KnowledgeBoat Logo

Mathematics

From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

  1. 60 cm2

  2. 65 cm2

  3. 30 cm2

  4. 32.5 cm2

Circles

1 Like

Answer

Given, the point P is 13 cm from O, the centre of the circle as shown in the figure below:

From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is (a) 60 cm2 (b) 65 cm2 (c) 30 cm2 (d) 32.5 cm2. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Radius of the circle (OQ) = 5 cm

PQ and PR are tangents from P to the circle.

PQ ⊥ OQ (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ OQP = 90°.

So, in △OQP,

OP2=OQ2+PQ2132=52+PQ2PQ2=13252PQ2=16925PQ2=144PQ=12 cmOP^2 = OQ^2 + PQ^2 \\[1em] 13^2 = 5^2 + PQ^2 \\[1em] PQ^2 = 13^2 - 5^2 \\[1em] PQ^2 = 169 - 25 \\[1em] PQ^2 = 144 \\[1em] PQ = 12 \text{ cm}

Area of △OPQ = 12×PQ×OQ\dfrac{1}{2} \times PQ \times OQ = 12×12×5=30\dfrac{1}{2} \times 12 \times 5 = 30 cm2.

Similarly,

PR ⊥ OR (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ ∠ORP = 90°.

So, in △ORP,

OP2=OR2+PR2132=52+PR2PR2=13252PR2=16925PR2=144PR=12 cmOP^2 = OR^2 + PR^2 \\[1em] 13^2 = 5^2 + PR^2 \\[1em] PR^2 = 13^2 - 5^2 \\[1em] PR^2 = 169 - 25 \\[1em] PR^2 = 144 \\[1em] PR = 12 \text{ cm}

Area of △POR = 12×PR×OR\dfrac{1}{2} \times PR \times OR = 12×12×5=30\dfrac{1}{2} \times 12 \times 5 = 30 cm2.

Area of quadrilateral PQOR = Area of △POR + Area of △OPQ = 30 + 30 = 60 cm2.

Hence, Option 1 is the correct option.

Answered By

1 Like


Related Questions