Question Video: Finding the Length of a Chord and the Radius | Nagwa Question Video: Finding the Length of a Chord and the Radius | Nagwa

Question Video: Finding the Length of a Chord and the Radius Mathematics • Third Year of Preparatory School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

The figure shows a circle with center 𝑀. Given that 𝐸𝐡 = 42, 𝐸𝐢 = 21, and 𝐸𝐴 = 42, find the length of segment 𝐸𝐷 and then the radius of the circle.

04:54

Video Transcript

The figure shows a circle with center 𝑀. Given that 𝐸𝐡 equals 42, 𝐸𝐢 equals 21, and 𝐸𝐴 equals 42, find the length of segment 𝐸𝐷 and then the radius of the circle.

From the information given, we can deduce that chord 𝐡𝐴 is bisected, because its two parts 𝐸𝐡 and 𝐸𝐴 both equal 42. It is therefore relevant to recall the first part of the chord bisector theorem. This theorem tells us that if a straight line passing through the center of a circle bisects a chord in the same circle, the line must be perpendicular to the chord. Therefore, the segment 𝐢𝐷, which passes through the center of our circle, is actually the perpendicular bisector of chord 𝐡𝐴.

Because chords 𝐡𝐴 and 𝐢𝐷 are perpendicular, we know that they form four right angles, one of which is the angle 𝐴𝐸𝑀. To answer this question, we must find the length of segment 𝐸𝐷 and the radius. We note that segment 𝑀𝐷 is a radius of the circle.

The only other piece of information we have not considered yet is the length of segment 𝐸𝐢. We are told that 𝐸𝐢 equals 21. We see from the diagram that segment 𝑀𝐢 is also a radius of the circle. And the length of 𝑀𝐢 equals the sum of 𝑀𝐸 plus 𝐸𝐢. Since we don’t know the length 𝑀𝐸, we will let 𝑀𝐸 equal π‘₯. Then, we can say that 𝑀𝐢 equals π‘₯ plus 21. Since we know that all radii in a circle have equal length, the length of any radius of our circle can be represented by the eπ‘₯pression π‘₯ plus 21. Therefore, radius 𝑀𝐷 equals π‘₯ plus 21.

We also recognize segment 𝑀𝐴 as a radius. So 𝑀𝐴 equals π‘₯ plus 21. At this point, we have an eπ‘₯pression for each side of a right triangle, specifically triangle 𝐴𝐸𝑀, which we have highlighted in green. The length of side 𝑀𝐸 equals π‘₯, 𝐸𝐴 equals 42, and 𝑀𝐴 equals π‘₯ plus 21.

From here, we can use the Pythagorean theorem, which states that the sum of the squares of the legs of a right triangle equal the square of the hypotenuse. Substituting the eπ‘₯pressions for each side length, we have π‘₯ squared plus 42 squared equals π‘₯ plus 21 squared. We will proceed to solve this equation for π‘₯. Once we know the value of π‘₯, we will be able to find the length of segment 𝐸𝐷 and the radius. We find that 42 squared is 1764 and π‘₯ plus 21 squared is π‘₯ squared plus 42π‘₯ plus 441.

Now it is necessary to simplify our quadratic equation into standard forms that it can be solved. However, after subtracting π‘₯ squared from each side of the equation, we realize that this is no longer a quadratic equation. What remains is the equation 1764 equals 42π‘₯ plus 441. Solving this equation, we find that π‘₯ equals 31.5.

Now we are ready to find the exact length of the radius, which we previously found to be represented by the expression π‘₯ plus 21. So we substitute 31.5 for π‘₯. And we find that the radius equals 52.5. Finally, to find the length of segment 𝐸𝐷, we will use the fact that 𝐸𝐷 equals 𝑀𝐸 plus 𝑀𝐷, which we see is true from the diagram. 𝑀𝐸 equals π‘₯. And because 𝑀𝐷 is a radius, it equals 52.5. After substituting 31.5 for π‘₯ and adding 52.5, we find that 𝐸𝐷 equals 84.

In conclusion, the length of segment 𝐸𝐷 is 84, and the radius equals 52.5.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy