Find This Length X | Looks Easy?

Added: 2026-05-25

[00:00:05]We will start our problem with a semicircle.  Then, using the center of the semicircle, we will construct a square so that its bottom right  corner is touching the center of the circle and the upper left corner is touching the perimeter  of the semicircle, as you can see on the sketch.

[00:00:27]Now let's say that the area of this square  is six. Then, from the bottom right corner of our semicircle, we will pull a line that goes  through the upper right corner of the square and all the way until the edge of the semicircle.  And therefore, this will represent the chord, and we will label the length of this chord as X.  Now the question is, can we somehow calculate the length of this chord X. This problem was sent to  me by my subscriber Anton. And if you guys want to try this one out for yourself, you can pause  the video right now and then come back and check the solution. All right. So let's solve this one  together. So first, I'm going to label the center of our semicircle as O., and I'm going to label  all the intersection points of our chord as A, B, and C. Then I will label the remaining vertices  of our square as P and Q. All right. So, first let's have a look at our square. Let's do a quick  reminder. If we have a square with a side length of S, we can calculate the area of the square as  S squared. And if we pull the diagonal of this square, we can get the value of the length of the  diagonal as the side times the square root of two.

[00:02:00]All right. So we're going to use those formulas  in our case. Since we know that the area of our square equals 6, we can say that the side squared  equals 6 as well. This means that the side of our square will be equal to √6. And we're just going  to substitute this value into our sketch. Next, let's take a look at the diagonal of our square  OP. And let's calculate its length. We can use the second formula. And if we substitute the value  for the side, we will get that this diagonal will have a length of √12. And we are not going to  rationalize this value. We're going to leave it as it is. And we're going to implement it into  Sketch. Now take a look at the diagonal OP. It goes from the center of the circle to the edge  of the circle. So this diagonal must represent the radius of the circle. So we can conclude that  the line OA is also the radius of the circle. So this line will have the length of √12 as well. All  right. So now let's clear our image and let's have a look at the right-angled triangle OBA. We know  that this triangle is a right-angle triangle due to the fact that one of its legs is the side  of our square OB and the bottom of the square aligns with the diameter of the semicircle.  So, since this is a right-angle triangle, we can use the Pythagorean theorem in order to  obtain the hypotenuse BA. So BA² equals the leg OB² plus the leg OA². And we will substitute  our values from the sketch. We're going to work through this and get that BA²= 18, which  means that BA equals √18. And again, I'm not going to rationalize this. I'm just going to put  this value into our sketch. Then let's simplify the image once again. Now, let's have a look at  the side of the square PB. This side equals √6.

[00:04:26]And if we just extend this line to the right to  the point of intersection with the semicircle and label this point of intersection as R, we can  notice that the line PB is equal in length to the line BR. This is due to the fact that the line  OB of the square is actually the line of symmetry since it passes through the center of the circle,  and it is perpendicular to the line PR. This leads us to conclude that the line BR also has a length  of √6. Now let's take this sketch and let's put it a little bit upward in order to complete the  circle. Then I'm going to get rid of the lines that we don't need for now. Now, let's remind  ourselves about one really interesting theorem that we can use in this case, and this is the  famous intersecting chords theorem. This theorem states that when two chords intersect inside a  circle, the product of the lengths of the segments of one chord equals the product of the lengths  of the segments of the other. So in our case, from the sketch, the product of the segments  CB and BA equals the product of the segments BP and BR. And you can notice that I have labeled  the sketch on the right the same as the sketch on the left to avoid any confusion. So we're just  going to use this equation, and we will substitute the values for those segments from our sketch.  We're going to work through this. And notice that I'm deliberately leaving the √36. I'm not  going to simplify it because CB =√ 36 over √18, which equals 2. And let's implement this value  into our sketch as well. Now let's clear our image and let's get back to the original position.  Now remember, we're looking for the length of the chord CA, and we can find this length as the sum  of the segments CB and BA. We are going to just substitute the values from the sketch, and this  time we will rationalize √18 to 3√2. In this way, we will get that the line x has the length of  4√2, and this is our awesome solution. Well, I really hope you guys enjoyed this problem.  If you did, do not hesitate to leave a like or subscribe to the channel for more content  like this. If you know a person who enjoys math problems like this, please share this video with  them. And if you found another way to solve it, please write it down in the comment section below.  And until next time, see you guys and take care.