Question

From a point P which is
tangents PQ and PR
(d) 1.4
P which is at a distance of 13 cm from the centre o of a ditde of radias Som the
and PR to the circle are drawn. Then the area of the quadrilateral POOR
(d) 125 cm
(a) 60 cm
(b) 65 cm
(c) 30 cm
12 cm
5 cm
Р.
13 cm
ST70
12 cm
5 cm
of the square that can be inscribed in a circle of radim​

Answer 1

Answer:

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

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