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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?
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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).
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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
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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?
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