A circle is inscribed in a quadrilateral ABCD in which B = 90o. If AD = 23 cm, AB = 29 cm and DS = 5 cm. Find the radius of the circle.
Since tangents drawn from an external point to a circle are equal.
DR = DS = 5cm
Now, AR = AD - DR = 23 - 5 = 18cm
AQ = 18cm
Also, BQ = AB - AQ = 29- 18 = 11cm
But, BP = BQ
Also, Q=P =90o.
In quadrilateral OQBP,
QOP = 360° - (P+Q+B)
= 360° - (90°+90°+90°) = 90°
Hence, OQBP is a square.
Hence, radius of the circle is 11cm.
You have rated this answer /10
- CBSE Sample Papers for Class 10 Mathematics
- R S Aggarwal and V Aggarwal Textbook Solutions for Class 10 Mathematics
- RD Sharma Textbook Solutions for Class 10 Mathematics
- NCERT Textbook Solutions for Class 10 Mathematics
- CBSE Syllabus for Class 10 Mathematics
- CBSE Previous year papers with Solutions Class 10 Mathematics