From a point P which is at a distance of 13 cm from the center 0 of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle is drawn. Then, the area of the quadrilateral PQOR is
Let us draw a circle of radius 5 cm having center O. P is a point at a distance of 13 cm from O. A pair of tangents PQ and PR are drawn.
Also, OQ = OR = radius = 5cm [1]
And OP = 13 cm
As OQ ⏊ PQ and OR ⏊PR [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]
△POQ and △POR are right-angled triangles.
In △PQO By Pythagoras Theorem [ i.e. (base)2+ (perpendicular)2= (hypotenuse)2 ]
(PQ)2 + (OQ)2= (OP)2
(PQ)2 + (5)2 = (13)2
(PQ)2 + 25 = 169
(PQ)2 = 144
PQ = 12 cm
And
PQ = PR = 12 cm [ tangents through an external point to a circle are equal] [2]
Area of quadrilateral PQRS, A = area of △POQ + area of △POR
[Using 1 and 2]