In the given fig. 9.36, the cross type figure is divided into five squares. The side of each square is 2 cm, then find the radius of the circle and also the area of portion between circle and cross type figure (Take )
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Diagonal AC of rectangle ABCD passes through the centre of the circle.
We can find the length of AC using Pythagoras theorem in right triangle ABC.
BC = BF + FG + GC = (2 + 2 + 2) cm = 6 cm
AB = 2 cm
In right ABC, we have:
Now, OA =
i.e., radius of the circle = cm
Now, area of square ABFE = (side)2 = (2)2 = 4 cm2
Similarly, area of square RSFG = area of square DCGH = area of square PQHE = area of square EFGH = 4 cm2
Area of 5 squares = 5 (4 cm2) = 20 cm2
and area of the circle =r2 = 3.14 × × = 31.4 cm2
Area of the portion between the circle and the cross-type figure = area of the circle − area of the 5 squares
= (31.4 – 20) cm2
= 11.4 cm2