Can you find area of the Yellow shaded Semicircle? | (Quarter Circle) | |

[00:00:00]Welcome to Premat. In this video, we have got this uh yellow shaded as semicircle with a center P fully inscribed in a quarter circle with a center O. As you can see in this given diagram such that the area of this uh blue shaded region has been given to us as 17 cm square. And bear in mind that this uh quarter circle area consists of uh this uh yellow shaded region area plus uh the area of this uh blue shaded region. And now our task is to calculate the area of this uh yellow shaded semicircle. And we are supposed to justify our answer as well. Please don't forget to give a thumbs up and subscribe. And please keep in mind that this figure may not be 100% true to the scale. Let's go ahead and get started.

[00:00:55]And here's our very first step. Let's focus on this uh yellow shaded semi circle. We know this uh O is the radius of this uh yellow semicircle. I'm going to label this one as lowerase R. Then this uh A length is going to be lowerase R radius as well. And now let's make an observation. We can see this whole AO is the radius of this quarter circle and this whole radius is going to be R + R is going to give us 2 * lowerase R radius. So thus the radius of this quarter circle turns out to be 2 * lowerase R. And now let's make an observation. We can see the area of the yellow shaded semiircle is going to be equal to the area of the whole quarter circle minus the area of the blue shaded region. As you can see in this given equation and now we can see the blue shaded region area has been given to us as 17 cm square. So therefore now our task is uh to focus on this uh quarter circle area and likewise we are going to focus on this uh yellow shaded semicircle area as well. And now let's focus on this uh yellow shaded uh semi circle. And now let's recall the area of a circle formula. Area is always equal to<unk> * r² where r is the radius. And since we are dealing with the semicircle so therefore the semicircle area is going to be this uh area pi r² and I'm going to divide it by 2. So therefore uh the yellow semicircle area is going to be pi r² / 2 since the radius of this uh semicircle is lowerase r. And now we are going to focus on this whole quarter circle and we are going to use the very same area of a circle formula as well.

[00:03:15]And since we are dealing with the quarter circle so therefore the quarter circle area is going to be this uh pi r² and I'm going to divide this time by 4.

[00:03:27]So therefore the quarter circle area is going to be<unk> / 4 * the radius of this quarter circle is 2 * r. So I'm going to write down 2 * lowerase r whole². And now we can see the square of this 2 r is going to be simply 4 * r². So therefore we can write uh<unk> / 4 * 4 * r² and now we can see this four and four they cancel each other out. So therefore we are ended up with simply p<unk> r².

[00:04:12]So thus the area of this uh quarter circle turns out to be<unk> * lowerase r². And now let's recall this equation once again. The yellow semicircle area is equal to the quarter circle area minus the blue shaded region area. And we already figured out this yellow semicircle area as p<unk> r²id by 2. And likewise the quarter circle area as pi r². And the area of the blue shaded region has been given to us as 17. So let's go ahead and fill in the blanks in this equation. So on the left hand side we got<unk> r² / 2 = to<unk> * r² - 17 and now I am going to move this uh left hand side pi r² / 2 on the other side and at the very same time I'm going to move this -17 in the opposite direction. So therefore this is going to become positive7 on the other side equ= to<unk> * r² -<unk> r² / 2 and now we are going to tweak this p<unk> r² this could be written as 2<unk>i r² / 2 as you can see in this uh next step.

[00:05:43]So therefore we can write uh 17 = 2<unk>i r² -<unk> r² all / 2 and we can see in the numerator 2 p<unk> r² -<unk> r² is going to give us simply p<unk> r². So therefore our pi r² / 2 value turns out to be equal to 17.

[00:06:14]And now let's make an observation. We can see that our yellow semi circle area is p<unk> r² / 2. And here we figured out our pi r²ide by2 value as 17. So therefore we conclude that our this yellow semicircle area is going to be simply 17 cm square. So thus after all the calculations and manipulations the yellow semicircle area turns out to be 17 cm square. So therefore the area of this yellow shaded semicircle is going to be 17 cm square. And now let's make an observation. You can see the area of this uh yellow shaded region is equal to the area of this uh blue shaded region.

[00:07:06]And that's our final answer. Thanks for watching and please don't forget to subscribe to my channel for more exciting videos. Bye.