 Hello and welcome to the session I am Deepika here. Let's discuss the question which says in given figure a triangle ABC is drawn to circumscribe a circle of radius 4 centimeter such that the segment BD and BC into which BC is divided by the point of contact DR of lengths 8 centimeter and 6 centimeters respectively finds the size AB and AC. Now we know that the lengths of tangents drawn from an external point to a circle are equal. Also we know that area of a right triangle 1 by 2 into base into height again we will use the hero's formula here for the area of a triangle root of S into S minus A into S minus B into S minus C where S is equal to A plus B plus C upon 2 and where ABC are the lengths of the sides of a triangle. So this is a key idea behind our question we will take the help of this key idea to solve the above question. So let's start the solution now we are given a triangle ABC which circumscribes a circle with center O and radius 4 centimeter also we are given BD is equal to 8 centimeter and DC is equal to 6 centimeter. Now we want to find sides AB and AC of a triangle. Constructions join AO, BO and CCO. Be the points of contact sides AB and AC respectively. See in the figure two tangents are drawn to the circle with center O from an external point B therefore BD is equal to is equal to BE and BD is given to us 8 centimeter because the lengths of tangent drawn from an external point to a circle are equal so BE is 8 centimeter. Again two tangents are drawn to the circle with center O from an external point E. So we have CD is equal to CF is equal to 6 centimeter. Again two tangents are drawn to the circle from an external point A let AF is equal to AE is equal to X centimeter. Therefore in triangle ABC AC is equal to AF plus FC and this is equal to X plus 6 centimeter and AB is equal to AE plus BE and this is equal to X plus 8 centimeter and BC is equal to BD plus DC this is 8 plus 6 that is 14 centimeter. Now we know the sides of a triangle ABC and we can find out the area of triangle ABC. So let us use the Vero's formula to find the area of triangle ABC. So first we will find out as S is equal to AB plus BC plus CA upon 2 and this is equal to now AB is given to us X plus 8 centimeter and BC is 14 centimeter plus CA is X plus 6 centimeter upon 2 and this is equal to 2X plus 28 by 2 which is again equal to 2 into X plus 14 upon 2. Now cancellation we have S is equal to X plus 14 area of a triangle ABC is under root of S into S minus A into S minus V into S minus C where AB and C are the sides of a triangle so here we have area of triangle ABC is equal to under root of X that is X plus 14 into S minus A X plus 14 minus X minus 8 into X minus C X plus 14 minus 14 into S minus C X plus 14 minus X minus 6 and this is equal to under root of X plus 14 into 6 into X into 8. Therefore area of triangle ABC is equal to under root of 48 X into X plus 14. So let us give this as number 1. Again from the figure we see area of triangle ABC is equal to area of triangle OBC plus area of triangle OAB plus area of triangle OAC. The radius of a triangle is given to us 4 centimeter that is OD is equal to OE is equal to OF is equal to 4 centimeter hence OE is perpendicular to AB and OF is perpendicular to AC and similarly OD is perpendicular to BC and area of triangle OBC is equal to half into base into height this is equal to half into 14 into 4 and this is equal to 28 centimeter area of triangle OAB is equal to half into base AB into height OE and this is equal to half into AB is the X plus 8 centimeter into OE is the radius of this circle which is 4 and this is equal to 2 into X plus a centimeter square. Again area of triangle AOC is equal to half into AC into OF and AC X plus 6 centimeter triangle OBC plus area of triangle OAB plus area of triangle AOC is equal to 28 plus 2 into X plus 8 plus 2 into X plus 16 centimeter square and this is equal to 28 plus 2X plus 16 plus 2X plus 32 and this is equal to 4X plus 56 let us give this as number 2. Now we know that area of triangle ABC is equal to sum of the areas of triangle OBC plus triangle OAB plus triangle AOC. Now from one we have area of triangle ABC is under root of 48X into X plus 14 and from two we have the sum of the areas of these triangle is 4X plus 56 this implies under root of 16 into 3X into X plus 14 is equal to 4 into X plus 14 this implies 4 into under root of 3X into X plus 14 is equal to 4 into X plus 14 now we will solve this equation on scaling both sides we get X into X plus 14 is equal to X plus 14 whole square this implies 3X square plus 42X is equal to X square plus 28X plus 196 this implies 3X square minus X square plus 42X minus 28X minus 196 is equal to 0 this implies 2X square plus 14X minus 196 is equal to 0 divide this equation by 2 we have X square plus 7X minus 98 is equal to 0 now we will factorize this equation this implies X square plus 14X minus 7X minus 98 is equal to 0 this implies this implies X into X plus 14 minus 7 into X plus 14 is equal to 0 X plus 14 into X minus 7 is equal to 0 therefore X plus 14 is equal to 0 or X minus 7 is equal to 0 that is X is equal to minus 14 or X is equal to 7 this implies X is equal to 7 because we can't take X is equal to minus 14 as length of side cannot be negative hence side AC is equal to X plus 6 centimetre and this is equal to 7 plus 6 13 centimetre and AB is equal to X plus 8 centimetre and this is equal to 15 centimetre hence the answer for the above question is AB is equal to 15 centimetre and AC is equal to 13 centimetre I hope the solution is clear to you bye and take care