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Question-28240




Question Number 28240 by $@ty@m last updated on 22/Jan/18
Answered by ajfour last updated on 22/Jan/18
DE×EA=CE×EF  6×2=6(√2)×EF  ⇒  EF=(√2)  so  CF=7(√2)  let foot of ⊥ from F on BC be G.  CG=7 , GB=1, FG=7  BF^(  2)  =GB^( 2) +FG^( 2)             =1+49 =50  BF =5(√2)  ≈ 7.07
$${DE}×{EA}={CE}×{EF} \\ $$$$\mathrm{6}×\mathrm{2}=\mathrm{6}\sqrt{\mathrm{2}}×{EF} \\ $$$$\Rightarrow\:\:{EF}=\sqrt{\mathrm{2}}\:\:{so}\:\:{CF}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$${let}\:{foot}\:{of}\:\bot\:{from}\:{F}\:{on}\:{BC}\:{be}\:{G}. \\ $$$${CG}=\mathrm{7}\:,\:{GB}=\mathrm{1},\:{FG}=\mathrm{7} \\ $$$${BF}^{\:\:\mathrm{2}} \:={GB}^{\:\mathrm{2}} +{FG}^{\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{49}\:=\mathrm{50} \\ $$$${BF}\:=\mathrm{5}\sqrt{\mathrm{2}}\:\:\approx\:\mathrm{7}.\mathrm{07} \\ $$
Commented by $@ty@m last updated on 22/Jan/18
Thank you very much.
$${Thank}\:{you}\:{very}\:{much}. \\ $$

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