A Nice challenging geometry math problem| solve for radius of the circle

You are welcome to another challenging geometry math question. Today we are asked to find the area. of this circle, okay, now this circle has inside it an inscribed triangle of sides five, five, and six I'm going to solve this using three methods, okay? Yeah, now let's go to the first method. First of all, remember that we have a triangle; let's draw it out. We have this triangle, okay? Now this is five, this is five, and this is six. Okay, from here to here is six. Now, if we call this a Call this B. Call this C. This is the center of the circle. If we draw a straight line. from the center of a circle to this point; now, at this point, according to the circle theorem This is a chord, okay, and when you draw a straight line from the center of the circle To touch your chord means it divides your chord into two equal parts. Okay, so this is a straight line. So it's definitely 90 degrees and 90 degrees, so it will divide this line into two equal parts. Now, if we call this place, um, let's call this O, then BO is equal to OC. Okay, this is CBO equals OC. Now it means that this should be equal to BC divided by two. b times c divided by two, which will give us b times c times six six divided by two, which will give us three. so it means that this place is three and this place is three as well, all right? Now I haven't got that? Now let us draw part of this triangle. So if we draw out part of this triangle, we are To have this is five, this is a right angle, right, and this is three, and we can get a zero, okay, using the Pythagorean theorem, according to Pythagoras, states that In a right-angled triangle, okay, the side opposite your right angle is your hypotenuse. Now let's call the Okay, now h squared equals a squared plus b squared according to Pythagoras. Now let's relate that to what we have here, so our hypotenuse here is five, so we have five squared is equal to a o a squared plus o squared plus three squared plus three squared, okay? Now let's subtract three. Squaring both sides, if we do that, we have five squared minus three squared is equal to a zero. three squared plus three squared minus three squared. Now, three squared minus three squared is zero. now we have five squared minus three squared, which is the same as 25 minus 9, is equal to a squared 25 minus nine gives us 16, so we have 16 equals a squared So it simply means that a zero will be the square root of 16, so a zero will be equal to 4. So it means that the distance from here to here is four; we have four from here to here, all right. And from here to here is six, okay. Now let us introduce another theorem, okay? The code, code. Power theorem. Now let me explain what this theorem is all about. Okay, according to the code. Code power theorem: it states that when two codes meet each other, guys, let's call it This code is code a, code b, and call this code c, d, okay, and this is the meeting point o according to the It states that o times o b is equal to c o times o d, all right? Yes, that is the power. Power code, the code, code power theorem. Now let us apply it here, because we need to get the radius of this circle; if we are able to get the diameter of this circle, we can Divide it into two to get the radius right. Yes, now let's go back to this, remember that. From here to here, let me stay here, so I'll draw it well from here to here—gave us four right now. All of this is six, but this is three and this is three, and from here to here we don't Know what it is, and this is our center. Okay, so let's call this—this is A, B, C. Okay, let's call it. Let's call this part P. Now, using the code theorem, code B C intersects or cuts code A P. at o so according to the code code theorem it says that three times three okay three times three should be equal to four times OP, right? Four times OP. Yeah, now three times three gives us Nine, we have nine is equal to four times o p. All right, now to get the magnitude of o p. We divide both sides of this equation by four, so we have nine divided by four is equal to o p. All right, so here is what nine over four, but I want you to look closely—you can see that this A line passing through the center of this circle—you might not draw it well, okay, according to our calculation. this line passes through the center of the circle and extends to the opposite side of the circle, which means This is our diameter, and if this is our diameter, remember that four plus nine over four should be because the letter is two times the radius, okay, so now we have four plus nine over four is equal to two r, okay. Now the LCM is four. Let us, um, make this one to be four by multiplying by four, and also multiplying the numerator by four, so four times four is sixteen, so we have sixteen four plus nine over four equals two r; now we have this, this should be over four; now 16 plus nine gives us 25, so we have 25 over 4 equal to 2r. Now, to get the value of r. We multiply both sides of this equation by one half, so we have 25 over 4 multiplied by one half. is equal to 2r multiplied by half, so you can see that two divided by two gives us that r is equal to 25 over 8, 25 over 8, and that is the value of our radius. This has taken a lot Time is already passing, okay, so I don't know, but let's try the second method, or we're going to shoot. Another video for that, okay? Thanks so much for watching. See you in my next video. Bye.