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Khan academy circle geometry5/31/2023 X in the triangle below? OK, so we can pull out the The answer makes sense, it is a little larger than the first chord length because it is closer to the center point of the circle. Solve for 1/2 the chord length and double it to get the second chord length. You then solve for the radius of the circle, and use that for the second side of your second triangle. 2 sides are given in the first triangle, distance from center and 1/2 the chord length. The idea was just that both cords form a right triangle with the hypotenuse equaling the radius of the circle. Just double that to get the length of the second cord So we already know 2 sides for this triangle and just need to solve for L and double it to get the second chord length. Moving on to the second cord, using the same logic, you have another triangle with these 3 sides: 17, r, and L. Now you have a triangle with sides: 20, 29, and r.ΔΆ0 is from the center to the chord, 29 is half the chord length, r is the hypotenuse of the triangle that is the same as the radius of the circle. If you draw a line from the center point of the circle to the first chord and intersect it at 90 degrees, the line will intersect the chord right in the middle, or bisect it. For this problem the chords do not intersect so you cant use the property in this video.
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