Can you find area of the Semicircle? | (Right Triangle) |

Welcome to Premath. In this video we have got this green semi-circle with the center O fully inscribed in a right triangle ABC as you can see in this given diagram such that the point D is the point of tangency and moreover this hypotenuse AB consists of two segments this segment AD length is 8 units whereas the segment BD length is 9 units and now our task is to calculate the area of this green shaded semi-circle. 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 and in this video I am going to share with you with two different approaches so therefore please watch the video till the very end and here's our very first method and since we are interested in calculating the area of this green shaded semi-circle so therefore we must calculate the radius lowercase r of this semi-circle first and I'm going to show you two different approaches to calculate the radius lowercase r and now let's focus on this hypotenuse AB that is going to be the sum of these two numbers 8 plus 9 is going to give us 17 units and now let's recall the two tangents theorem according to this theorem if two tangent segments are drawn to one circle from the same external point then they are congruent as you can see in this given diagram here we got these points A and B the points of tangency and O is the external point then the segment AO is going to be equal to the segment BO in other words these two segments are congruent and now let's revert our attention back to our this original problem and here we can see this segment AC is going to be equal to this segment AD because these are our two tangents and now we know that this tangent AD length is 8 units so therefore this side AC length has got to be 8 units as well and now we are going to focus on this big right triangle ABC and I'm going to label this BC side length as x units and now our task is to calculate the value of x so therefore we are going to apply the Pythagorean theorem on this triangle and here's our Pythagorean theorem a square plus b square equal to c square and in our case our hypotenuse is 17 whereas our two other legs are x and 8. Let's go ahead and fill in the blanks in this Pythagorean formula so we got x square plus 8 square equal to 17 square and now we are going to subtract 64 from both sides this is gone so therefore x square value turns out to be equal to 225 and now I'm going to undo this square by taking a square root on both sides so therefore we could see that our x value simply turns out to be positive 15 units so thus our this BC side length which is represented by x turns out to be 15 units and now in this next step I'm going to connect this center O with this point D as you can see in this next step and now we could see that this OD is the radius of this semicircle so I'm going to label this one as lowercase r and likewise this segment OC is the radius lowercase r as well and now let's recall the circle theorem according to this theorem the angle between the radius and the tangent line will always be exactly 90 degrees so no wonder this angle has got to be a 90 degrees angle since this is our radius and this hole is our tangent line and here's our next step now I'm going to connect this center O with this vertex A as you can see in this next step and now let's make an observation we can see the area of this big right triangle ABC is going to be equal to the area of the triangle ACO plus the triangle AOB area and here's our triangle ABC and this is our right triangle ACO and finally this triangle is our AOB triangle and now we are going to calculate the area of all of these three triangles and now let's recall the area of a triangle formula area is always equal to half times base times the height of the triangle and now let's focus on this triangle ABC its base is 15 and the height is 8 so therefore triangle ABC area is going to be a half times 15 times 8 is going to be equal to and now the area of this triangle ACO is going to be we got the base lowercase r and the height is 8 so therefore we could write a half times lowercase r times 8 plus and now let's focus on this triangle AOB area our base of this triangle is 17 whereas the height is lowercase r so therefore the area of this triangle AOB is going to be a half times 17 times the lowercase r and now we can see 1 divided by 2 1 divided by 2 and 1 divided by 2 they are common so therefore 1 divided by 2 1 divided by 2 and 1 divided by 2 they are gone and if we multiply that is going to give us 120 equals to 8 times r plus 17 times r let's divide both sides by 25 to isolate radius r this is gone so therefore if we reduce this one that is going to give us 24 divided by 5 so thus the radius of this green semicircle turns out to be 24 divided by 5 by using the very first method and now let me show you the second method as well to calculate this radius lowercase r and now let's assume alpha and beta are our two complementary angles in other words the sum of these two angles alpha plus beta must be equal to 90 degrees if i label this angle as our angle alpha and this is our 90 degree angle in this triangle ABC then this angle has got to be our angle beta and now in this right triangle BDO this angle is beta we know this angle is a 90 degree angle so therefore this angle has got to be our angle alpha so therefore we could see that these two right triangles the right triangle BDO and this bigger right triangle ABC are our similar triangles according to angle angle similarity theorem and here's our right triangle BDO and this whole big right triangle is our triangle ABC and since these two triangles are similar so we are going to have a proportion in other words the ratio of OD and BD must be equal to the ratio of AC and BC side lengths and here we can see our OD length is lowercase r i'm going to replace that one with lowercase r our BD segment length is 9 units and now our segment AC length is 8 units so i'm going to replace that one with 8 and finally our BC side length is 15 so i'm gonna substitute 15 over here and now i'm going to multiply both sides by 9 to isolate lowercase r and here we can see 9 and 9 is gone so therefore lowercase r turns out to be 8 times 9 is 72 divided by 15 and if we reduce this fraction our lowercase r turns out to be equal to 24 divided by 5 units so thus our radius lowercase r turns out to be 24 divided by 5 by using the second method as well and here's our final step now we are going to calculate the area of this green semicircle and now let's recall the area of a circle formula area is always equal to pi r square where lowercase r is the radius so thus the semicircle area is going to be pi divided by 2 and our radius is lowercase r is 24 divided by 5 so i'm going to write down 24 divided by 5 whole square let's simplify so we are going to get pi divided by 2 times 576 divided by 25 and if we simplify that is going to give us 288 pi divided by 25 square units the area of this green semicircle so thus the area of this semicircle turns out to be 288 pi divided by 25 square units and that's going to be approximately equal to 36.19 square units as well and that's our final answer thanks for watching and please don't forget to subscribe to my channel for more exciting videos bye