The SIMPLE Circle Puzzle That Confuses EVERYONE!

Hello everyone, and welcome back to my channel. In today's video, we are going to solve another interesting geometric problem. In this question, A circle contains a horizontal diameter, and an inclined diameter. From a point on the horizontal diameter, located exactly 1 unit away from the circle's circumference, two line segments, are drawn to the endpoints of the inclined diameter. If these segments have lengths of 7 units, and 11 units. Our goal, is to find the area of the circle. Before we proceed, please take a moment to hit the like button, and subscribe to my channel. Your support means a lot to me, and enables me to create more content. Thank you. Also, feel free to pause the video and give it a try, and tell us your answer in the comment section. To solve this problem. First. Let P, be the given point on the horizontal diameter. Let A, and B, be the endpoints of the inclined diameter. Since A B, is a diameter of the circle, its midpoint, is the center of the circle, which we will call O. The line segment P O, connects point P, to the midpoint of the base A B, in triangle P A B. This means that segment P O, is a median of triangle P A B. Next. Let R, be the radius of the circle. Since O, is the center, and A B, is a diameter. O A, is equal to R. And O B, is equal to R. Since the horizontal line is also a diameter. And point P, lies on this diameter, and is exactly one unit away from the circle's edge. Therefore, the distance from the center O, to point P, is the total radius minus one unit. So. P O, is equal to R, minus one. Next, let's recall Apollonius's theorem. It states that, for any triangle, the sum of the squares of any two sides, equals twice the square of half the third side, plus twice the square of the median intersecting that third side. For triangle P A B, with median P O, the formula is. P A squared, plus P B squared, is equal to two times. O A squared, plus P O squared. P A, is equal to seven. P B, is equal to eleven. O A, is equal to R. And P O, is equal to R, minus one. Substituting these values, into the formula gives us. Seven squared, plus eleven squared, is equal to two times. R squared, plus R, minus one squared. To solve for the radius. First, let's simplify the left side of the equation. Seven squared, is forty nine. And eleven squared, is one hundred and twenty one. So. Forty nine, plus one hundred and twenty one, is equal to two times. R squared, plus R, minus one squared. Forty nine, plus one hundred and twenty one, is equal to one hundred and seventy. So, one hundred and seventy, is equal to two times. R squared, plus R, minus one squared. Next, let's divide both sides by two. Two will cancel out two. And one hundred and seventy divided by two, is eighty five. And we are left with. Eighty five, is equal to. R squared, plus R, minus one squared. Now, let's expand the squared binomial on the right side. Doing this will give us. Eighty five, is equal to. R squared, plus R squared, minus two R, plus one. R squared, plus R squared, is equal to two R squared. And this will give us. Eighty five, is equal to two R squared, minus two R, plus one. Rearranging this will give us. Two R squared, minus two R, minus eighty five, plus one. Is equal to zero. Negative eighty five, plus one. Is negative eighty four. And we are left with. Two R squared, minus two R, minus eighty four is equal to zero. And this forms a standard quadratic equation. Now, let's make this easier to solve by dividing the entire equation by two. Two, will cancel out two. And eight four divided by two, is forty two. And we are left with. R squared, minus R, minus forty two is equal to zero. Factoring this quadratic expression gives us. R, minus seven, times R, plus six, is equal to zero. Simplifying this gives us two possible values for the radius. R, is equal to seven, or R, is equal to negative six. Since a geometric radius must always be a positive value, we discard negative six. This leaves us with a radius of seven units. With the radius established as seven, the area calculation is straightforward. Since the area of a circle is equal to pi. Times the radius squared. It follows that. The area of the circle is equal to pi. Times seven squared. And this gives us a final exact answer of forty nine pi. Which is approximately one hundred and fifty four square units. Hence, the area of the circle is equal to one hundred and fifty four square units. Thanks for watching. Don't forget to like and subscribe for more videos.