 Hello and welcome to the session. In this session we discuss the following question which says P2R is a right triangle, right angle at Q. A circle is inscribed in it. The length of the two sides containing the right angle are 3 cm and 4 cm. Find the radius of the circle. We must recall one fact which says that the lengths of the tangents drawn from an external point to a circle are equal. This is the key idea to be used in this question. Let's see the solution now. Consider this figure. So here we are given that PQR is a right angled triangle where we have angle Q is equal to 90 degrees and we are given that the lengths of the two sides containing the right angle that is angle Q are 3 cm and 4 cm. So we take QR equal to 3 cm and PQ equal to 4 cm. It's also given that a circle is inscribed in the triangle. The circle touches the side PQ of triangle PQR at point A, the side QR at point B and side PR at point C and we have taken O to be the center of the circle. The F0 let the radius of the in circle be x cm. That is we have OA is equal to OB is equal to OC is equal to x cm. Now since we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore this angle that is angle OAQ is equal to angle OBQ is equal to 90 degrees. So obviously this angle would also be equal to 90 degrees that is angle AOB is also equal to 90 degrees. So in the figure OAQB all angles are of measure 90 degrees. So therefore we can say that OAQB is a parallelogram and thus opposite sides of parallelogram OAQB are equal. So this would mean that OA is equal to OB is equal to QB is equal to QA is equal to x cm. Now that it's given PQ equal to 4 cm and AQ or QA is equal to x cm. So this would mean that PA is equal to 4 minus x cm. Since from the figure we have that PA is equal to PQ minus AQ. Also we have QR is equal to 3 cm and QB is equal to x cm. Therefore VR is equal to 3 minus x cm. Since from the figure we have that VR is equal to QR minus QB. Now PA would be equal to PC equal to 4 minus x cm. Since they are the tangents drawn from the point P also VR would be equal to RC equal to 3 minus x cm. Since they are also the tangents drawn from the point R. Now from the figure we have PR is equal to PC plus RC. Now this PC is equal to 4 minus x plus RC which is 3 minus x. So we get PR is equal to 7 minus 2 x cm. So now in right triangle PQR we have PR square is equal to PQ square plus QR square by the Pythagoras theorem. So from here we get 7 minus 2 x the whole square is equal to 4 square plus 3 square. This gives us 49 minus 28x plus 4x square is equal to 16 plus 9. That is we get 4x square minus 28x plus 49 minus 25 is equal to 0. This gives us 4x square minus 28x plus 24 is equal to 0. This further gives us x square minus 7x plus 6 is equal to 0. Now splitting the middle term we get x square minus x minus 6x plus 6 is equal to 0. So from the first pair we take out x common inside the bracket we have x minus 1. From the second pair we take out minus 6 common inside the bracket we have x minus 1 this is equal to 0. So this gives us x minus 1 into x minus 6 is equal to 0. This gives us x minus 1 equal to 0 or x minus 6 equal to 0. That is we have x is equal to 1 or x is equal to 6. Now we cannot take x equal to 6 since if we take x equal to 6 this would mean that this AQ is also of measure 6 cm but this cannot be possible since PQ is of measure 4 cm and AQ is part of PQ. So this cannot be larger in length than PQ. So therefore neglecting x equal to 6 we get x is equal to 1 and we had assumed x to be the radius of the in circle. So we say radius of the in circle is equal to 1 cm. So this is our final answer. This completes the session. Hope you have understood the solution of this question.