The SEMICIRCLE Puzzle That Confuses Everyone! Can You Find The Area?|

[00:00:00]Hello everyone and welcome back to my channel. In today's video, we are going to solve another interesting geometric problem. In this question, we are given a semicircle with diameter A B center O and a chord A C. A point D lies on the chord A C such that the length of segment A D is equal to three. The length of segment D C is equal to 8. The line segment connecting the center O to point D, which is segment O D, has a length of four.

[00:00:37]Our goal is to find the area of the semicircle. Before we proceed, please take a moment to hit the like button and subscribe to my channel. Your support means a lot to me and enables me to create more content. Thank you. Also, feel free to pause the video and give it a try and tell us your answer in the comment section.

[00:00:59]To solve this problem, first let's connect OC with a straight line. Let the radius of the semicircle be R.

[00:01:09]The segment O A is a radius. So the length of O A is equal to R. The segment O C is also a radius. So the length of O C is equal to R. This makes triangle AC O an isosles triangle where the length of segment O A is equal to the length of segment O C which is equal to R. Next let theta be the angle A D O. Since points A D and C lie on a straight line the adjacent angle CD O must be supplementary meaning its value is equal to 180° minus theta.

[00:01:50]Now let's apply the law of cosiness to the two adjacent triangles. Triangle A D O and triangle CD O. In triangle AO, the square of O A is equal to the square of A D plus the square of O D minus 2 * A D * O D * the cosine of theta.

[00:02:23]O A is equal to R A D is equal to 3 and O D is equal to 4.

[00:02:38]Substituting these values in the formula will give us the square of r is equal to 3^ 2 + 4^ 2 - 2 * 3 * 4 * the cosine of theta.

[00:02:56]This simplifies to the square of r is equal to 9 + 16 - 24 * the cosine of theta.

[00:03:10]And this will give us the square of r is equal to 25 - 24 times the cosine of theta.

[00:03:22]Let this be equation one.

[00:03:25]In triangle CD O, the square of O C is equal to the square of CD plus the square of O D - 2 * C D * O D * the cosine of 180°us theta.

[00:03:49]O C is equal to R, C D is equal to 8 and O D is equal to 4.

[00:03:59]Now let's recall the trigonometric identity where the cosine of 180°us theta is equal to the cosine of theta.

[00:04:10]Substituting this along with other values in the above expression will give us the square of r is equal to 8^ 2 + 4^ 2 - 2 * 8 * 4 * the cosine of theta.

[00:04:33]This simplifies to the square of r is equal to 64 + 16 + 64* the cosine of theta.

[00:04:47]And this will give us the square of r is equal to 80 + 64* the cosine of theta.

[00:04:57]Let this be equation two.

[00:05:00]Now we have a system of two equations with two variables which are the square of r and the cosine of theta. To eliminate the term containing the cosine of theta, we can scale both equations.

[00:05:12]So the cosine terms cancel out when added together. To do this, we multiply equation 1 by 8 and multiply equation 2 by 3.

[00:05:24]Multiplying equation 1 by 8 will give us 8 * the square of r is equal to 200 - 192 time the cosine of theta.

[00:05:40]Let this be equation 3 and multiplying equation 2 by 3 will give us 3 * the square of r is equal to 240.

[00:05:54]+ 192 time the cosine of theta. Let this be equation 4.

[00:06:03]Now let's add equation 3 and equation four together. 8 * the square of r + 3 * the square of r is equal to 11 * the square of r.

[00:06:15]200 + 240 is equal to 440.

[00:06:23]-192 * the cosine of theta + 192 * the cosine of theta is equal to 0. And we are left with 11 * the square of r is equal to 440.

[00:06:38]Dividing both sides of the equation by 11.

[00:06:42]11 will cancel out 11 and 440 / 11 is 40 and we are left with the square of r is equal to 40.

[00:06:54]Now to calculate the area of the semicircle.

[00:06:59]Let's recall that the formula for the area of a semicircle is 12 *<unk>* the square of the radius.

[00:07:08]If we substitute the square of the radius with 40, the area of the semicircle is equal to 12 *<unk> * 40.

[00:07:19]40 / 2 is 20.

[00:07:22]And we are left with 20 *<unk>.

[00:07:26]Therefore, the exact area of the semicircle is 20 pi units squared. And this is the first method of solving the question.

[00:07:35]Second method, instead of using trigonometry, we can solve this purely with circle geometry by extending the line segment O D. To do this, first let's complete the semicircle into a full circle with center O and radius R. Next, let's draw a straight line that passes through the center O and point D. This line is a diameter of the circle. It intersects the circumference of the circle at two points. Let this points be E and F.

[00:08:09]This gives us a straight diameter E O F of length 2 R.

[00:08:15]Because O is the center of the circle, the distance from O to any point on the circle's edge is the radius R. This follows that the distance O F is equal to R and the distance O E is also equal to R. Since the point D lies on this line at a distance of O D equal to 4 from the center.

[00:08:41]It follows that the length of segment D is the radius minus 4.

[00:08:48]Now notice that the chord AC and the constructed diameter E F are two chords of a circle that intersect at the interior point D. According to the intersecting chords theorem, the products of the segments of intersecting chords are equal. From this theorem, a D * D C is equal to DE * DF.

[00:09:17]A D is equal to 3.

[00:09:21]D C is equal to 8.

[00:09:24]D E is equal to R - 4.

[00:09:30]And DF is equal to R + 4.

[00:09:36]Substituting these values into the equation will give us 3 * 8 is equal to R - 4.

[00:09:47]* r + 4 3 * 8 is 24.

[00:09:55]Using the difference of squares identity, r - 4 * r + 4 is equal to the square of r - 16. Rearranging this will give us the square of r is equal to 24 + 16.

[00:10:17]24 + 16 is equal to 40.

[00:10:22]Therefore the square of r is equal to 40. To calculate the area of the semicircle we will use the standard formula for the area of a semicircle.

[00:10:33]The area is equal to 12 *<unk> * the square of the radius r.

[00:10:40]Substituting 40 for the square of r, the area is equal to 12 *<unk> * 40.

[00:10:52]40 / 2 is 20 and we are left with 20 *<unk>.

[00:10:58]Hence the area of the semicircle is equal to 20 pi square units.

[00:11:05]Thanks for watching. Don't forget to like and subscribe for more videos.