Question : Point O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, point P is taken. From this point, two tangents, PQ and PR, are drawn to the circle. Then, the area of quadrilateral PQOR is:
Option 1: $60\ \text{cm}^2$
Option 2: $32.5\ \text{cm}^2$
Option 3: $65\ \text{cm}^2$
Option 4: $30\ \text{cm}^2$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $60\ \text{cm}^2$
Solution :
Given: Point O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, point P is taken.
We know that the area of triangle is $\frac{1}{2} \times (\text{base}) \times (\text{height})$.
$\angle OQP=\angle ORP=90°$(Angle at a point of contact of the tangent)
Applying Pythagoras' theorem in $\angle OQP$,
$(QP)^2+(OQ)^2=(OP)^2$
⇒ $QP=\sqrt{(OP)^2–(OQ)^2}$
⇒ $QP=\sqrt{(13)^2–(5)^2}$
⇒ $QP=\sqrt{169–25}$
⇒ $QP=\sqrt{144}$
⇒ $QP=12\ \text{cm}$
The area of $\triangle OQP$ = $\frac{1}{2} \times 12 \times 5=30\ \text{cm}^2$
The area of quadrilateral $PQOR = 2 × 30 = 60\ \text{cm}^2$
Hence, the correct answer is $60\ \text{cm}^2$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
