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Chords and a Circle's Center - Problem 3
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
In some classes you’ll apply the Pythagorean Theorem to what you know about circles. In this problem we’re told that line segment AB, which is a chord, is 24. We see that CD is perpendicular that passes through the centre, which means it has to bisect that segment AB, that chord. We are also told that CD is equal to 5
So I’m going to write that CD must be 5 and DB must be half of 24, which 12. We’re being asked to find what’s the radius of the circle. So just looking at it right here, it’s a little confusing to figure out how you’re going to determine the answer.
What I’m going to do is I’m going to realize that if I swung this radius and I switched it all the way up and if I drew a radius from C, which is the centre of the circle, to point B now I’ve created a right triangle.
So the trick to this problem is realizing that you could draw in a radius forming a right triangle. So I’m going to redraw that right triangle, where we have point D, we have point B and we have the center C.
We know that this red segment has to be our radius because the radius is the same no matter where you are in the circle. So I’m going to label the hypotenuse as r, hypotenuse is opposite the 90 degree angle. We see that this side has to be 12 and DC is 5. Now one of the things that you should hopefully have memorized is your Pythagorean triples and I see that this is a Pythagorean triple because we have 5, 12 which means r has to be 13. So our radius here is 13.
Now let’s Mr. McCall I don’t have them memorized. Here’s how you’d solve it, you’d say that the 5² plus 12² is equal to r². Then you say that 5² is 25, 12² is 144 and that equals R². 25 and 144 is 169, which is equal 169 which is equal to r² and then you’d take the square root of both sides and the square root of 169 is positive or negative 13. But since we’re talking about a distance here, we’re going to take the positive number, square root of r² is equal to r and we’d end up with the same answer.
I think it’s just a little bit easier to have your Pythagorean triples memorized. The key to the problem, the key to this problem is drawing in that radius so you have a right angle.
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