A Nice geometry math question | how to find the area of the unshaded portion.

Hello, lovely friends, welcome back to my channel. First, I have to apologize. Yes, this is intended to be a semicircle. But since I'm drawing it freehand, Just assume this is a semicircle. This is a semicircle and this is a quadrant. That is one fourth of a circle. Now the equation states that we should find the area of the shaded region. If this is a semicircle, then this is also a semicircle. This is the center of the first semicircle. This is the center of the second semicircle. First, we need to organize everything. That is, we need to consider how we obtain the value first. Now let the area of the quadrant, That is the area of this quarter of the circle, equal to a cube. So the area of the quarter of the circle... Sorry. The area, everything that we are asked to solve the area of the shaded portion, let it be equal to a is going to be equal to... If we obtain the area of this quadrant, the entire quadrant Let's call it the area of, with subscript q. Now aq minus... It's going to be minus the sum of the areas of these two semicircles. So we have one large semicircle and a smaller one. So let's call the big one AP. Area of the large one plus area of the small semicircle. So it means that the area of the shaded region that we are looking for is equal to the area of the quadrant, the whole of this minus times this is the area of the large semicircle, minus times this is also minus area of the smaller semicircle Now at least we have an idea of what we are about to do. Now the next thing is, what is the area? How do we find the area? Remember that this is a quarter of a circle. Now the area of a quarter of a circle, that is aq is what we are looking for now, should be equal to pi times square divided by 4. Since this is a quarter of a circle, It should be pi times square divided by 4. Now according to this drawing, the radius of this quarter of a circle, The radius is 12 meters. So it means that the area is equal to pi times 12 squared divided by 4 So the area of a quarter of a circle is pi times 144 divided by 4 So I will write this somewhere. Let me write this down somewhere. The area is equal to a q minus a b minus a s. Now we have that aq is equal to pi. When 4 divides this, we get 33 remaining. When 2 divides 24, we get 6. So it will be pi 36 or 36π. So the area of the quadrant is 36π square meters. So let's set this aside. Now let's solve it. the area of the large semicircle. Now, since this is a semicircle, the area of the large semicircle will be equal to pi arrow square divided by 2. This is a semicircle. That is half of a circle. circle. Now we have the area of the big semi-circle is pi times. Now what is the radius? The radius is half of this. That is from here to here. That is the radius. That is half of 12, which will give us 6. So we have 6 squared. divided by 2. So the area of the large semicircle is equal to pi times 36 divided by 2. So the ab will be equal to 18. So 18π square meters. Now very easy, right? Yes. We are asked to find also; we need to find the area. of this as well. And this is where the problem lies, because we don't know the radius. of this. So how do we go about it? So I'll now Write out the area of the quarter circle. gave us. Where is it? Area of the quarter circle Is 36π, right? 36π. Then The area of the large semicircle gave us 18π. 18 pi. Now we are looking for the area of the smaller semicircle. Now let us explain that. Now let us draw these two things out. Let's draw it out first. So we have this and we have a larger one joining each other. Let us say That's how they are, but that's not how they are. I'm trying to understand it. So we have something like this. They are joining each other. at a single point Let's handle it this way. And this is the center of this. Now when two circles touch each other at a certain point, that means that there is a straight line passing through this location. If we draw a straight line, this is a tangent. So the radius of each of them is perpendicular to the tangent line at the point of contact. will form a 90-degree angle with this tangent. And when that happens, you'll notice they will form a straight line together. So we're going to have this straight line connecting this place to this place. So we can draw it like this. Even though it's not a straight line. but you understand what I'm trying to say. Now remember that this is 90 degrees. And from here to here is the radius. of the small semicircle Now let's call it R. That is the radius of the small semicircle. Let me come here. Now, if this area from here to here is the radius of the smaller semicircle, then from here to here is also the radius of the smaller semicircle R. Now remember that from here Here is the radius of the larger semicircle, which is 6. Now from here to here is the radius of the large semicircle, which is 12. So if from here to here If it's 12, then this portion will be 12 minus r, so we have 12 minus r now. Let me draw this particular triangle that we just found amazing, so we have Something like this, all right? So here is 12 minus r. Okay, here will be r plus 6, r plus 6. This is our hypotenuse, and here will be half of this, which is six. Okay, now, um, using the The Pythagorean theorem states that if you have a right-angled triangle All right, the hypotenuse, which is the side facing the hypotenuse, okay? This hypotenuse is the side facing the right angle, if squared, is going to be equal to the sum of this side squared plus this side squared, okay, so that means that r plus 6 squared, so we have r plus 6 squared, which happens to be our hypotenuse