98% Students FAILED to Solve this Beautiful Geometry Pr oblem?

Hello everyone, you are welcome. Today we have a new interesting geometry math problem. Let's say we have given a circle and there is a triangle inside the circle whose three sides length is given. We have to find out the radius of this circle. Now how can we solve this beautiful geometry math problem. First of all let us suppose this is our triangle ABC. So suppose this is point A, this is point B and this is point C. Now first of all here we will draw a perpendicular from point A to this line segment BC. Now let us suppose this is the height of this triangle. Let us suppose this is H. Now this total length BC is 8 units. What will be this length? Let us suppose this length is equal to X units. What will be the remaining length? This will be 8 minus X units. Here our target is to find the radius of this circle. So for that here we will try to find out the value of X and the value of H. So for that let us suppose this is point D. So here we have two right angle triangles. Triangle ABD and triangle ADC. So here in both the triangles we will apply the Pythagoras theorem and we will find out the value of X and H. Therefore in the first triangle, triangle ABD, we can write in triangle ABD hypotenuse will be equal to base square plus perpendicular. So this will become hypotenuse is 7. So 7 square is equal to base square plus perpendicular square. And we will move this to the left hand side. So the value of H square will become 7 square minus X square. Let us suppose this is our equation number first. Now in triangle number second, triangle ADC. So here in triangle ADC we can write 9 square is equal to 8 minus X square plus H square. Again we will move this to the left hand side. So this will become H square will become 9 square minus 8 minus X whole square. Let us suppose this is equation number second. Here we will compare equation number first and second. So this will become, this implies 7 square minus X square is equal to 9 square minus 8 minus X whole square. Here we will explain this right hand side using A minus B whole square identity. So this equation will become, this is 7 square which is 49 minus X square is X square is equal to 7 square minus X square, X is 2. This will become 2 square. So X square will become this is 49 minus 4 which is 45. Now we will take square root on both sides. so this will become this and this will be cancel so the value of h will become we can write this as 9 times 5 now square root of 9 is simply 3 times square root of 5 so here this is the value of h so this is the value of height so here we substitute the value of x and h here so here our h is 3 times square root of 5 and the value of x is 2 so this will become 8 minus x which will become 6 now what we will do next next let us suppose this is our center O now we will join this A with O after joining A with O I also expand this line segment so let us suppose this is point E and I also join this point E with point C now as we know that in any semicircle of a circle the angle in front of the diameter is always a right angle so here this is A is the diameter of this circle and this angle will be a right angle 90 degree angle we also know that that any chord making the same two angles in any region in the circle so this is a chord AC and these two angles angle at point B and E are in the same region in the same side so these two angles must be equal so let us suppose this is angle theta so this angle will be also theta now look to these two triangles triangle ABD and triangle AEC in these two triangles these two angles are right angle 90 degree angles and these two angles are same angle theta and we also noted in any two triangles if two angles are same then the third angle will be also same so here in both the triangles this angle will be also same now in these two triangles they have the same corresponding angles equiangular triangles so here these two triangles are equiangular triangles similar triangles so therefore we can write so here triangle ABD is similar with triangle AEC if these two triangles are similar then the ratio between two sides of one triangle will be equal to the ratio of the corresponding sides of the second triangle so here the ratio of these two sides of this triangle will be equal to the ratio of these two sides of this one triangle so therefore we can write the ratio between AB with AD will be equal to the ratio between AE with AC so let's substitute the values here from the figure the value of AB is 7 the value of AD is h which is 3 times root of 5 AE is this is diameter of the circle which is 2r 2 times of the radius and this AC is 9r so this will become AB is 7 this will become 7 divided by AD is 3 times root of 5 is equal to A is the diameters which is 2r and AC is 9r so we will cross multiply both the numbers this will become 2r times 3 times root of 5 is equal to 9 times 7 but there is 3 in both sides 3 times 1 is 3 times 3 is 9 so this will become 2 times square root of 5 r is equal to this will become 3 times 7 which is 21 we will divide both side by 2 times square root of 5 this and this will be cancelled so the final value of r will become r is equal to 21 divided by 2 times square root of 5 this is the value of r in terms of fraction if we further simplifying this number this fraction so it gives an above 4.7 up to one decimal places so finally r is equal to 4.7 is our answer and radius of the given circle so finally the radius of this circle is about 4.7 units which is our required target now in last don't forget to subscribe my youtube channel for such a more interesting and beautiful geometry map problems thanks for watching