1. From a point which is at a distance of 13 cm from centre O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is
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By: anil on 05 May 2019 01.55 am
$$ecause$$ PQ is a tangent to circle at Q,
$$ herefore angle{OQP}=90$$
In $$ riangle{POQ}$$,
$$PO^{2}=OQ^{2}+PQ^{2}$$
$$13^{2}=5^{2}+PQ^{2}$$
$$PQ=12 cm$$
Similarly, $$PR=12 cm$$
$$ecause riangle{POQ}$$ is a right-angled triangle,
$$ herefore$$ Area of $$ riangle{POQ}$$=$$frac{1}{2} imes{PQ} imes{OQ}$$
$$=frac{1}{2} imes5 imes12$$
$$=30$$
Area of quadrilateral POQR=$$2 imes$$ Area of $$ riangle{POQ}$$
$$=2 imes30$$
$$=60 cm^{2}$$
Hence, Option B is correct.
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$$ecause$$ PQ is a tangent to circle at Q,
$$ herefore angle{OQP}=90$$
In $$ riangle{POQ}$$,
$$PO^{2}=OQ^{2}+PQ^{2}$$
$$13^{2}=5^{2}+PQ^{2}$$
$$PQ=12 cm$$
Similarly, $$PR=12 cm$$
$$ecause riangle{POQ}$$ is a right-angled triangle,
$$ herefore$$ Area of $$ riangle{POQ}$$=$$frac{1}{2} imes{PQ} imes{OQ}$$
$$=frac{1}{2} imes5 imes12$$
$$=30$$
Area of quadrilateral POQR=$$2 imes$$ Area of $$ riangle{POQ}$$
$$=2 imes30$$
$$=60 cm^{2}$$
Hence, Option B is correct.