SSC CGL Exam > SSC CGL Questions > From a point P which is at a distance of 13 c...
Start Learning for Free
From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)
- a)65 cm2
- b)60 cm2
- c)30 cm2
- d)90 cm2
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
From a point P which is at a distance of 13 cm from center O of a circ...
In ΔOQP and ΔORP,
∠OQP = ∠ORP = 90°
OR = OR {= radius}
PQ = PR {tangent}
∴ ΔOQP ≌ ΔORP
Area of ΔOQP = Area of ΔORP
Most Upvoted Answer
From a point P which is at a distance of 13 cm from center O of a circ...
To find the area of quadrilateral PQOR, we can break it down into two triangles - ΔPQO and ΔPRO, and a rectangle OQRP.
1. Finding the length of the tangents:
From the given information, we know that OP = 13 cm (distance between point P and center O) and the radius of the circle is 5 cm. We can use the Pythagorean theorem to find the length of the tangents PQ and PR.
Using Pythagorean theorem,
OP² = OQ² + PQ²
13² = 5² + PQ²
169 = 25 + PQ²
PQ² = 144
PQ = 12 cm (since lengths are positive)
Similarly, we can find PR using the same method:
PR² = OQ² + QR²
13² = 5² + QR²
169 = 25 + QR²
QR² = 144
QR = 12 cm
2. Finding the area of triangles ΔPQO and ΔPRO:
The area of a triangle can be found using the formula: Area = (base * height) / 2.
For triangle ΔPQO:
Base = PQ = 12 cm
Height = OQ = 5 cm (radius of the circle)
Area of ΔPQO = (12 * 5) / 2 = 60 cm²
Similarly, for triangle ΔPRO:
Base = PR = 12 cm
Height = OR = 5 cm (radius of the circle)
Area of ΔPRO = (12 * 5) / 2 = 60 cm²
3. Finding the area of rectangle OQRP:
The area of a rectangle can be found using the formula: Area = length * width.
Length = OR = 2 * radius = 2 * 5 = 10 cm
Width = PQ = 12 cm
Area of OQRP = 10 * 12 = 120 cm²
4. Finding the area of quadrilateral PQOR:
To find the area of the quadrilateral PQOR, we add the areas of triangles ΔPQO and ΔPRO, and subtract the area of rectangle OQRP.
Area of quadrilateral PQOR = (Area of ΔPQO + Area of ΔPRO) - Area of OQRP
= (60 + 60) - 120
= 120 - 120
= 0 cm²
Therefore, the area of quadrilateral PQOR is 0 cm².
Note: It seems there might be a mistake in the options provided for the answer. The correct answer should be 0 cm², not 60 cm² (option B).
1. Finding the length of the tangents:
From the given information, we know that OP = 13 cm (distance between point P and center O) and the radius of the circle is 5 cm. We can use the Pythagorean theorem to find the length of the tangents PQ and PR.
Using Pythagorean theorem,
OP² = OQ² + PQ²
13² = 5² + PQ²
169 = 25 + PQ²
PQ² = 144
PQ = 12 cm (since lengths are positive)
Similarly, we can find PR using the same method:
PR² = OQ² + QR²
13² = 5² + QR²
169 = 25 + QR²
QR² = 144
QR = 12 cm
2. Finding the area of triangles ΔPQO and ΔPRO:
The area of a triangle can be found using the formula: Area = (base * height) / 2.
For triangle ΔPQO:
Base = PQ = 12 cm
Height = OQ = 5 cm (radius of the circle)
Area of ΔPQO = (12 * 5) / 2 = 60 cm²
Similarly, for triangle ΔPRO:
Base = PR = 12 cm
Height = OR = 5 cm (radius of the circle)
Area of ΔPRO = (12 * 5) / 2 = 60 cm²
3. Finding the area of rectangle OQRP:
The area of a rectangle can be found using the formula: Area = length * width.
Length = OR = 2 * radius = 2 * 5 = 10 cm
Width = PQ = 12 cm
Area of OQRP = 10 * 12 = 120 cm²
4. Finding the area of quadrilateral PQOR:
To find the area of the quadrilateral PQOR, we add the areas of triangles ΔPQO and ΔPRO, and subtract the area of rectangle OQRP.
Area of quadrilateral PQOR = (Area of ΔPQO + Area of ΔPRO) - Area of OQRP
= (60 + 60) - 120
= 120 - 120
= 0 cm²
Therefore, the area of quadrilateral PQOR is 0 cm².
Note: It seems there might be a mistake in the options provided for the answer. The correct answer should be 0 cm², not 60 cm² (option B).
Free Test
FREE
| Start Free Test |
Community Answer
From a point P which is at a distance of 13 cm from center O of a circ...
In ΔOQP and ΔORP,
∠OQP = ∠ORP = 90°
OR = OR {= radius}
PQ = PR {tangent}
∴ ΔOQP ≌ ΔORP
Area of ΔOQP = Area of ΔORP
Attention SSC CGL Students!
To make sure you are not studying endlessly, EduRev has designed SSC CGL study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in SSC CGL.
|
Explore Courses for SSC CGL exam
|
|
Similar SSC CGL Doubts
Top Courses for SSC CGLView all
From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer?
Question Description
From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct 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 From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct 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 From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer?.
From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct 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 From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct 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 From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer?.
Solutions for From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC CGL.
Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free.
Here you can find the meaning of From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)a)65 cm2b)60 cm2c)30 cm2d)90 cm2Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice SSC CGL tests.
|
|
Explore Courses for SSC CGL exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup