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Question Number 65128 by Tawa1 last updated on 25/Jul/19

Answered by ajfour last updated on 25/Jul/19

Commented by ajfour last updated on 25/Jul/19

$$OP=8$$$$withOasorigin$$$$P(4,\sqrt{64-16})\equiv (4,4\sqrt{3})$$$$OQ=8\sqrt{2}$$$$eq.ofOQisy-x=0$$$$\mathrm{\perp }distanceofPfromlineOQ$$$$p=\frac{4\sqrt{3}-4}{\sqrt{2}}$$$$shadedarea=Area(△OPQ)$$$$=\frac{1}{2}\times 8\sqrt{2}\times \frac{4(\sqrt{3}-1)}{\sqrt{2}}$$$$=16(\sqrt{3}-1).$$

Commented by Tawa1 last updated on 25/Jul/19

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

Commented by Tawa1 last updated on 25/Jul/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Commented by Tawa1 last updated on 25/Jul/19

Commented by Tawa1 last updated on 25/Jul/19

$$\mathrm{This}\mathrm{sir}$$

Commented by mr W last updated on 25/Jul/19

Commented by mr W last updated on 25/Jul/19

$$DC=DB=DE=r$$$$CF=\frac{CB}{2}=2$$$$\frac{CF}{CD}=\frac{CD}{AC}$$$${CD}^2=CF\times AC$$$$\Rightarrow r^2=2\times 9=18$$$$\Rightarrow r=3\sqrt{2}$$

Answered by mr W last updated on 25/Jul/19

Commented by mr W last updated on 25/Jul/19

$$OC=OP=CP=r=8$$$$\Rightarrow \mathrm{\angle }OCP=60°=\frac{\pi }{3}$$$$\Rightarrow \mathrm{\angle }PCQ=30°=\frac{\pi }{6}$$$$A_{shade}={Cap}_{OQ}-{Cap}_{OP}-{Cap}_{PQ}$$$$=\frac{r^2}{2}(\frac{\pi }{2}-\sin \frac{\pi }{2})-\frac{r^2}{2}(\frac{\pi }{3}-\sin \frac{\pi }{3})-\frac{r^2}{2}(\frac{\pi }{6}-\sin \frac{\pi }{6})$$$$=\frac{r^2}{2}(\frac{\pi }{2}-\frac{\pi }{3}-\frac{\pi }{6}-\sin \frac{\pi }{2}+\sin \frac{\pi }{3}+\sin \frac{\pi }{6})$$$$=\frac{64}{2}(-1+\frac{1}{2}+\frac{\sqrt{3}}{2})$$$$=16(\sqrt{3}-1)$$

Commented by mr W last updated on 25/Jul/19

$$Anotherway:$$$$A_{shade}=\frac{1}{2}\times OP\times OQ\times \sin \mathrm{\angle }POQ$$$$=\frac{1}{2}\times 8\times 8\sqrt{2}\times \sin (60°-45°)$$$$=32\sqrt{2}\times (\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}-\frac{1}{2}\times \frac{1}{\sqrt{2}})$$$$=16(\sqrt{3}-1)$$

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