From a point P which is at a distance of 10 cm from the centre O of a circle of radius 6 cm, a pair of tangents PQ and PR to the circle at point Q and respectively, are drawn. Then the area of the quadrilateral PQOR is equal to
Answer & explanation
Correct answer: option 4
In \(\Delta \)OQP
\( {(OP) }^{2 } \) = \( {(OQ) }^{2 } \) + \( {(QP) }^{2 } \)
= \( {(10) }^{2 } \) = \( {(6) }^{2 } \) + \( {(QP) }^{2 } \)
= \( {(QP) }^{2 } \) = 100 - 36 = 64
= QP = \(\sqrt {64 }\) = 8
Area of \(\Delta \)OPQ = \(\frac{1}{2}\) x Base x height = \(\frac{1}{2}\) x 6 x 8 = 24 cm
Area of quadrilateral PQOR = 2 x Area of \(\Delta \)OPQ = 2 x 24 = 48 cm.