Ensino Médio ⇒ Triângulo

[AngelitaB]

En um triângulo ABC traça as cevianas interiores AN e BK que se interseptam em L, tal que NC=2(BN), AK=3(KC) e área de LNCK menos metade de ABL igual 4 [tex3]cm^{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]cm^{2}[/tex3]

.Calcule a área da região triangular ABC.
a)20
b)18
c)16
d)14
e)12

Resposta

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e

[geobson]

........up............

[petras]

geobson,
A área pedida deveria ser do triângulo ABN

[tex3]\frac{S_1+S2}{3a} = \frac{S_3+S_4}{a}\implies S_1+S_2=3(S_3+S_4)(I)\\
\frac{S_1+S_3}{2b}=\frac{S_2+S_4}{b} \implies S_1+S_3=2(S_2+S_4)(II)\\
S_3-\frac{S_2}{2}=4 \implies 2S_3-S_2=8(III)\\
Unir~CL\\
\frac{S_{\triangle CLK}}{a} = \frac{S_2}{3a} \implies S_{\triangle CLK} = \frac{S_2}{3}(IV)\\
\frac{S_{\triangle CLN}}{2b}=\frac{S_4}{b} \implies S_{\triangle CLN} = 2S_4(V)\\
S_3 = IV+V = \implies S_3 = \frac{S_2}{3}+2S_4 \implies 3S_3=S_2+6S_4(VI)\\
De(I),(II),(III) e (VI): S_1=27, S_2 = 18, S_3 = 13, S_4 =2\\
\therefore \boxed{S_2+S_4 = 20}(\triangle ABN)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{S_1+S2}{3a} = \frac{S_3+S_4}{a}\implies S_1+S_2=3(S_3+S_4)(I)\\
\frac{S_1+S_3}{2b}=\frac{S_2+S_4}{b} \implies S_1+S_3=2(S_2+S_4)(II)\\
S_3-\frac{S_2}{2}=4 \implies 2S_3-S_2=8(III)\\
Unir~CL\\
\frac{S_{\triangle CLK}}{a} = \frac{S_2}{3a} \implies S_{\triangle CLK} = \frac{S_2}{3}(IV)\\
\frac{S_{\triangle CLN}}{2b}=\frac{S_4}{b} \implies S_{\triangle CLN} = 2S_4(V)\\
S_3 = IV+V = \implies S_3 = \frac{S_2}{3}+2S_4 \implies 3S_3=S_2+6S_4(VI)\\
De(I),(II),(III) e (VI): S_1=27, S_2 = 18, S_3 = 13, S_4 =2\\
\therefore \boxed{S_2+S_4 = 20}(\triangle ABN)[/tex3]

(Solução:anipascual-adaptada)

[geobson]

petras, obrigado, meu amigo!