Three circles of radius 7 cm are placed in such a way that each circle touches the other two. What will be the are of the portion enclosed by these circles ?
Answer & explanation
Correct answer: option 1
We can conclude that the centres of circles form an equilateral triangle with = r + r = 2r
The area enclosed between circles =
Area of a triangle - 3 (Area of sector)
The angle subtended by sector in an equilateral triangle = 60
Area of a sector of circle = \(\theta \)/360 x \(\Pi \)\( {r }^{2 } \)
⇒ \(\frac{60}{360}\) x \(\frac{22}{7}\) x \( {7 }^{2 } \) = \(\frac{77}{3}\)
3 (Area of sector)= 77 ..(1)
Area of equilateral triangle = (√3/4)\( {a }^{2 } \)
a = Length of the side = 2r = 2(7) = 14 cm
⇒ (√3/4) x \( {14}^{2 } \) = 49\(\sqrt {3 }\) ..(2)
Subtracting equation 2 - equation 1
Therefore, the area enclosed between circles is $ 49\sqrt{3} - 77 cm^2$