Only 1 Out of 10 People Get This Right

Added: 2026-05-06

[00:00:00]In this figure we are required to find the area of the shaded region. So we have the radius of this circle has been given to us as three and this angle here subended from these two tangent lines is 60°.

[00:00:18]So with that what is the area of this shaded region?

[00:00:24]First of all, what you're going to do, let us let these points of tangency. Let this be A and let this be B. Let this point be C and let this center be O.

[00:00:37]The next step is we're going to connect the center O to the points of tangency B, point B, and O to point A.

[00:00:48]Just like this. Okay. So, we've connected O to B and O to A.

[00:00:55]As you can notice OB and O A are both radius or radi of this circle here and the radius was already given to us. So OB is three and also O A is three. Now we know that from tangent to circle theorem the angle between the radius and the tangent is always 90°. So this angle between the radius and tangent is 90. So this is 90° also this is 90°.

[00:01:27]Then also from from the two tangent theorem which states that if two tangent segments are drawn to a circle from the same external point just like this one.

[00:01:39]These two tangent lines BC and AC are drawn from the same external point C.

[00:01:46]then it implies that those two segments are congruent. That implies that AC is congruent to BC.

[00:02:02]Now let's first focus on the quadrilateral OCB.

[00:02:07]This one here. Okay. OCB quadrilateral OCB.

[00:02:23]We know that the total interior angle sum is 360°.

[00:02:28]Since since this angle here OB C this is 90° and this is 60° this is 90° and we know that sum of the interior angles of the quadrilateral is 360° that implies that this one here will be so angle B O A will be equals to 360° ° - 90° + 60° + 90° which is equals to 360°US 240° which gives us 120°.

[00:03:14]So this angle is 120°.

[00:03:20]So that implies now that to find the shaded area.

[00:03:25]So our shaded area will be equals to quadrilateral area of O A to be area of O C B minus the area of this sector here minus area of the sector O O B A. Okay, so this implies that now we have to find the area of the quad of the quadrat O A CB and the area of the sector OA to be able to find the area of the shaded region. So the next step we're going to do is we're going to connect point O to point C just like this. So we've drawn this line. So as a result you can notice that triangle OBC is congruent to triangle O A. This one here this upper triangle is congruent to this down triangle.

[00:04:39]That implies that this angle here is equals to this angle here. So if if the whole of this angle is 120° that implies that this one will be 120 by 2 which is 60°. Also this one will be 60°.

[00:04:59]Okay.

[00:05:01]Also this angle here BC is equals to angle O C A. So this one is equals to this one. So if the whole of it is 60 this will be 60 by 2 which is 30° also this will be 30°.

[00:05:20]Now focusing on triangle OC it's like this.

[00:05:32]So this is S. This is O.

[00:05:38]This is 90°. This is 60°. This is 30°.

[00:05:46]As you can notice, this triangle here is a special 30 to 60 to 90 triangle.

[00:05:58]Okay? It's a special triangle.

[00:06:02]Now since this is 30° and it's our smallest angle so it oppos opposite side will be the smallest side. So this will be smallest side and the angle which has the largest side which in this case is 90° it opposite side will be the longest side okay which is O C. So, O A is the smallest side and OC is the longest side. And in such a triangle which is which is a 30 to 60 to 90 triangle, the smallest the longest side is twice the size of the smallest side. So if the smallest side is say one. Okay, that means the longest side will be two. So the longest side is just double of the smallest side. So if this is one, this will be two. So if this is one and this is two, we can find this side what it's equals to.

[00:07:10]So for SC it will using Pythagoras since it's a right triangle a² + b= c² and our a is a c. So a c² plus our b is 1. So 1 square= to our c which is 2^ 2.

[00:07:33]Therefore a c 2 = 2^ 2 which is 4 - 1 2 which is 1. Therefore a c² = 4 - 1 is 3.

[00:07:44]Find the square root on both sides. This cancels. Therefore a c is equals to<unk> 3. Therefore this side AC is if this is one this is two this is <unk>3 but if we if we are to go back to our original diagram we can see that O A is three and in this case we are saying O A is one so to make it three we're going to multiply all sides by three okay multiply each side by three to balance everything so this side will be equals 2 3 * 1 which is 3. This side will be 3 <unk>3 and this side will be 6.

[00:08:31]So therefore our SC is 3 <unk>3 and our OC is 6.

[00:08:42]Okay. But remember AC is congruent to BC. Therefore if this is 3 <unk>3 also this side is 3 <unk>3.

[00:08:52]So now let's find the area of triangle OC. Okay. Area is equals to a half * base time height.

[00:09:05]So it will be a half. The base is 3 <unk>3 and the height is O A which is 3. So this will be equals to 3 * 3 is 9. So 9<unk>3 / 2. So this is the area of O A C. But remember area I mean rect triangle O A C is congruent to to triangle O CB.

[00:09:38]Okay.

[00:09:39]Therefore area of quadrilateral O A CB will be equals to 9<unk>3 / 2 + 9<unk>3 over 2. So this will be equals to.

[00:10:03]So here using crisscross 2 2 * this is 18. So 18 <unk>3 + 2 * this is 18 <unk>3 over 2 * 2 is 4. This gives us we factor out 18<unk>3 into 1 + 1 over 4. So this is 18 <unk>3 into 2 over 4. This two goes into four two times and this two goes into 18 to give 9. So this will be equals to 9 <unk>3.

[00:10:42]So the area therefore area of quadilateral o a c b is equals to 9 <unk>3 square units. Now for the area of sector OBA this one here area of a sector is theta / 360° *<unk> r². This is how we find area of a sector. So this is equals to our angle here is 120°. So it will be 120 / 360 *<unk> into the radius of the circle is 1 sorry is 3. So into 3 squared.

[00:11:34]So 120 goes into 360 3 times. So we have 1 out of 3 *<unk> 3 2 is 9. So * 9 this is equals to 9<unk> / 3 3 goes into 9 three times. So area of the sector is equals to 3 pi.

[00:12:02]And remember we said to get area of shaded region this area is equals to area of quadrilateral A or A CB minus area of sector O B A. Okay. So this is equals to this area is we found area of this to be 9 <unk>3 and we found area of sector OB to B3 pi.

[00:12:37]So area will be approximately 6.16 square units.

[00:12:49]So this is the area of the shaded region. this one here. Thank you so much for watching. Kindly subscribe to this channel if you haven't done so yet. Like this video if you enjoyed it. Comment your thoughts about the video. If you have things you'd like to add, you can share them in the comment section. And do not forget to share this video with your friends and family for an enlightening experience. See you in the next video.