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Question
Four equal circles, each of radius a units, touch each other. Show that the area between them is `(6/7"a"^2)` sq units.
Solution
When four circles touch each other, their centres form the vertices of a square. The sides of the square are 2a units.
Area of the square = (2a)2 = 4a2 sq. units
Area occupied by the four sectors
`= 4xx90/360xxpixx"a"^2`
= πa2 sq. units
Area between the circles = Area of the square - Area of the four sectors
`=4xx90xx360xx"a"^2`
= πa2 sq. units
Area between the circles = Area of the square -- Area of the four sectors
`= (4 - 22/7)"a"^2`
`=6/7"a"^2 "sq". "units"`
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