anonymous
  • anonymous
AO=12 and BC=25. What is AB? I think it's 37, but I'm not sure.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
whpalmer4
  • whpalmer4
Close...
whpalmer4
  • whpalmer4
What is the length of the hypotenuse (OB)?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ParthKohli
  • ParthKohli
|dw:1369717643930:dw|
ParthKohli
  • ParthKohli
\(OC = OA = 12\) because they are the radii of the same circle.
ParthKohli
  • ParthKohli
Also, as whpalmer asked, what is the hypotenuse, or \(OB\)? A good start might be noticing that \(OB = OC + CB\).
whpalmer4
  • whpalmer4
or, just tell us how you came up with 37...then we can see if it is a simple arithmetic error, or how you approached the problem, or whatever...
KenLJW
  • KenLJW
OB^2=OA^2+AB^2 if angle A = 90 37^2=12^2+AB^2 other wise you have the sine angle formula
whpalmer4
  • whpalmer4
I assumed (and ParthKohli too, I'm sure) that because only one intersection point is shown, AB is tangent to the circle, which means that OAB is a right angle.
KenLJW
  • KenLJW
imagine A approaching C
KenLJW
  • KenLJW
Not enough information given without assumption angle A = pi/2
KenLJW
  • KenLJW
When you assume you make a retriceof u and me
KenLJW
  • KenLJW
retriceof u and me
KenLJW
  • KenLJW
I guess it doesn't like that word
anonymous
  • anonymous
That didn't help at all. You never even told me how to get that answer....
whpalmer4
  • whpalmer4
you also didn't tell us how you got your answer...
anonymous
  • anonymous
adding them...Duh..
whpalmer4
  • whpalmer4
adding what, exactly, duh?
KenLJW
  • KenLJW
Solve 37^2=12^2+AB^2
anonymous
  • anonymous
The numbers..DUH
anonymous
  • anonymous
And uh...I need to know AB to do that. DUH!!
anonymous
  • anonymous
emily the ans for ab is 35. its a right angle triangle
anonymous
  • anonymous
Now wait...you're going "duh" at everyone else and yet you can't figure out what AB is from the equation 37^2=12^2+AB^2 ???
KenLJW
  • KenLJW
I used the Pythagorean theorem a^2+b^2=h^2 h = sqrt(a^2+b^2) or in out case b=sqrt(h^2-a^2)
anonymous
  • anonymous
@Kapt_Crazy So I was right? The answer is 35? :)
KenLJW
  • KenLJW
Generally you shouldn't just give quantity but show all work so it's easy checked or to show how it's done
KenLJW
  • KenLJW
not give
anonymous
  • anonymous
yeah u right :)
anonymous
  • anonymous
No, you didn't say 35. You said "I think it's 37" which you absurdly arrived at by adding AO and BC together. Also, 35 is only correct if AB is tangent to the circle at A. But since that information isn't given, then its only an assumption.
anonymous
  • anonymous
qweqwe if it was not a right angle triangle they would give u more info, like one of the angles or sumthing
anonymous
  • anonymous
Oh, thankyou for going back and checking if I said 35. :) @qweqwe123123123123111
KenLJW
  • KenLJW
Yes they would have to give 2 angles to solve
anonymous
  • anonymous
The assumption is probably a safe one, @Kapt_Crazy, But UNLESS THAT INFO IS GIVEN EXPLICITLY, it's still just an assumption. And the number of poorly formed and incomplete problems that are posted here every single day are legion, so there's no good reason to assume that this problem is stated any better than any other.

Looking for something else?

Not the answer you are looking for? Search for more explanations.