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In figure arcs are drawn by taking vertices `A`, `B` and `C` of an equilateral triangle of side `10` `cm` , To intersect the sides `BC`, `CA` and `AB` at their respective mid- points `D`, `E` and `F`. Find the area of the shaded region. (use `pi = 3.14`)

Text Solution

Verified by Experts

Since, ABC is an equilateral triangle.
`:. angleA = angleB = angleC = 60^(@)`
and AB = BC = AC = 10cm
So, E, F and D are mid - points of the sides .
`:.` AE = EC = BD = BF = FA = 5cm
Now, area of sector CDE = `(thetapir^(2))/(360)=(60 xx 3.14)/(360)(5)^(2)`
= `(3.14xx25)/(6)=(78.5)/(6)=13.0833cm^(2)`
`:.` Area of shaded region = 3 (Area of sector CDE)
= `3xx13.0833`
= `39. 25cm^(2)`
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