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TopNewestEveryone 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. – Shubham Srivastava · 4 years, 4 months ago

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– Shubham Srivastava · 4 years, 4 months ago

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.\[ Area Of Sector SOQ = \frac {90}{360} \pi r^2\] \[Area Of Semicircle STO = \frac {1}{2} \pi (\frac {r}{2})^2 = Area Of Semicircle QTO\] \[Sector SOQ = a_1 + Semicircle STO + Semicircle QTO - a_2\] \[\frac {1}{4} \pi r^2 = a_1 - a_2 + 2 \times (\frac {1}{2} \pi \frac {r^2}{4})\] \[a_1 - a_2 = \frac {1}{4} \pi r^2 - \frac {1}{4} \pi r^2 = 0\]Log in to reply

0 – Dharmik Panchal · 4 years, 4 months ago

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

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– Anshul Chauhan · 4 years, 4 months ago

4a1 + 4π(1/4)^2 - 4a2 = π(1)^2Log in to reply

– Adam Silvernail · 4 years, 4 months ago

\(4a_1+4π(1/2)^2 - 4a_2 = π(1)^2\), implies \(a_1 - a_2 = 0\)Log in to reply

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– Do-Hyung Kim · 4 years, 4 months ago

hey isn't 0 is the right answer? by using similarity??Log in to reply