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PA and PB are two tangents drawn from an external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be (SSC Sub. Ins. 2012)
- a)a square
- b)concyclic
- c)a rectangle
- d)a rhombus
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
PA and PB are two tangents drawn from an external point P to a circle ...
OAPB is concyclic because ∠A + ∠B = 180°
& ∠O + ∠P = 180°
Most Upvoted Answer
PA and PB are two tangents drawn from an external point P to a circle ...
Given:
PA and PB are two tangents drawn from an external point P to a circle with center O, where the points A and B are the points of contact.
To prove:
The quadrilateral OAPB is concyclic.
Proof:
Definition:
Concyclic points are points that lie on the same circle.
Construction:
Let's draw the diagram to understand the given information.
Proof by contradiction:
Assume that the quadrilateral OAPB is not concyclic.
1. Angle OPA:
Since PA is a tangent to the circle, angle OPA is a right angle. (Tangent is perpendicular to the radius at the point of contact)
2. Angle OPB:
Similarly, since PB is a tangent to the circle, angle OPB is also a right angle.
3. Angle OAP:
Since triangle OPA is a right-angled triangle, angle OAP is the complement of angle OPA. Therefore, angle OAP is also a right angle.
4. Angle OB:
Since triangle OPB is a right-angled triangle, angle OPB is the complement of angle OPB. Therefore, angle OPB is also a right angle.
5. Angle APB:
Since angle OPA and angle OPB are right angles, the sum of the angles in triangle APB is 180 degrees. Therefore, angle APB is a straight angle.
6. Conclusion:
From the above steps, we can see that all the angles in the quadrilateral OAPB are right angles (OPA, OPB, OAP, and OPB).
If all the angles in a quadrilateral are right angles, then the quadrilateral is a rectangle.
Therefore, the quadrilateral OAPB is a rectangle.
Hence, option 'c' is the correct answer.
PA and PB are two tangents drawn from an external point P to a circle with center O, where the points A and B are the points of contact.
To prove:
The quadrilateral OAPB is concyclic.
Proof:
Definition:
Concyclic points are points that lie on the same circle.
Construction:
Let's draw the diagram to understand the given information.
Proof by contradiction:
Assume that the quadrilateral OAPB is not concyclic.
1. Angle OPA:
Since PA is a tangent to the circle, angle OPA is a right angle. (Tangent is perpendicular to the radius at the point of contact)
2. Angle OPB:
Similarly, since PB is a tangent to the circle, angle OPB is also a right angle.
3. Angle OAP:
Since triangle OPA is a right-angled triangle, angle OAP is the complement of angle OPA. Therefore, angle OAP is also a right angle.
4. Angle OB:
Since triangle OPB is a right-angled triangle, angle OPB is the complement of angle OPB. Therefore, angle OPB is also a right angle.
5. Angle APB:
Since angle OPA and angle OPB are right angles, the sum of the angles in triangle APB is 180 degrees. Therefore, angle APB is a straight angle.
6. Conclusion:
From the above steps, we can see that all the angles in the quadrilateral OAPB are right angles (OPA, OPB, OAP, and OPB).
If all the angles in a quadrilateral are right angles, then the quadrilateral is a rectangle.
Therefore, the quadrilateral OAPB is a rectangle.
Hence, option 'c' is the correct answer.
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PA and PB are two tangents drawn from an external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be (SSC Sub. Ins. 2012)a)a squareb)concyclicc)a rectangled)a rhombusCorrect answer is option 'B'. Can you explain this answer?
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PA and PB are two tangents drawn from an external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be (SSC Sub. Ins. 2012)a)a squareb)concyclicc)a rectangled)a rhombusCorrect answer is option 'B'. Can you explain this answer? for SSC CGL 2024 is part of SSC CGL preparation. The Question and answers have been prepared according to the SSC CGL exam syllabus. Information about PA and PB are two tangents drawn from an external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be (SSC Sub. Ins. 2012)a)a squareb)concyclicc)a rectangled)a rhombusCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC CGL 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PA and PB are two tangents drawn from an external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be (SSC Sub. Ins. 2012)a)a squareb)concyclicc)a rectangled)a rhombusCorrect answer is option 'B'. Can you explain this answer?.
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