Question

O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O a point P is taken. From this point, two tangent PQ and PR are drawn to the circle. Then, the area of quad. PQOR is  (a) 60 cm2   (b) 32.5 cm2  (c) 65 cm2  (d) 30 cm2

Solution

GIVEN: PQ & PR are 2 tangents and QO & OR are 2 radius at contact point Q & R. Angle PQO=90° [A TANGENT TO A CIRCLE IS PERPENDICULAR TO THE RADIUS THROUGH THE POINT OF CONTACT] By Pythagoras theorem PQ²= OP² - OQ² PQ² = 13²- 5² = 169- 25= 144 PQ= √ 144= 12 PQ=12cm PQ= PR =12cm [The Lengths of two tangents drawn from an external point to a circle are equal] In ∆OPQ & ∆ OPR OQ= OR (5cm) given OP = OP ( Common) PQ= PR( 12cm) Hence ∆OPQ =~ ∆OPR ( by SSS congruence) Area of ∆OPQ =Area ∆OPR Area of quadrilateral QORP= 2×(area of ∆ OPR) Area of quadrilateral QORP= 2× 1/2 × base × altitude Area of quadrilateral QORP= OR× PR Area of quadrilateral QORP=12× 5= 60 cm² _____________________________ Area of quadrilateral QORP=60cm² ___________________________  MathematicsSecondary School Mathematics XStandard X

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