Question Video: Finding a Missing Length Using Equidistant Chords from the Center of a Circle | Nagwa Question Video: Finding a Missing Length Using Equidistant Chords from the Center of a Circle | Nagwa

Question Video: Finding a Missing Length Using Equidistant Chords from the Center of a Circle Mathematics • Third Year of Preparatory School

Given that 𝑀𝐶 = 𝑀𝐹 = 3 cm, 𝐴𝐶 = 4 cm, line segment 𝑀𝐶 ⊥ line segment 𝐴𝐵, and line segment 𝑀𝐹 ⊥ line segment 𝐷𝐸, find the length of the line segment 𝐷𝐸.

03:11

Video Transcript

Given that 𝑀𝐶 is equal to 𝑀𝐹, which is equal to three centimeters, 𝐴𝐶 is equal to four centimeters, line segment 𝑀𝐶 is perpendicular to line segment 𝐴𝐵, and line segment 𝑀𝐹 is perpendicular to line segment 𝐷𝐸, find the length of the line segment 𝐷𝐸.

We are told in the question that 𝑀𝐶 is equal to 𝑀𝐹. And this means that the two chords 𝐴𝐵 and 𝐷𝐸 are equidistant from the center 𝑀. They are both three centimeters away from the center. We recall the theorem that states that if two chords are equidistant from the center, they are also equal in length. This means that in this question, the length 𝐴𝐵 is equal to the length 𝐷𝐸. We also know that 𝑀𝐶 is the perpendicular bisector of 𝐴𝐵. And this means that 𝐴𝐵 is equal to two multiplied by 𝐴𝐶. Since 𝐴𝐶 is equal to four centimeters, 𝐴𝐵 is equal to eight centimeters. We can therefore conclude that the length of the chord 𝐷𝐸 is eight centimeters.

So far in this video, we have discussed the lengths of chords depending on their distance from the center of the circle. We will now consider the converse relationship. In the diagram shown, we have two congruent circles. We will consider the case where the chords 𝐴𝐵 and 𝐶𝐷 are equal in length and, more importantly, what this says about the distance of the chords from their respective centers. In this example, these are the lengths 𝑂𝐸 and 𝑃𝐹, respectively, Adding the radii 𝑂𝐴 and 𝑃𝐶, we know that these lengths must be equal, as the circles are congruent. As the chords are equal in length, the half chords must also be equal in length such that 𝐴𝐸 is equal to 𝐶𝐹. This is because 𝐴𝐸 is equal to a half of 𝐴𝐵 and 𝐶𝐹 is equal to a half of 𝐶𝐷.

Using our knowledge of the Pythagorean theorem, the third sides of our right triangles must also be equal in length. The length 𝑂𝐸 is equal to the length 𝑃𝐹. In other words, the distance of the chords from their respective centers are equal. This can be summarized as follows. Two chords of equal lengths in the same circle or congruent circles are equidistant from the center of the circle or the respective centers.

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