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
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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
We can use what we know about chords and their relationship with the center of the circle to solve for missing for missing side lengths. If we start off we’re told that CD which is one of our chords is equal to 20. We’re looking for FD which is part of that segment and we’re looking for OF.
Well, as it’s marked I see that F is the midpoint of CD. Which means FD is going to be half of that. So FD is going to be 10 since the whole distance is 20.
Now, I know that chords that are congruent, I know that they have the same number of congruence markings, are the same distance from the center of the circle. I see that chord AB is 10 units away from the center of the circle, which means OF, which measures the distance from the center of the circle, these two segments must be congruent which means OF must be 10 as well.
So the key here is remembering that the perpendicular bisector will create two congruent line segments and that congruent chords are the same distance away from the center of the circle.
Unit
Circles