In fig., a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively, If AB = 29 cm, AD = 23 cm, B = 90o and DS = 5 cm, then the radius of the circle (in cm) is
Asked by nanda kumar | 22nd Feb, 2014, 07:57: PM
Given that DS = 5 cm;
Since DS and DR are tangents from the same external point to the circle, DS = DR = 5 cm
Since AD = 23 cm, AR = AD - DR = 23 - 5 = 18 cm.
Similarly, AR and AQ are the tangents from the same external point to the circle and hence AR = AQ = 18 cm.
Since AB = 29 cm, BQ = AB - AQ = 29 - 18 = 11 cm.
Since CB and AB are the tangents to the circle, angle OPB and angle OQB is equal to 90 degrees.
Given that angle B is 90 degrees and hence angle POQ is also equal to 90 degrees and hence OQBP is a square.
Since BQ is 11 cm, the side of the square OQBP is 11 cm
From the figure it is clear that the side of the square is the radius of the circle and hence radius of the circle is 11 cm.
Answered by | 25th Feb, 2014, 09:24: AM
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