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Question Number 64778 by aliesam last updated on 21/Jul/19

Commented by som(math1967) last updated on 21/Jul/19

join O,Q  O,N O,M  M,Q N,Q  now ON=NQ=OQ=OM=MQ  ∴∠NOM=60+60=120°  ∠MLN=(1/2)∠NOM=60°  ∠NKM=∠NOM=120°  [subtend on same segment]  now ext∠NKM=∠NLK+∠LNK  ∴∠LNK=120−60=60  so △KLN equilateral

$${join}\:{O},{Q}\:\:{O},{N}\:{O},{M}\:\:{M},{Q}\:{N},{Q} \\ $$$${now}\:{ON}={NQ}={OQ}={OM}={MQ} \\ $$$$\therefore\angle{NOM}=\mathrm{60}+\mathrm{60}=\mathrm{120}° \\ $$$$\angle{MLN}=\frac{\mathrm{1}}{\mathrm{2}}\angle{NOM}=\mathrm{60}° \\ $$$$\angle{NKM}=\angle{NOM}=\mathrm{120}°\:\:\left[{subtend}\:{on}\:{same}\:{segment}\right] \\ $$$${now}\:{ext}\angle{NKM}=\angle{NLK}+\angle{LNK} \\ $$$$\therefore\angle{LNK}=\mathrm{120}−\mathrm{60}=\mathrm{60} \\ $$$${so}\:\bigtriangleup{KLN}\:{equilateral} \\ $$

Commented by aliesam last updated on 21/Jul/19

god bless you

$${god}\:{bless}\:{you} \\ $$

Answered by som(math1967) last updated on 21/Jul/19

△KLN equilateral  area=((√3)/4)×6^2 =9(√3) sq unit

$$\bigtriangleup{KLN}\:{equilateral} \\ $$$${area}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{6}^{\mathrm{2}} =\mathrm{9}\sqrt{\mathrm{3}}\:{sq}\:{unit} \\ $$

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