Question Video: Finding the Range of Values of an Unknown That Satisfy the Given Conditions | Nagwa Question Video: Finding the Range of Values of an Unknown That Satisfy the Given Conditions | Nagwa

Question Video: Finding the Range of Values of an Unknown That Satisfy the Given Conditions 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!

If 𝑀𝐹 > 𝑀𝐸, find the range of values of π‘₯ that satisfies the data represented.

05:45

Video Transcript

If 𝑀𝐹 is greater than 𝑀𝐸, find the range of values of π‘₯ that satisfies the data represented.

We begin by examining the diagram to see what information we’ve been given. We have a radius, which is highlighted in orange, whose length 𝐴𝑀 equals 33 centimeters. We also have two chords which are highlighted in yellow. On the left, we have the chord of length 𝐷𝐢 equal to 24 centimeters. And on the right, we have length 𝐴𝐡 equal to π‘₯ plus four centimeters. We also have the center of the circle, which is point 𝑀. We recall that the distance of a chord from the center is measured by the length of the segment from the center intersecting perpendicularly with the chord. Therefore, we recognize 𝑀𝐹 as mentioned in the directions is the distance of chord 𝐷𝐢 from the center point 𝑀. And 𝑀𝐸, which we are told is a shorter length than 𝑀𝐹, represents the distance of chord 𝐴𝐡 from the center point 𝑀.

Now we will move these two important facts to the bottom of the screen so that we have space for our next step. To answer this question, we need some more information about the possible values of π‘₯. To do this, we will need some way to compare the lengths of the two chords. So we recall a theorem which is about the relationship between the lengths of chords and their distances from the center. From this theorem, we know that for two chords of the same circle, the chord closer to the center has a greater length than the other. In this example, we have two chords 𝐷𝐢 and 𝐴𝐡. Whichever of these two chords is closer to the center has the greater length. We are told that 𝑀𝐹 is greater than 𝑀𝐸, so this means that chord 𝐴𝐡 is closer to the center of the circle.

This leads to the fact that chord 𝐴𝐡 has a greater length than chord 𝐢𝐷. In the given diagram, we note that 𝐴𝐡 equals π‘₯ plus four centimeters and 𝐢𝐷 equals 24 centimeters. Therefore, the inequality 𝐴𝐡 greater than 𝐢𝐷 can be written as π‘₯ plus four is greater than 24. Solving this inequality leads us to π‘₯ is greater than 20. Our goal is to find the range of values of π‘₯. And we have successfully found the lower bound. We need to do some further reasoning in order to find the upper bound of π‘₯. Now that we have cleared some space, we will contemplate what might be the maximum length of chord 𝐴𝐡. We are especially interested in this chord because it involves the variable π‘₯.

We recall that the length of a chord is larger when it is closer to the center, so the longest chord should occur when the distance from the center is zero. If the distance of a chord from the center is zero, the chord should contain the center. In this case, the chord is the diameter of the circle. This means the maximum length of chord 𝐴𝐡 would be the same as the length of the diameter of the same circle. Since the radius of the circle is 33 centimeters, its diameter is two times 33. This tells us that the length of chord 𝐴𝐡 cannot exceed 66 centimeters. Additionally, since 𝐴𝐡 in the given diagram does not contain the center 𝑀, we know that the length of chord 𝐴𝐡 must be strictly less than 66 centimeters.

By substituting π‘₯ plus four as the length of 𝐴𝐡, we have π‘₯ plus four is less than 66, which leads to π‘₯ is less than 62. This gives us the upper bound for π‘₯. By combining both lower and upper bounds, we have 20 is less than π‘₯ is less than 62. This compound inequality means that the range of values of π‘₯ includes any number between 20 and 62, not including 20 and not including 62. In interval notation, our answer may be written as 20 comma 62 with round brackets or with square brackets facing outwards.

In review, we found the lower bound of 20 by comparing the distance of our two chords from the center, and the chord with the shortest distance from the center was the longer chord. Then, to find the upper bound of less than 62, we considered what the longest possible chord might be, which was the length of the diameter. Combining our lower bound and upper bound gave us a final answer, which we could write as a compound inequality or in interval notation.

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