Question
Download Solution PDFTwo circles of radius 7 units each, intersect in such a way that the common chord is of length 7 units. What is the common area in square units of the intersection?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Radius of each circle = 7 units
Length of the common chord = 7 units
Formula Used:
Area of intersection of two circles = 2×(Area of sector - Area of triangle)
Area of sector = (θ/360) × πr2
Area of equilateral triangle = √3/4 × side2
Calculation:
In the ΔAOB,
All sides are equal, so it is an equilateral triangle.
i.e. θ = ∠AOB = 60°
Now,
Area of intersection of two circles = 2×(Area of sector - Area of triangle)
⇒ Area = 2 [(θ/360) × πr2 - √3/4 × side2]
⇒ Area = 2 [(60/360) × π × 72 - √3/4 × 72]
⇒ Area = 2 × 72 [(1/6) × π - √3/4]
⇒ Area = 2 × 49 [π/6 - √3/4]
⇒ Area = 98 [π/6 - √3/4] unit2
∴ The correct answer is option (2).
Last updated on Jun 26, 2025
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