Question

The area of an equilateral triangle is `1732. 05 c m^2` . About each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. `(U s e pi=3. 14)` .

Answer 1

Let each side of equilateral triangle be x cm
Area of equilateral triangle `= (sqrt3)/(4) x^(2)`
Given area of equilateral triangel `= 17320.5 cm^(2)`
`:. 17320.5 = (sqrt3)/(4) x^(2)`
or `x^(2) = (17320.5 xx 4)/(sqrt3)`...(1)
Area of sector AMN `= (pi r^(2) theta)/(360^(@)) = 3.14 xx ((x)/(2))^(2) xx (60^(@))/(360^(@))`

Area of 3 sectors `= 3.14 xx (x^(2))/(4) xx (60^(@))/(360^(@)) xx 3`
`= 3.14 xx (1)/(4) xx (17320.5 xx 4)/(sqrt3) xx (60^(@))/(360^(@)) xx 3` [from (1)]
`= 3.14 xx (17320.5)/(2sqrt3)`
`= 3.14 xx (17320.5)/(2 sqrt3) xx (sqrt3)/(sqrt3)` (Multiplying with `(sqrt3)/(sqrt3)`)
`= (1.57 xx 17320.5 xx sqrt3)/(3) = 15700 cm^(2)`
Hence, the area not included in the circle = Area of `Delta -` Area of three sectors
`= 17320.5 - 15700 = 1620.5 cm^(2)`