hey this is presh talwalker start with the circle whose radius length is equal to 12. in each semi-circle inscribe a circle that passes through the center of the large Circle Let each of these circles have radius length equal to B then inscribe a circle that's tangent to these two circles and the large Circle and let its radius length be equal to a finally inscribe a circle in the semi-circle that's tangent to the diameter tangent to the large Circle and tangent to the circle with radius length equal to B let this radius length be equal to C the question is what is a plus b plus c equal to I think David for the suggestion pause the video if you'd like to give this problem a try and when you're ready keep watching to learn how to solve this problem [Music] what's fun about this problem is the three radius lengths are an increasing order of difficulty to calculate them let's start with the radius length B construct a diameter of this circle which has length equal to 2B this is exactly a radius of the large Circle so 2B is equal to 12 which means B is equal to 6. starts out easy enough now let's calculate the value of a construct the following right triangle where one leg is equal to B the hypotenuse is equal to the sum of the radii so its length will be a plus b what's the length of the other leg of this right triangle we know that this length is equal to a and a radius of the large circle is equal to 12. so the other leg will have length equal to 12 minus a since we have a right triangle we have the square of a plus b is equal to B squared plus the square of 12 minus a substituting in that b is equal to 6 we can simplify this equation and we can work out that a is going to be equal to 4. so now let's work out the value of C we will first construct the following radius of the large Circle this will have length equal to 12. next construct this radius which has length equal to C now we have a right triangle here where one leg is equal to C the length of the hypotenuse will be the length of the radius of the large Circle minus the length of the radius of the small circle so its length will be equal to 12 minus C but the length of the other leg be equal to X now let's solve for x squared since we have a right triangle x squared is equal to the square of 12 minus C minus c squared we can simplify this equation to get x squared is equal to 144 minus 24c now let's construct the following right triangle one leg will be equal to X the other leg will be the difference in the lengths of the radii so it will be equal to B minus C the hypotenuse will be the sum of the radii lengths so it will be equal to B plus c now let's solve for x squared in this right triangle we have x squared is equal to the square of B plus C minus the square of B minus C we can simplify this equation to get x squared is equal to four BC now we can substitute in that b is equal to 6. so if x squared is equal to 24c so we can now focus on these two equations and we can set them equal to each other so if 24c is equal to 144 minus 24c simplifying will give that c is equal to 3. so we solve for the values of a b and c and this gives that a plus b plus C is equal to four plus six plus three and that gives the answer of 13. thanks for making us one of the best communities on YouTube see you next episode of mind your decisions where we solve the world problems one video at a time