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Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is
- a)9π - 18
- b)18
- c)9π
- d)9
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let B...
Let AB = a (a = 6)
CQB is a semicircle of radius a/√2
CPB is a quarter circle (quadrant) of radius a
So, area of semicircle = π*a2/4
Area of quadrant =π*a2/4
So, area of region enclosed by BPC, BQC = Area of Δ(ABC) = 18.
Option (B)
CQB is a semicircle of radius a/√2
CPB is a quarter circle (quadrant) of radius a
So, area of semicircle = π*a2/4
Area of quadrant =π*a2/4
So, area of region enclosed by BPC, BQC = Area of Δ(ABC) = 18.
Option (B)
Most Upvoted Answer
Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let B...
Understanding the Triangle and Shapes
Given a right-angled isosceles triangle ABC with \( AB = AC = 6 \, \text{cm} \) and \( BC \) as the hypotenuse, we first calculate the length of \( BC \) using the Pythagorean theorem.
- Calculate Length of BC:
\[
BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \, \text{cm}
\]
Area of the Semi-Circle BQC
The semi-circle \( BQC \) has a diameter \( BC \). Thus, its radius \( r \) is half of \( BC \):
- Radius of Semi-Circle:
\[
r = \frac{BC}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \, \text{cm}
\]
- Area of Semi-Circle:
\[
\text{Area}_{\text{BQC}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3\sqrt{2})^2 = \frac{1}{2} \pi \cdot 18 = 9\pi \, \text{sq cm}
\]
Area of Sector BPC
The arc \( BPC \) is a sector of a circle centered at \( A \) with radius \( AB = 6 \, \text{cm} \).
- Area of Sector:
The angle \( \angle BAC \) is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.
\[
\text{Area}_{\text{BPC}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 6^2 \cdot \frac{\pi}{2} = \frac{1}{2} \cdot 36 \cdot \frac{\pi}{2} = 9\pi \, \text{sq cm}
\]
Area of the Region Enclosed by BPC and BQC
To find the enclosed area between arc \( BPC \) and semi-circle \( BQC \):
- Total Area:
\[
\text{Enclosed Area} = \text{Area}_{\text{BPC}} - \text{Area}_{\text{BQC}} = 9\pi - 18
\]
Thus, the area of the region enclosed by \( BPC \) and \( BQC \) is \( 9\pi - 18 \, \text{sq cm} \), confirming option B as the correct answer.
Given a right-angled isosceles triangle ABC with \( AB = AC = 6 \, \text{cm} \) and \( BC \) as the hypotenuse, we first calculate the length of \( BC \) using the Pythagorean theorem.
- Calculate Length of BC:
\[
BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \, \text{cm}
\]
Area of the Semi-Circle BQC
The semi-circle \( BQC \) has a diameter \( BC \). Thus, its radius \( r \) is half of \( BC \):
- Radius of Semi-Circle:
\[
r = \frac{BC}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \, \text{cm}
\]
- Area of Semi-Circle:
\[
\text{Area}_{\text{BQC}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3\sqrt{2})^2 = \frac{1}{2} \pi \cdot 18 = 9\pi \, \text{sq cm}
\]
Area of Sector BPC
The arc \( BPC \) is a sector of a circle centered at \( A \) with radius \( AB = 6 \, \text{cm} \).
- Area of Sector:
The angle \( \angle BAC \) is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.
\[
\text{Area}_{\text{BPC}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 6^2 \cdot \frac{\pi}{2} = \frac{1}{2} \cdot 36 \cdot \frac{\pi}{2} = 9\pi \, \text{sq cm}
\]
Area of the Region Enclosed by BPC and BQC
To find the enclosed area between arc \( BPC \) and semi-circle \( BQC \):
- Total Area:
\[
\text{Enclosed Area} = \text{Area}_{\text{BPC}} - \text{Area}_{\text{BQC}} = 9\pi - 18
\]
Thus, the area of the region enclosed by \( BPC \) and \( BQC \) is \( 9\pi - 18 \, \text{sq cm} \), confirming option B as the correct answer.
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Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC isa)9π - 18b)18c)9πd)9Correct answer is option 'B'. Can you explain this answer?
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Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC isa)9π - 18b)18c)9πd)9Correct answer is option 'B'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC isa)9π - 18b)18c)9πd)9Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC isa)9π - 18b)18c)9πd)9Correct answer is option 'B'. Can you explain this answer?.
Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC isa)9π - 18b)18c)9πd)9Correct answer is option 'B'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC isa)9π - 18b)18c)9πd)9Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC isa)9π - 18b)18c)9πd)9Correct answer is option 'B'. Can you explain this answer?.
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