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Question Number 12894 by tawa last updated on 06/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17

$$S=\pi \frac{D^2}{4}-3\times \pi \frac{D_1^2}{4}=\frac{\pi }{4}({84}^2-3\times {42}^2)=441\pi .◼$$

Commented by tawa last updated on 06/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by RasheedSoomro last updated on 06/May/17

$$\mathrm{Let}A\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{outer}\mathrm{circle},$$$$a\mathrm{is}\mathrm{an}\mathrm{area}\mathrm{of}\mathrm{smaller}\mathrm{circle}\mathrm{and}c$$$$\mathrm{is}\mathrm{an}\mathrm{common}\mathrm{area}\mathrm{of}\mathrm{any}\mathrm{two}\mathrm{circles}.$$$$\mathrm{If}S\mathrm{is}\mathrm{shaded}\mathrm{area}\mathrm{then}$$$$$$$$S=A-4a+4c$$$$A=\pi {\left(42\right)}^2=1764\pi$$$$a=\pi {\left(21\right)}^2=441\pi$$$$\mathrm{If}sis\mathrm{the}\mathrm{segment}\left(\mathrm{half}\mathrm{of}c\right)\mathrm{associated}$$$$\mathrm{with}\mathrm{smaller}\mathrm{circle}\mathrm{then}$$$$s=\mathrm{Quarter}\mathrm{of}\mathrm{small}\mathrm{circle}-\underset{\mathrm{whose}\mathrm{each}\mathrm{leg}\mathrm{is}21}{\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle}}$$$$=\frac{441\pi }{4}-\frac{21\times 21}{2}=\frac{441\pi -2\times 441}{4}=\frac{441(\pi -2)}{4}$$$$$$$$c=2s=\frac{441(\pi -2)}{2}$$$$S=A-4a+4c=1764\pi -4\left(441\pi \right)+4\left(\frac{441(\pi -2)}{2}\right)$$$$=1764\pi -1764\pi +882(\pi -2)=882(\pi -2)$$

Commented by tawa last updated on 06/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by RasheedSoomro last updated on 07/May/17

$$\mathrm{Pl}\mathrm{confiirm}\mathrm{the}\mathrm{answer}.$$

Commented by sandy_suhendra last updated on 09/May/17

Commented by sandy_suhendra last updated on 09/May/17

$${\mathrm{A}}_1=\frac{1}{4}\pi .{42}^2-\frac{1}{2}\pi .{21}^2-{21}^2$$$$=220.5\pi -441=220.5(\pi -2)$$$$\mathrm{S}=4\times {\mathrm{A}}_1=4\times 220.5(\pi -2)=882(\pi -2)$$

Commented by RasheedSindhi last updated on 11/May/17

$$\mathrm{Thanks}\mathrm{sandy}(\mathrm{my}\mathrm{old}\mathrm{friend}!)$$$$\mathrm{Your}\mathrm{method}\mathrm{is}\mathrm{simpler}!$$

Commented by sandy_suhendra last updated on 18/May/17

$$\mathrm{hallo}\mathrm{Rasheed},\mathrm{why}\mathrm{have}\mathrm{you}\mathrm{changed}\mathrm{your}\mathrm{user}\mathrm{name}?$$$$\mathrm{And}\mathrm{also}\mathrm{long}\mathrm{time}\mathrm{no}\mathrm{see}\mathrm{you}$$

Commented by RasheedSindhi last updated on 19/May/17

$$\mathrm{ThanX}!\mathrm{Actually}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}$$$$\mathbf{weak}\mathbf{point}\mathbf{of}\mathbf{this}\mathbf{fine}\mathbf{forum}$$$$\mathbf{that}\mathbf{one}\mathbf{cannot}\mathbf{use}\mathbf{unique}$$$$\mathbf{ID}\mathbf{on}\mathbf{different}\mathbf{devices}!$$$$\mathbf{Even}\mathbf{on}\mathbf{same}\mathbf{device},\mathbf{after}$$$$\mathbf{reinstallation}\mathbf{changing}\mathbf{ID}$$$$\mathbf{is}\mathbf{must}!$$$$\mathrm{So}\mathrm{I}\mathrm{use}\mathrm{two}\mathrm{devices}\mathrm{use}\mathrm{two}$$$${\mathrm{ID}}^{\prime }\mathrm{s}.$$

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