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Question Number 49774 by ggny last updated on 10/Dec/18

Answered by afachri last updated on 10/Dec/18

$${\mathrm{ABC}}_{\mathrm{perimeter}}=\mathrm{AE}+\mathrm{CF}+\mathrm{AC}+(\mathrm{EB}+\mathrm{BF})$$$$27=8+8+9+(x+x)$$$$x=1cm$$$$\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{whose}$$$$\overline{\mathrm{A}B}=\mathrm{BC}=\mathrm{CA}=9\mathrm{cm}$$$$\mathrm{so}\mathrm{\angle }\mathrm{ABC}=\mathrm{\angle }\mathrm{BCA}=\mathrm{\angle }\mathrm{ACB}=\frac{\pi }{3}$$$$\mathrm{so},$$$${\mathrm{CE}}^2={\mathrm{AE}}^2+{\mathrm{AC}}^2-2\left(\mathrm{AE}\right)\left(\mathrm{AC}\right)\cos \mathrm{\angle }\mathrm{BACq}$$$$=8^2+9^2-2\left(8\right)\left(9\right)\left(\frac{1}{2}\right)$$$$=73$$$$\mathrm{CE}=\sqrt{{73}^{}}\mathrm{cm}$$

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