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Two circles with radius 2R and √2R intersect each other at points A and B. The centers of both the circles are on the same side of AB. 0 is the center of the bigger circle and ∠AOB is 60°. Find the area of the common region between two circles.
- a)(√3- π - 1) R2
- b)(√3 - π) R2
- c)(13π/6 + 1 - √3 ) R2
- d)(13π/6 +√3) R2
- e)None of the above
Correct answer is option 'C'. Can you explain this answer?
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Two circles with radius 2R and √2R intersect each other at points A a...
Let us draw the diagram according to the given info,
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We can see that AD = A0*cos60° = 2R* 1/2 = R
In triangle, ACD
=>sinACD = AC/AC
=>sinACD = R/√2 * R = 1/√2
=>∠ACD = 45°
By symmetry we can say that ∠BCD = 45°
Therefore we can say that ∠ACB = 90°
Hence, the area colored by green color = 270°/360° * π * (√2R)2 = 3/2 * π * R2... (1)
Area of triangle ACB = 1/2 * R* 2R = R2... (2)
Area shown in blue color = 60°/360° *π* (2R)2-√3/4 * (2R)2 = 2/3 *π* R2-√3 * R2... (3)
By adding (1) + (2) + (3)
Therefore, the area of the common region between two circles = 3/2 * π * R2 + R2 + 2/3 * π * R2 - √3 * R2
=> (13π/6 + 1 +√3)R2Hence, option C is the correct answer.
Most Upvoted Answer
Two circles with radius 2R and √2R intersect each other at points A a...
To find the area of the common region between the two circles, we can break it down into three parts: the sector of the larger circle, the sector of the smaller circle, and the triangle formed by the centers of the circles and point B.
1. Sector of the larger circle:
The angle at the center of the larger circle is 60° (given), so the angle at the circumference is also 60°. The area of the sector is given by (60/360) * π(2R)^2 = (1/6) * 4πR^2 = (2/3)πR^2.
2. Sector of the smaller circle:
The angle at the center of the smaller circle is also 60° (since the circles intersect at point B). The radius of the smaller circle is √2R, so the area of the sector is (1/6) * π(√2R)^2 = (1/6) * 2πR^2 = (1/3)πR^2.
3. Triangle formed by the centers and point B:
We can see that this triangle is an equilateral triangle, because the angles at the centers are equal (60°) and the sides are equal (2R and √2R). The area of an equilateral triangle with side length s is given by (√3/4) * s^2. In this case, the side length is 2R, so the area of the triangle is (√3/4) * (2R)^2 = (√3/4) * 4R^2 = √3R^2.
Therefore, the total area of the common region is the sum of the areas of the three parts:
(2/3)πR^2 + (1/3)πR^2 + √3R^2
= (3/3)πR^2 + √3R^2
= (π + √3)R^2
So, the correct answer is option C: (13π/6 - 1 - √3)R^2.
1. Sector of the larger circle:
The angle at the center of the larger circle is 60° (given), so the angle at the circumference is also 60°. The area of the sector is given by (60/360) * π(2R)^2 = (1/6) * 4πR^2 = (2/3)πR^2.
2. Sector of the smaller circle:
The angle at the center of the smaller circle is also 60° (since the circles intersect at point B). The radius of the smaller circle is √2R, so the area of the sector is (1/6) * π(√2R)^2 = (1/6) * 2πR^2 = (1/3)πR^2.
3. Triangle formed by the centers and point B:
We can see that this triangle is an equilateral triangle, because the angles at the centers are equal (60°) and the sides are equal (2R and √2R). The area of an equilateral triangle with side length s is given by (√3/4) * s^2. In this case, the side length is 2R, so the area of the triangle is (√3/4) * (2R)^2 = (√3/4) * 4R^2 = √3R^2.
Therefore, the total area of the common region is the sum of the areas of the three parts:
(2/3)πR^2 + (1/3)πR^2 + √3R^2
= (3/3)πR^2 + √3R^2
= (π + √3)R^2
So, the correct answer is option C: (13π/6 - 1 - √3)R^2.
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Two circles with radius 2R and √2R intersect each other at points A and B. The centers of both the circles are on the same side of AB. 0 is the center of the bigger circle and ∠AOB is 60°. Find the area of the common region between two circles.a)(√3- π - 1) R2b)(√3 - π) R2c)(13π/6 + 1 - √3 ) R2d)(13π/6 +√3) R2e)None of the aboveCorrect answer is option 'C'. Can you explain this answer?
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