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Question-49774




Question Number 49774 by ggny last updated on 10/Dec/18
Answered by afachri last updated on 10/Dec/18
ABC_(perimeter)  =  AE + CF + AC + (EB + BF)           27   =   8 + 8 + 9 + (x + x)              x  =  1 cm  ABC is an equilateral triangle whose  A^� B = BC = CA  =  9 cm  so ∠ABC  =  ∠BCA  =  ∠ACB  =   (π/3)  so,  CE^2   =  AE^2  + AC^2   −  2(AE)(AC) cos∠BACq               =      8^2   +    9^2      −   2(8)(9)((1/2))               =       73     CE    =    (√(73^ ))  cm
$$\mathrm{ABC}_{\mathrm{perimeter}} \:=\:\:\mathrm{AE}\:+\:\mathrm{CF}\:+\:\mathrm{AC}\:+\:\left(\mathrm{EB}\:+\:\mathrm{BF}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{27}\:\:\:=\:\:\:\mathrm{8}\:+\:\mathrm{8}\:+\:\mathrm{9}\:+\:\left({x}\:+\:{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\:=\:\:\mathrm{1}\:{cm} \\ $$$$\mathrm{ABC}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{whose} \\ $$$$\bar {\mathrm{A}B}\:=\:\mathrm{BC}\:=\:\mathrm{CA}\:\:=\:\:\mathrm{9}\:\mathrm{cm} \\ $$$$\mathrm{so}\:\angle\mathrm{ABC}\:\:=\:\:\angle\mathrm{BCA}\:\:=\:\:\angle\mathrm{ACB}\:\:=\:\:\:\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{so}, \\ $$$$\mathrm{CE}^{\mathrm{2}} \:\:=\:\:\mathrm{AE}^{\mathrm{2}} \:+\:\mathrm{AC}^{\mathrm{2}} \:\:−\:\:\mathrm{2}\left(\mathrm{AE}\right)\left(\mathrm{AC}\right)\:\mathrm{cos}\angle\mathrm{BACq} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\:\:\mathrm{8}^{\mathrm{2}} \:\:+\:\:\:\:\mathrm{9}^{\mathrm{2}} \:\:\:\:\:−\:\:\:\mathrm{2}\left(\mathrm{8}\right)\left(\mathrm{9}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\:\:\:\mathrm{73}\:\:\: \\ $$$$\mathrm{CE}\:\:\:\:=\:\:\:\:\sqrt{\mathrm{73}^{} }\:\:\mathrm{cm} \\ $$

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