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Question Video: Finding the Length of a Segment Using the Properties of Chords Mathematics

Points 𝑋 and π‘Œ are midpoints of segments 𝐴𝐡 and 𝐢𝐷 respectively, and point 𝑀 is at the center of the circle. If 𝐴𝐡 = 60, what is πΆπ‘Œ?

03:20

Video Transcript

Points 𝑋 and π‘Œ are midpoints of segments 𝐴𝐡 and 𝐢𝐷, respectively, and point 𝑀 is at the center of the circle. If 𝐴𝐡 equals 60, what is πΆπ‘Œ?

We can begin by highlighting that we’re told that 𝑋 and π‘Œ are the midpoints of segments 𝐴𝐡 and 𝐢𝐷, respectively. Line segments 𝐴𝐡 and 𝐢𝐷 are in fact chords of the circle because in each line segment, the endpoints both lie on the circle. The point 𝑀 is the center of the circle.

Now, consider that since 𝑋 and π‘Œ are midpoints of chords 𝐴𝐡 and 𝐢𝐷, respectively, then we can say that they have been bisected by points 𝑋 and π‘Œ. Since line segments 𝑋𝑀 and π‘Œπ‘€ pass through the center 𝑀, then we recall that this means that 𝑋𝑀 and π‘Œπ‘€ are both perpendicular bisectors of the respective chords 𝐴𝐡 and 𝐢𝐷. From the markings on the diagram, we can observe that the line segments 𝑋𝑀 and π‘Œπ‘€ are marked as congruent. This means that the distance from the center 𝑀 to the chord 𝐴𝐡 is the same as the distance from 𝑀 to the chord 𝐢𝐷. We can say then that the chords are equidistant from the center.

Let’s consider for a moment why these distances have to be the perpendicular distance from the chord to the center, by taking a different circle with center 𝑃. Just because we can draw two chords and draw any two congruent line segments from each chord to the center would not mean that the chords are equidistant. For example, these two pink chords are clearly not equidistant from the center. We must use the perpendicular distance from the chords to the center to conclude if the chords are equidistant or not. And we have already demonstrated that in our circle with center 𝑀, we have two perpendicular line segments.

We can then recall the equidistant chords theorem. This theorem tells us that if two chords in the same circle are equidistant from the center of the circle, then their lengths are equal. So, because chords 𝐴𝐡 and 𝐢𝐷 are equidistant from the center 𝑀, then we can write that 𝐴𝐡 equals 𝐢𝐷. We are given that 𝐴𝐡 is equal to 60, or 60 length units. Therefore, we know that 𝐢𝐷 must also be 60 length units. We need to find the length of πΆπ‘Œ, but we know that π‘Œ is the midpoint of 𝐢𝐷. And that means the length of πΆπ‘Œ must be half of 60, which is 30.

We can therefore give the answer that πΆπ‘Œ is 30, and if we needed to give units, they would be length units.

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