 Hello and welcome to the session. In this session we discuss the following question which says in figure, that is this figure. A circle is inscribed in a quadrilateral A, B, C, D in which angle V is equal to 90 degrees. If A, D is equal to 23 centimeters, A, B equal to 29 centimeters and D is equal to 5 centimeters, find the radius R of the circle. We know that the tangent to a circle is perpendicular to the radius through the point of contact. The lengths of two tangents from an external point to a circle are equal. This is the key idea for this question. Now let's move on to the solution. This is the given figure. Let's see what all is given to us. We are given that angle B is equal to 90 degrees, that is this angle, then A, D is equal to 23 centimeters, A, B is equal to 29 centimeters and D is equal to 5 centimeters. So we have marked the lengths of A, D, A, B and D is. We need to find the radius R of the circle. Now as you can see in the figure, that is this B, C is the tangent to the circle. So we have since B, C is tangent to the circle, so from the key idea which says that the tangent to a circle is perpendicular to the radius through the point of contact. So we have that angle OPB would be equal to 90 degrees since this OP is the radius of the circle. So using the statement of the key idea, this OP would be perpendicular to the tangent BC through the point of contact that is P. Hence angle OPB is 90 degrees. Also since AB is the tangent to the circle, so we have angle OQB would be equal to 90 degrees since AB is the tangent and OQ is the radius of the circle. So OQ would be perpendicular to the tangent AB through the point of contact that is Q. We also have that OP is equal to OQ is equal to R that is the radius of the circle. Then we have found out that angle OQB is equal to angle OPB is equal to 90 degrees. Then it is given that angle B is equal to 90 degrees. So obviously angle QOP would also be equal to 90 degrees that is this angle. Then we also have BP equal to BQ since these are the tangents from the external point B and we know that the lengths of the two tangents from an external point to a circle are equal. So BP would be equal to BQ. Therefore from all these things we say that OPBQ is a square. Now we have DR would be equal to DS since these are the tangents from the external point D and we know that lengths of the tangents from an external point to a circle are equal. Then again we have AR is equal to AQ since these are the tangents from an external point A also we have that BP is equal to BQ. These are also the tangents from an external point B. Now DR would be equal to 5 centimeters since DR is equal to DS and it is given that DS is equal to 5 centimeters. So we have got DR equal to 5 centimeters. Now from this figure you can see that DR is equal to AD minus AR. So we have AD minus AR is equal to 5 centimeters and it is given that AD is 23 centimeters thus we say that 23 centimeters minus AR is equal to 5 centimeters. So from here we have AR is equal to 23 minus 5 centimeters which is equal to 18 centimeters is the length of AR. Now since we have that AR is equal to AQ so this implies that AQ is equal to 18 centimeters. From this figure we have AQ is equal to AB minus BQ so we write here AB minus BQ is equal to 18 centimeters. We are given that AB is 29 centimeters so we have 29 centimeters minus BQ is equal to 18 centimeters. This gives us BQ is equal to 29 minus 18 centimeters so from here we have BQ is equal to 11 centimeters. Now since we have that OP BQ is a square therefore we have OP is equal to OQ is equal to PB is equal to BQ and we already have that BQ is 11 centimeters so we have OP is equal to OQ which is equal to R is equal to 11 centimeters. Since we have OP and OQ are the radius of the circle that is R thus our final answer is radius R of the circle is equal to 11 centimeters. So this completes the session hope you have understood the solution for this question.