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In the given figure, a circle is inscribed in a quadrilateral $ABCD$ in which $\angle B = 90^\circ$. If $AD=17$ cm, $AB = 20$ cm and $DS = 3$ cm, then find the radius of the circle.
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$DR = DS = 3$ cm
$\therefore AR = AD – DR = 17 – 3 = 14$ cm
$\Rightarrow AQ = AR = 14$ cm
$\therefore QB = AB – AQ = 20 – 14 = 6$ cm
Since $QB = OP = r \therefore$ radius = 6 cm
$\therefore AR = AD – DR = 17 – 3 = 14$ cm
$\Rightarrow AQ = AR = 14$ cm
$\therefore QB = AB – AQ = 20 – 14 = 6$ cm
Since $QB = OP = r \therefore$ radius = 6 cm