This will blow away your mind

This geometry problem will take you straight to Cloud 9. We are given a triangle. Let the top left vertex be A, the top right vertex be B, and the bottom vertex be C. The top side AB, which we will denote using C, has length 20. The left side AC, which we denote using B, has length 11. The right side BC, which we denote using A, has length 13. A semicircle is placed inside the triangle in such a way that its diameter lies along the top side AB, and its curved boundary just touches the two other sides, AC and BC, like this. Our job is to find the radius of this semicircle. So, can you solve it? Before we start, just note one thing. If a line is tangent to a circle, then the perpendicular distance from the center of the circle to that line is exactly equal to the radius of the circle because a tangent always touches the circle at one point and is at right angle to the radius at that point. Great! Now, here is an important observation that we can make from this problem. Let us call the center of this semicircle as some point O. We do not know where exactly this O is, but the point is that it doesn't matter. Wherever it lies, we know for sure that since the semicircle is perfectly tangent to both sides, AC and BC, the perpendicular distance from its center O to each of those two sides must be equal to the radius of this semicircle, which we call R. Now, here comes the magical trick. Because we know the values of all three sides of the triangle ABC, let us compute the total area of triangle ABC using Heron's formula, which is given like this, where S is the semi-perimeter calculated using this formula. It will become clear later why we are calculating the total area. So, the semi-perimeter S equals 11 plus 13 plus 20, all divided by 2. 11 plus 13 plus 20 equals 44. 44 divided by 2 equals S, equals 22. Now, subtract each side from 22. 22 minus 11 equals 11. 22 minus 13 equals 9. 22 minus 20 equals 2. Now, multiply all four numbers together. That gives 22 times 11 times 9 times 2. Rewrite 22 as 2 times 11. So, both of them give 11 squared. Then, both these give 2 squared, and 9 is 3 squared. So, we get the square root of 11 times 2 times 3 whole squared, or 66 squared. Thus, the total area A of the triangle is 66 square units. Noice! Now, it will become clear to you why we calculated the total area of the triangle ABC. Draw a line from point C to the center O. This divides the big triangle into two smaller triangles, namely, triangle AOC and triangle BOC. The main thing is that the total area of triangle ABC is equal to the sum of the areas of these two smaller triangles. Look at triangle AOC. Its base is side AC, which is 11. Its height is the perpendicular from O to AC, which is exactly R. So, its area is half times 11 times R, which equals 11R divided by 2. Similarly, triangle BOC has base BC equals 13 and height R. So, its area is half times 13 times R. Therefore, the total area of the big triangle is 11R divided by 2 plus 13R divided by 2. Add them, and the total becomes 24R divided by 2, which simplifies to 12R. But, we already found that the total area is 66. So, we set 12R equal to 66. Divide both sides by 12. 66 divided by 12 simplifies by dividing both numerator and denominator by 6. That gives 11 divided by 2. So, we get R equals 11 divided by 2, which is 5.5. So, the radius of the semicircle is 5.5 units. Like, share, and subscribe. So good!