Can you find area of the Yellow shaded Semicircle? | (Equilateral Triangle) |

[00:00:00]Welcome to PreMath. In this video, we have got this yellow shaded a semi-circle with a center O fully inscribed in an equilateral triangle ABC, as you can see in this given diagram, such that D and E are our points of tangency. And moreover, the area of this equilateral triangle has been given to us as 3 * √ 3 square units. And now our task is to calculate the area of this yellow shaded semi-circle. Please don't forget to give a thumbs up and subscribe.

[00:00:40]And please keep in mind that this figure may not be 100% true to the scale. Let's go ahead and get started. And here's our very first step. Since we are dealing with this equilateral triangle, that means all these side lengths of this triangle have equal length. Let's assume the side length of this equilateral triangle is lower case A, then this side length is going to be lower case A and lower case A across the board. And now let's recall this crucial fact. The interior angles of an equilateral triangle are 60° each. So therefore, this angle has got to be 60°. Likewise, this angle is 60° and 60°.

[00:01:28]And now we know that the area of this equilateral triangle ABC has been given to us as 3 * √ 3. And now let's recall the area of an equilateral triangle formula. Area of this equilateral triangle is always equal to √ 3 * A² / 4. Where this lower case A is the side length of this triangle. And now we know that the area of this equilateral triangle has been given to us as 3 * √3. I'm going to substitute that value over here. So, therefore, we can write this one as √3 / 4 * a² is going to be equal to 3 * √3.

[00:02:22]And now we can see this √3 and this √3, they are gone. So, therefore, we are ended up with a² / 4 = 3.

[00:02:35]And now I'm going to multiply both sides by 4 to remove this fraction. And here we can see this 4 and 4 is gone. So, therefore, a² value turns out to be 12.

[00:02:51]And now I'm going to undo this square by taking a square root on both sides. So, therefore, our lowercase a value turns out to be 2 * √3 units.

[00:03:07]So, thus the side length of this equilateral triangle turns out to be 2 * √3 units.

[00:03:13]So, therefore, this side BC length is going to be 2 * √3.

[00:03:23]In other words, uh this uh segment length BO is going to be half of that one is going to be √3.

[00:03:32]And likewise, this OC length is going to be √3 as well. And now in this next step, I am going to connect this center O with this point of tangency D.

[00:03:46]As you can see in this next step.

[00:03:49]And now let's make an observation. We can see that this OD length is our radius of this my circle. So, I'm going to label that one as a lower case r.

[00:04:00]And now our task is to find the value of this radius lower case r.

[00:04:06]And now let's recall the circle theorem.

[00:04:08]According to this theorem, the angle between the radius and the tangent line will always be 90°.

[00:04:15]So, no wonder this angle has got to be our 90° angle. Since this is our radius and this is our tangent line.

[00:04:24]And now let's focus on this tiny right triangle OBD.

[00:04:30]We know this angle is 60°, this angle is 90°. So, therefore this remaining angle has got to be 30°.

[00:04:41]And now let's focus once again on this right triangle ODB. And now let's recall this famous trigonometric ratio. Sine of theta is always equal to opposite divided by hypotenuse. So, therefore for our this angle 60°, this side is the opposite side of this angle. Whereas this is our hypotenuse. So, therefore for sine of our angle 60°, our side is radius lower case r divided by our hypotenuse is square root of 3.

[00:05:28]And now we know that sine of 60° is square root of 3 divided by 2. So, therefore I'm going to replace that one with its value square root of 3 divided by 2.

[00:05:41]So, therefore this could be written as square root of 3 divided by 2 is going to be equal to lowercase r divided by square root of three. And now I'm going to cross multiply. So therefore we can write this one as two times our radius lowercase r equals to square root of three times square root of three is simply becomes a three. And now I'm going to divide both sides by two to isolate radius r. So therefore our radius lowercase r turns out to be equal to three divided by two units. So does the radius lowercase r of this yellow shaded semicircle turns out to be three divided by two.

[00:06:34]And here's our final step. Now we are going to calculate the area of this yellow shaded semicircle.

[00:06:43]And now let's recall the area of a circle formula. Area is always equal to pi times r squared where lowercase r is the radius. And since we are dealing with this semicircle so therefore the area of the semicircle is going to be pi lowercase r squared and we are going to divide this one by two. So therefore the area of this yellow semicircle is going to be pi divided by two times the radius of this semicircle is three divided by two whole squared.

[00:07:19]And now we know that the square of this three divided by two is going to be nine divided by four.

[00:07:27]So therefore we can write pi divided by two times nine divided by four. Or if we multiply and simplify that is going to give us nine times pi divided by eight.

[00:07:45]square units.

[00:07:49]So, this after all the calculations and manipulations, the area of this yellow shaded semi-circle turns out to be 9 pi / 8 square units. And that could be approximately equal to 3.5343 square units as well. In other words, the area of this yellow shaded semi-circle is going to be 9 pi / 8 square units. And that's our final answer.

[00:08:23]Thanks for watching and please don't forget to subscribe to my channel for more exciting videos. Bye.