Is it really possible to get the area of the shaded part below because i have tried to divide the figure into several parts but my answer comes to 7.85cm².please someone help me solve it.
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Makau Elijah
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Is the 4m the diameter of the larger circle? If it is then you can find the shaded area. don't be distracted by the inner circle (apart from noting its radius). Can you find the area of the sector from the middle of the large circle to the right hand edge of the shaded area? – Paul Oct 30 '18 at 14:35
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@Paul, Yes the diameter of the outer semicircle is 4m – Makau Elijah Oct 30 '18 at 14:43
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Mark on the sector described and see if you can find the sectorial angle then go from there. – Paul Oct 30 '18 at 14:54
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Connect the center of the bigger circle with the ends of the chord. You will see that the required area is a combination of two sectors of the big circle and a triangle with the sides $2,2,2\sqrt{3}$. Both circular segments are $30$ degree so the combined area is $1/6$ of the big circle. Thus, the area in question is $\large{\frac{4\pi}{6}+\sqrt{3}}=\frac{2\pi}{3}+\sqrt{3}\approx 3.83$
Vasili
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The total area of the bigger semi-circle is 6.28 and the shaded portion is 5.56. Don't you think it is way off by eye balling even? – Satish Ramanathan Oct 30 '18 at 15:33
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@SatishRamanathan: thanks for spotting my mistake, forgot to divide by $2$ when calculated the area of a triangle :) – Vasili Oct 30 '18 at 16:05
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Area of the shaded portion$$={2\pi} - \int_{1}^{2} 2.\sqrt{4-y^2}dy=3.826$$
Satish Ramanathan
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