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Question Number 183663 by AROUNAMoussa last updated on 28/Dec/22

Commented by mathocean1 last updated on 28/Dec/22

$$Perimetre=2\pi +12$$

Answered by a.lgnaoui last updated on 28/Dec/22

$$P=P_1+P_2+P3$$$$P_1=perumetre\left(OPQ\right)$$$$P_2=perimetre\left(QBR\right)$$$$P_3=perimetre\left(RAP\right)$$$$P_4=perimetre\left(OABO\right)$$$$P=\mathrm{Arc}\left(PQ\right)+Arc\left(QR\right)+Arc\left(PR\right)+P_4$$$$P=2\pi r+12+2\pi (r=raynducercleinscrit)$$$$\sin 30=\frac{r}{{OO}_1}=\frac{r}{6-r}=\frac{\sqrt{3}}{2}$$$$2r=6\sqrt{3}-r\sqrt{3}\Rightarrow r=\frac{6\sqrt{3}}{2+\sqrt{3}}$$$$perimetreducercle=2\pi r=\frac{12\pi \sqrt{3}}{2+\sqrt{3}}$$$$P=2\pi r+(2\pi +12)$$$$=12+2\pi (12\sqrt{3}-17)$$$$P=35,7794$$

Commented by a.lgnaoui last updated on 28/Dec/22

$$plaiseereeur\sin 30=\frac{1}{2}\left(not\frac{\sqrt{3}}{2}\right)$$$$\sin 30=\frac{1}{2}r=2$$$$P=2\pi r+(2\pi +12)=6\pi +12$$$$Perimetre=30,8495$$$$$$

Commented by a.lgnaoui last updated on 28/Dec/22

Answered by manxsol last updated on 28/Dec/22

$$nemotecnicLOR$$$$L=\mathrm{\Theta }R$$$$L=\frac{\pi }{3}\times 6=2\pi$$$$trianglerectangle30(r,\sqrt{3}r,2r)$$$$R=OT=2r+r=6\Rightarrow r=2$$$$perimeter=6+6+2\pi +2\pi \left(2\right)$$$$perimeter=12+6\pi$$

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