Solution to Captivating Circles 6

Captivating Circle #6

We wanted to find the area of the largest piece of the small circle. The radius of the small circle is √2 / 2. To find the distance between 1 chord dividing the small circle and it’s centre, we use the equation below. This gives us 0.2649 units. The pythagorean theorem gives us the length of half the part of the chord in the small circle, in the second equation. This gives us 0.4298. asin(0.4298 / (√2 / 2) ) gives us half the section angle, making the section of the small circle encompassing the missing segment equal to 74.8653°. The triangle making the difference between the segment and sector has area 0.4298 * 0.2649 = 0.1139. The sector has area 74.8653π / 720 = 0.3267. π/2 – 0.3267 * 2 = 0.917. And there we have it: the area of the biggest piece in the smallest circle has area 0.9174 units squared

1-\int_{-\sqrt{h(2-h)}}^{\sqrt{h(2-h)}}\sqrt{1-x^2}\,dx – 2(1-h)\sqrt{h(2-h)}=\frac{\pi}{3}\\[16pt] x=\sqrt{(\sqrt{2}/2)^2-0.2649^2}=0.4298

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