A Nice Geometry Problem - 98% FAILED to Solve this Beautiful Geometry Problem

Hello everyone, you are welcome. Today we have a very interesting geometry map problem. That is here we have given a circle and there is a scalene triangle inside the circle whose three different side length is given. Our target our job is to find out the radius of this circle. Now how can we solve this beautiful geometry map problem. First of all let us suppose this is our triangle ABC. So let us suppose this is point A, this is point B and this is point C on this triangle. So first of all here we will draw a perpendicular from this point A to this side BC. So this figure will become. Now let us suppose this is point D. We look to this side this total length BC is 14 units. So what about this length and this length. Let us suppose this length BD is So what will be this length. This length will be 14 minus x units. So let us suppose this AD which is perpendicular to this side BC is the height of this triangle. So let us suppose this is H. Now we want to find out the radius of this circle. Here we will try to find out the width of X and the width of H. So if I take it with one triangle ABC. Here we have two right angles triangle inside this triangle. Triangle ABD and triangle ADC. So here in both the right angle triangles we will apply the Pythagoras theorem and with that we will find out the width of H. So first we will apply the Pythagoras theorem in triangle ABD. So by Pythagoras theorem in any right angle triangle the hypotenuse square is equal to base square plus perpendicular square. So here our hypotenuse is 30 base is x and perpendicular is H. So this figure will become. So here in triangle in triangle ABD we will apply the Pythagoras theorem. So by Pythagoras theorem hypotenuse square that is 13 square is equal to base square plus perpendicular square. And we will take this x square to the left hand side. So this will become 13 square is simply 169 minus x square is equal to h square. Now let us suppose this is our equation number first. Then we will apply the Pythagoras theorem in the second right angle triangle ADC. So again here the square of hypotenuse will be equal to base square plus perpendicular square. Hypotenuse is 15 base is 14 minus x and perpendicular is h. So this will become. So here in triangle ADC hypotenuse is 15. So this will become 15 square is equal to base is 14 minus x whole square plus perpendicular is h. So this will become h square. Again we will take this expression to the left hand side. So this will become 15 square is simply 225 minus 14 minus x whole square is equal to h square. Let us suppose this is our equation number second. Relative to both equation, equation number first and second both had the same right hand side. So here we will compare the left hand sides. So from here we will get this implies the left hand side of equation number first is 169 minus x square is equal to left hand side of equation number second is 225 minus 14 minus x whole square. Let us simplify this equation for the value of x. So first we will expand this one identity. So this will become 169 minus x square is equal to 225 minus. We will expand this using a minus b whole square identity. So this will become 14 square plus x square minus 2 times 14 x which is 28 x. Next we multiply this negative sign inside the parenthesis. So this will become this is 169 minus x square is equal to 225. This will become minus 14 square is above 196 minus x square and negative time negative is positive so it's become positive 28 x. So let's further simplify this one equation. So this is 169 minus x square is equal to 225 minus 196 it is above 29 minus x square plus 28 x. Looking to both sides of this equation there is negative x square in both sides. We can cancel them with each other. This one equation will become this is 28 x and here we take this 29 to the left hand side. So this will become 169 minus 29 is equal to 28 x. So we can write this as 28 x is equal to 169 minus 29 it is above 140. We will divide both sides by 28. So the value of x will become 28 and 28 will be cancelled and 28 times minus 28. 28 times minus 28 is above 140. So here the value of x will become x is equal to 5 units. I have substituted this value of x in any one equation equation number first or second to find out the value of h. So for that we substitute the value of x in equation number first. So our equation number first will become equation number first implies our equation number first is h square is equal to 169 minus x square. So let's substitute x is equal to 5. This equation equation will become x square is equal to 169 minus x is 5 this will become 5 square. So h square is equal to 169 minus 5 is simply 5 square is 25. 169 minus 25 it is above 144. So h square is equal to 144. Here we will take square root on both sides. So square root will be cancelled. So the final value of h will become so root of 144 is above this is 12. So the value of h which is the height of the above triangle that is 12 units. So finally here in this figure the value of this x it is above 5 and the value of this height h is it is 12 units. Now what is our next target? Our next target is to find out the radius of this circle. So to find out the radius of this circle here we will draw a diameter on this circle from point A. So this figure will become now let us suppose this is point E and let's suppose this is the center of this circle O. To find out the radius of this circle here we will try to find out the value of this segment AO or OE. So for that here we will connect this point E with point C. So this figure will become let us connect this point with point C. Now look into the figure this figure triangle ABD. This is a right angle triangle where the angle at point D is a right angle 90 degrees angle. So what do you think about the triangle AEC? Is this is a right angle triangle or not? By Taylor's theorem if there is any triangle inside the semicircle or half circle upper circle then that triangle will be a right angle triangle. So by Taylor's theorem here the triangle AEC is a right angle triangle where the angle at point C is a right angle. Look to both the triangles triangles ABD and triangle AEC. These two angles at the same corresponding points are right angles 90 degree angles same angles. But if we extend this side of the triangle ABD so these two triangles triangles ABD and triangle AEC has the same common external arc and we also know that if two triangles have the same common external arc then the angles subtended by the same arc are always equal or same. So here these two angles are subtended by this arc the angle at point B and the angle at point E. So these two triangles will be equal these two angles will be equal. So let us suppose this is angle alpha so this will be also angle alpha. Look into the triangles triangle ABD and triangle AEC. These two triangles have the same these two corresponding angles and by angle angle properties of two triangles these triangles are similar triangles and we also know that in similar triangles the ratio among any two sides of one triangle are proportional to the ratio among the same corresponding sides of the second triangle. Then we will take the ratio among their perpendiculars and their hypotenuse. So this figure will become so here we will take the ratio of perpendicular to hypotenuse of this triangle and this triangle. So here in this triangle our perpendicular is AC and hypotenuse is AE and in this triangle our perpendicular is AD hypotenuse is AB. So this will become AC divided by AE is equal to here perpendicular is AD divided by hypotenuse is AB. Now let's substitute the values from the figure so here the value of AC is 15 units AE is not given that is the diameter of this circle AD is 12 units and AB is 13 units. Let's substitute these values here so the value of AC is 15 units divided by AE is not given AD is 12 units and AB is 13 units. Let's simplify this one equation for the value of AE which is the diameter of the above circle. So let's cross multiply these numbers so this will become 12 times AE is equal to this is 13 times 15. Let's divide both sides by 12 so this number this equation will become 12 and 12 will be cancel. So the value of AE will become that is 13 times 15. Now 13 times 15 it is about 195 divided by 12 and 195 divided by 12 it gives him about 16.25. Here AE is the diameter of this circle which is 16.25. Now our target is to find out the radius of the circle so the radius is the half of the diameter so therefore for that we will divide the value of AE by 2. So therefore our radius will become radius is half of the diameter so this will become AE divided by 2. So here the value of AE is 16.25 divided by 2 and 16.25 divided by 2 it gives him about radius that is equal to it gives him about 8.125 units. So finally our radius is 8.125 units that is our final answer and final target. So finally the radius of this circle it is about 8.125 units that is our final target. And last don't forget to subscribe my YouTube channel for such a more interesting and step-by-step geometry solutions. Thanks for watching.