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Question Number 67102 by TawaTawa last updated on 22/Aug/19

Answered by mr W last updated on 23/Aug/19

Commented by mr W last updated on 23/Aug/19

$$\tan \alpha =\frac{30}{60}=\frac{1}{2}$$$$\beta =\frac{\pi }{2}-\alpha$$$$\gamma =\beta -\alpha =\frac{\pi }{2}-2\alpha$$$$\frac{PR}{\sin \beta }=\frac{PQ}{\sin \gamma }$$$$\Rightarrow PR=\frac{\sin \beta }{\sin \gamma }\times PQ=\frac{\cos \alpha }{\cos 2\alpha }\times 30$$$$A_{green}=\frac{PQ\times PR\times \sin \alpha }{2}$$$$=\frac{30\times \cos \alpha \times 30\times \sin \alpha }{2\times \cos 2\alpha }$$$$=\frac{900\times \sin 2\alpha }{4\times \cos 2\alpha }$$$$=225\tan 2\alpha$$$$=225\times \frac{2\times \frac{1}{2}}{1-\frac{1}{2}\times \frac{1}{2}}$$$$=225\times \frac{4}{3}$$$$=300$$$$$$$$A=30\times 60-2\times 300=1200{cm}^2$$

Commented by mr W last updated on 23/Aug/19

$$alternative:$$$$letP=origin$$$$eqn.linePR:$$$$y=\frac{x}{2}$$$$eqn.oflineQR:$$$$\frac{y-0}{60-0}=\frac{x-30}{60-30}$$$$\Rightarrow y=2x-60$$$$intersectionR:$$$$\frac{x}{2}=2x-60$$$$\Rightarrow x_R=x=40$$$$\Rightarrow y_R=\frac{40}{2}=20$$$$A_{green}=\frac{PQ\times y_R}{2}=\frac{30\times 20}{2}=300$$$$A=30\times 60-2\times 300=1200{cm}^2$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{but}\mathrm{answer}\mathrm{says}1800.$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{That}\mathrm{means}\mathrm{they}\mathrm{are}\mathrm{wrong}.\mathrm{I}\mathrm{will}\mathrm{use}\mathrm{your}\mathrm{answer}\mathrm{sir}.$$$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Answered by mr W last updated on 23/Aug/19

Commented by mr W last updated on 23/Aug/19

$$trytosolvethisonebyyourself.$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Ohh},\mathrm{good}\mathrm{sir},\mathrm{i}\mathrm{will}\mathrm{try}\mathrm{it}.$$

Commented by TawaTawa last updated on 23/Aug/19

$$(0,0)\mathrm{and}(60,30)\mathrm{equation}:\frac{\mathrm{y}-0}{\mathrm{x}-0}=\frac{60-0}{30-0}=1,\mathrm{y}=2\mathrm{x}$$$$(30,0)\mathrm{and}(0,60)\mathrm{equation}:\frac{\mathrm{y}-0}{\mathrm{x}-30}=\frac{60-0}{0-30}=-2$$$$\mathrm{y}=-2(\mathrm{x}-30),\mathrm{y}=-2\mathrm{x}+60,2\mathrm{x}=-2\mathrm{x}+60,\mathrm{x}=15,\mathrm{y}=30$$$$\mathrm{intersection}(15,30)$$$$\mathrm{A}=\frac{1}{2}\times 30\times 30=450$$$$6\times \mathrm{A}=2700$$$$$$$${\mathrm{A}}_{\mathrm{green}}=3600-2700$$$${\mathrm{A}}_{\mathrm{green}}=900{\mathrm{cm}}^2$$$$$$

Commented by mr W last updated on 23/Aug/19

$${it}^{\prime }sanoctagon.$$$$A_{green}=8\times \frac{{15}^2\sin \frac{2\pi }{8}}{2}=450\sqrt{2}$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Haa},\mathrm{this}\mathrm{one}\mathrm{is}\mathrm{another}\mathrm{one}.$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Please}\mathrm{sir}\mathrm{teach}\mathrm{me}.$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{I}\mathrm{only}\mathrm{used}\mathrm{the}\mathrm{approach}\mathrm{of}\mathrm{geometry}\mathrm{you}\mathrm{used}.$$$$\mathrm{and}\mathrm{i}\mathrm{got}\mathrm{it}\mathrm{wrong}.(⌣..⌣)$$

Commented by Rasheed.Sindhi last updated on 24/Aug/19

$$\mathbf{S}\mathrm{ir}\mathrm{is}\mathrm{it}\mathrm{a}\mathrm{regular}\mathrm{octagon}?\mathrm{If}\mathrm{it}\mathrm{is}\mathrm{then}\mathrm{all}\mathrm{its}$$$$\mathrm{vertices}\mathrm{must}\mathrm{be}\mathrm{concentric}.\mathrm{But}\mathrm{a}\mathrm{picture}$$$$\mathrm{by}\mathrm{GeoGebra}\mathrm{raises}\mathrm{doubt}.\mathrm{S}eethe$$$$followingpicture.\mathrm{I}\mathrm{think}{\mathrm{you}}^{\prime }\mathrm{ve}\mathrm{used}$$$$\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{regular}\mathrm{octagon}\mathrm{for}\mathrm{area}.$$

Commented by Rasheed.Sindhi last updated on 24/Aug/19

Commented by mr W last updated on 24/Aug/19

$$youarerightsir!$$$${it}^{\prime }sslightlydifferentfroma$$$$regularoctagon.$$$$thedistancefromitsvertexesto$$$$itscenteris15or10\sqrt{2}$$$$\Rightarrow A_{green}=8\times \frac{15\times 10\sqrt{2}}{2}\times \sin \frac{\pi }{4}=600$$

Commented by mr W last updated on 24/Aug/19

Commented by TawaTawa last updated on 24/Aug/19

$$\mathrm{Sir},\mathrm{that}\mathrm{means}\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{now}600?$$

Commented by mr W last updated on 24/Aug/19

$$600{cm}^{2}iscorrect.$$

Commented by TawaTawa last updated on 24/Aug/19

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

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