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∆ABC ~ ∆DEF ( ˂B = ˂E, ˂A = ˂D, ˂C = ˂F ) => S(orange) ~ S(blue) => S(∆ABC) = 9.18
We have ∠ (EDF)= ∠ (DAB)+ alpha=∠ (BAC)∠ (EFD)= ∠ (FCA)+alpha= ∠ (ACB)So triangle DEF similar to ABC ( case AA)Ratio of similarity= circum(DEF)/circum(ABC)= .9/2.7= 1/3So Area(ABC)= Area(DEF)/(1/9)= 9.18
Easily Tr.s ABC & DEF are similar and the ratio of the circumradii = 2.7/0.9 = 3So the area of TR. ABC = (1/9)*1.02 = 9.18Sumith PeirisMoratuwaSri Lanka
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∆ABC ~ ∆DEF ( ˂B = ˂E, ˂A = ˂D, ˂C = ˂F )
ReplyDelete=> S(orange) ~ S(blue)
=> S(∆ABC) = 9.18
We have ∠ (EDF)= ∠ (DAB)+ alpha=∠ (BAC)
ReplyDelete∠ (EFD)= ∠ (FCA)+alpha= ∠ (ACB)
So triangle DEF similar to ABC ( case AA)
Ratio of similarity= circum(DEF)/circum(ABC)= .9/2.7= 1/3
So Area(ABC)= Area(DEF)/(1/9)= 9.18
Easily Tr.s ABC & DEF are similar and the ratio of the circumradii = 2.7/0.9 = 3
ReplyDeleteSo the area of TR. ABC = (1/9)*1.02 = 9.18
Sumith Peiris
Moratuwa
Sri Lanka