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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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Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let ...
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AB = AC = 6cm. Thus, BC = 
= 6√2 cm
= 6√2 cmThe required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC
Area of semicircle BQC
Diameter BC = 6√2cm
Radius = 6√2/2 = 3√2 cm
Area = πr2/2 = π*(3√2) /2 = 9π
Area of quadrant BPC
Area = πr2/4 = π * (6)2 /4 = 9π
Area of triangle ABC
Area = 1/2 * 6 * 6 = 18
The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC
= 9π - 9 π + 18 = 18
Most Upvoted Answer
Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let ...
Given information:
- Triangle ABC is right-angled and isosceles, with hypotenuse BC.
- BQC is a semi-circle with diameter BC.
- BPC is an arc of a circle centered at A, lying between BC and BQC.
- AB has length 6 cm.
To find: The area enclosed by BPC and BQC.
Solution:
1. Draw the diagram of triangle ABC with points B, C, and Q on the circumference of the semi-circle BQC, and point P on the arc BPC.
2. Since triangle ABC is isosceles, we can conclude that angle ABC = angle BCA. Let's call this angle x.
3. Since triangle ABC is right-angled, angle BAC = 90 degrees. Therefore, angle BCA = 90 - x degrees.
4. Since BQC is a semi-circle, angle BQC = 180 degrees.
5. In triangle BQC, angle B + angle Q + angle C = 180 degrees. Since BQC is a semi-circle, angle BQC is a right angle. Therefore, angle B + angle Q + 90 = 180 degrees. Simplifying, we get angle B + angle Q = 90 degrees. Since triangle ABC is isosceles, angle B = angle C. Therefore, angle C + angle Q = 90 degrees.
6. In triangle ABC, angle A + angle B + angle C = 180 degrees. Substituting the values we have, angle A + x + x = 180 degrees. Simplifying, we get angle A + 2x = 180 degrees. Therefore, angle A = 180 - 2x degrees.
7. In triangle BPC, angle B + angle P + angle C = 180 degrees. Since angle B = angle C, angle B + angle P + angle B = 180 degrees. Simplifying, we get 2(angle B + angle P) = 180 degrees. Therefore, angle B + angle P = 90 degrees.
8. In triangle ABC, AB = AC = 6 cm. Therefore, BC = 6√2 cm (using Pythagoras theorem).
9. The area enclosed by BPC and BQC can be divided into two parts: the area of triangle BQC and the area of sector BPC.
10. The area of triangle BQC can be calculated using the formula: Area = 1/2 * base * height. The base of triangle BQC is BC = 6√2 cm and the height is BC/2 = 3√2 cm.
11. The area of sector BPC can be calculated using the formula: Area = (angle/360) * π * r^2. The radius of the circle is AB = 6 cm. To find the angle, we can subtract angle A from angle BPC. Since angle A = 180 - 2x degrees and angle BPC = 90 degrees, angle BPC - angle A = 90 degrees - (180 - 2x) degrees = 2x - 90 degrees.
12. Substituting the values into the formula, the area of sector BPC = ((2x - 90)/360) * π * (6^2) square cm.
13. Adding the area of triangle BQC and the area of sector BPC, we get the total
- Triangle ABC is right-angled and isosceles, with hypotenuse BC.
- BQC is a semi-circle with diameter BC.
- BPC is an arc of a circle centered at A, lying between BC and BQC.
- AB has length 6 cm.
To find: The area enclosed by BPC and BQC.
Solution:
1. Draw the diagram of triangle ABC with points B, C, and Q on the circumference of the semi-circle BQC, and point P on the arc BPC.
2. Since triangle ABC is isosceles, we can conclude that angle ABC = angle BCA. Let's call this angle x.
3. Since triangle ABC is right-angled, angle BAC = 90 degrees. Therefore, angle BCA = 90 - x degrees.
4. Since BQC is a semi-circle, angle BQC = 180 degrees.
5. In triangle BQC, angle B + angle Q + angle C = 180 degrees. Since BQC is a semi-circle, angle BQC is a right angle. Therefore, angle B + angle Q + 90 = 180 degrees. Simplifying, we get angle B + angle Q = 90 degrees. Since triangle ABC is isosceles, angle B = angle C. Therefore, angle C + angle Q = 90 degrees.
6. In triangle ABC, angle A + angle B + angle C = 180 degrees. Substituting the values we have, angle A + x + x = 180 degrees. Simplifying, we get angle A + 2x = 180 degrees. Therefore, angle A = 180 - 2x degrees.
7. In triangle BPC, angle B + angle P + angle C = 180 degrees. Since angle B = angle C, angle B + angle P + angle B = 180 degrees. Simplifying, we get 2(angle B + angle P) = 180 degrees. Therefore, angle B + angle P = 90 degrees.
8. In triangle ABC, AB = AC = 6 cm. Therefore, BC = 6√2 cm (using Pythagoras theorem).
9. The area enclosed by BPC and BQC can be divided into two parts: the area of triangle BQC and the area of sector BPC.
10. The area of triangle BQC can be calculated using the formula: Area = 1/2 * base * height. The base of triangle BQC is BC = 6√2 cm and the height is BC/2 = 3√2 cm.
11. The area of sector BPC can be calculated using the formula: Area = (angle/360) * π * r^2. The radius of the circle is AB = 6 cm. To find the angle, we can subtract angle A from angle BPC. Since angle A = 180 - 2x degrees and angle BPC = 90 degrees, angle BPC - angle A = 90 degrees - (180 - 2x) degrees = 2x - 90 degrees.
12. Substituting the values into the formula, the area of sector BPC = ((2x - 90)/360) * π * (6^2) square cm.
13. Adding the area of triangle BQC and the area of sector BPC, we get the total
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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?.
Solutions 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? in English & in Hindi are available as part of our courses for CAT.
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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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice CAT tests.
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