = \sqrt{2r^2}\]\[AB = \sqrt{2} \sqrt{r^2}\]And what is the square root of r^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We 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\[AB = r \sqrt{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh okay thank u soo much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats what i had too!!!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.