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# Geometry!!

Note by Shubham Bagrecha
4 years, 4 months ago

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Everyone has assumed that a1 and area of other "similar looking" sectors are equal, so that 4a1 is their total area. Similar assumption have been made for a2 also. But one needs to prove these assumptions. · 4 years, 4 months ago

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Here is the another method. Though the answer is same but the reasoning involved in this method is such that one doesn't require the above stated two assumptions, so there is no need to prove the assumptions.$Area Of Sector SOQ = \frac {90}{360} \pi r^2$ $Area Of Semicircle STO = \frac {1}{2} \pi (\frac {r}{2})^2 = Area Of Semicircle QTO$ $Sector SOQ = a_1 + Semicircle STO + Semicircle QTO - a_2$ $\frac {1}{4} \pi r^2 = a_1 - a_2 + 2 \times (\frac {1}{2} \pi \frac {r^2}{4})$ $a_1 - a_2 = \frac {1}{4} \pi r^2 - \frac {1}{4} \pi r^2 = 0$ · 4 years, 4 months ago

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0 · 4 years, 4 months ago

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the answer is 3π/16 · 4 years, 4 months ago

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4a1 + 4π(1/4)^2 - 4a2 = π(1)^2 · 4 years, 4 months ago

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$$4a_1+4π(1/2)^2 - 4a_2 = π(1)^2$$, implies $$a_1 - a_2 = 0$$ · 4 years, 4 months ago

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Comment deleted Apr 01, 2013

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hey isn't 0 is the right answer? by using similarity?? · 4 years, 4 months ago

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