A Nice Geometry Problem ‐ Can you Find the Height?

Hello everyone, you are welcome. Today we have a nice geometric math problem, that is here we have given a right angled triangle and there is a circle inside the triangle. We have to find out the height and perpendicular of this triangle. What we have given in this problem? Here we have given the base of this triangle which is 20 units and also given the radius of this circle which is 4 units. So we will try to find out the width of edge. How can we solve this math problem? First of all let us suppose this is our triangle ABC. Let us suppose this is point A, this is point B and this is point C. Let us suppose this is point B, this is point Q and this is point R where the circle touches the sides of this triangle. Now to find out the width of edge, first of all let us connect this center with this point V and this point Q, so this figure will become. Now here this is the radius of this circle which is 4 units, so this is also radius, so this will be also 4 units and this will be also 4 units. Now if this, let us suppose this is our center O, if OQ is 4 units, so this PB will be also 4 units and this is 4 units, so this will be also 4 units. Now look at this side of the triangle BC, here it has total length of 20 units but here this length BQ is 4 units, so what will be the remaining length QC? This will be 20 minus 4 which is 16, so this will be 16 units. What about this one, what will be this one then? Here we will use tangent to a circle theorem. That is if we have two tangents from an external point to a circle then the length of that two tangents will be equal congruent. So we have two tangents from point C to this circle tangent QC and tangent RC. Now QC is 16 units, so this tangent will be also 16 units, so this will be also 16 units. We have also two tangents from point A to this circle tangent AP and tangent AR which will be same in length but here the length of any one tangent is not given. So let us suppose this tangent is equal to x, so this will be also x. Now look into the figure here, what is the length of AB? This is simply x plus 4, so from here from this figure we can write our side length AB will be, our AB will become, this will become x plus 4. What is BC? BC is total 20 units, so the length of BC is 20 units. What will be the length of AC? AC is simply x plus 16, so this will become AC is equal to x plus 16. Now what is our target here? Since we have to find out the height of this triangle, the perpendicular of this triangle, so for that here we need the width of x. Now to find out the width of x as this is a right angled triangle, so here we will apply the Pythagoras theorem and we will find out the width of x. So by using Pythagoras theorem, by Pythagoras theorem, Pythagoras theorem states that hypotenuse square is equal to perpendicular square plus base square. So it will become hypotenuse square is equal to perpendicular square plus base square. Here our hypotenuse is AC and our perpendicular and height is AB and base is BC. So this will become AC square is equal to perpendicular AB, so this will become AB square plus base is BC, this will become BC square. So we substitute the width of AC, AB and BC from here, so this will become the width of AC is simply x plus 16, so this will become x plus 16 whole square is equal to AB is x plus 4, so it will become x plus 4 whole square plus BC is 20, so this will become 20 square. Now here in these two expressions we will use A plus B whole square identity. So here in these two expressions we will use this one identity A plus B whole square is equal to A square plus B square plus 2 times AB. So using this algebraic identity here these two terms will become, this will become x square plus 16 square plus 2 times 16 times x is equal to and this will become, this will become x square plus 4 square plus 2 times 4 times x plus 20 square is simply 400, this will become 400. Let's simplify both sides, here there is x square in both side which has positive sign, so we can cancel them with each other and let's simplify the other terms, the 16 square simply this is 256 plus 2 times 16 is simply 32, x is equal to 4 square is 16, so this is 16 and this is 400, so this will become 416 plus 2 times 4 is 8, so this will become 8x. Then we will move and take this variable to the left hand side and this constant number to the right hand side, so this will become 32x minus 8x is equal to 416 minus 256, 32 minus 8 is simply 24, so it will become 24x and this will become 6 minus 6 is 0, 41 minus 25 it is above 16, so this is 160. Now we will divide both sides by 24, so this 24 and this 24 will be cancel, so this will become x is equal to 160 divided by 24, if we divide 160 by 24 it gives me about 6.67 up to two decimal places, so finally x is equal to 6.67 which is the length of AB. Now here our target is to find out the height of this triangle, the length of AB, so therefore from here we can write the length of AB our AB is simply the height of the triangle which is x plus 4, but here the width of x is simply 6.67 plus 4, the 6.67 plus 4 it is simply 10.67, so finally this is our required height. So finally the height and perpendicular of this right angled triangle it is above this is simply 10.67 units which is our final target. And last don't forget to subscribe my youtube channel for such a more interesting and step-by-step geometry problems. Thanks for watching.