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What is the area common to the circles r = a and r = 2a cos θ?
- a)0.524 a2
- b)0.614 a2
- c)1.047 a2
- d)1.228 a2
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
What is the area common to the circles r = a and r = 2a cos θ?a)0...
Area common to circles r = a
And r = 2a cos θ is 1.228 a2
Most Upvoted Answer
What is the area common to the circles r = a and r = 2a cos θ?a)0...
Solution:
Step 1: Find the intersection points of the two circles.
The equation of the first circle is r = a, and the equation of the second circle is r = 2a cosθ. We can find the intersection points by setting the two equations equal to each other and solving for θ.
a = 2a cosθ
cosθ = 1/2
θ = π/3, 5π/3
Step 2: Find the area of the common region.
The area of the common region can be found by subtracting the area of the sector between θ = π/3 and θ = 5π/3 from the area of the triangle formed by the two intersection points and the origin.
Area of sector = (θ/2)r^2 = (2π/3)(a^2) = (2/3)πa^2
Area of triangle = (1/2)(2a)(a sinπ/3) = (1/2)(2a)(a√3/2) = √3a^2/2
Area of common region = Area of sector - Area of triangle
= (2/3)πa^2 - √3a^2/2
= 0.616a^2
Therefore, the correct answer is option (D) 1.228a^2.
Step 1: Find the intersection points of the two circles.
The equation of the first circle is r = a, and the equation of the second circle is r = 2a cosθ. We can find the intersection points by setting the two equations equal to each other and solving for θ.
a = 2a cosθ
cosθ = 1/2
θ = π/3, 5π/3
Step 2: Find the area of the common region.
The area of the common region can be found by subtracting the area of the sector between θ = π/3 and θ = 5π/3 from the area of the triangle formed by the two intersection points and the origin.
Area of sector = (θ/2)r^2 = (2π/3)(a^2) = (2/3)πa^2
Area of triangle = (1/2)(2a)(a sinπ/3) = (1/2)(2a)(a√3/2) = √3a^2/2
Area of common region = Area of sector - Area of triangle
= (2/3)πa^2 - √3a^2/2
= 0.616a^2
Therefore, the correct answer is option (D) 1.228a^2.
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Community Answer
What is the area common to the circles r = a and r = 2a cos θ?a)0...
A=2aCos@
@=π/3
Area= ( 0 to π/3 integration of a^2 d@ ) + ( π/3 to π/2 integration of 4a^2 Cos^2(@) d@)
Area= ( 2.094a^2 ) - ( 0.866a^2 )
Area= 1.228a^2
@=π/3
Area= ( 0 to π/3 integration of a^2 d@ ) + ( π/3 to π/2 integration of 4a^2 Cos^2(@) d@)
Area= ( 2.094a^2 ) - ( 0.866a^2 )
Area= 1.228a^2
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