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Question
In the following figure, the square ABCD is divided into five equal parts, all having same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22 cm, find:
the perimeter of the part ABEF.
Solution
We have a square ABCD.
We have to find the perimeter of ABEF. Let O be the centre of the circular region.
Use Pythagoras theorem to get,
`2(AE+r)^2=22^2`
`AE+r=15.56`
`AE=(15.56-5.56)cm`
`= 10c,`
Similarly,
`BF=10cm`
Now length of arc EF,
`= "perimeter of circular region"/4`
`= 34.88/4 cm`
So, perimeter of ABFE,
`=AB+BF+EF+AE`
`=(22+10+8.64+10)`
`= 50.64`
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