Ensino Fundamental ⇒ Semelhança de triângulos. Tópico resolvido

[geobson]

Calcule " x" na figura.
A)5
B)2
C)3
D)4
E)1

Resposta

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D

[petras]

geobson,

[tex3]\mathsf{\triangle FCB \sim \triangle ECA\rightarrow \frac{a}{h}=\frac{3}{5}\rightarrow \boxed{h = \frac{5a}{3}}\\
\frac{b}{3}=\frac{b+a}{5}\therefore\boxed{b = \frac{3a}{2}}\\
\boxed{EC = a + \frac{3a}{2}=\frac{5a}{2}}\\
Pit.~em ~ \triangle AEC: h^2+5^2 = (\frac{5a}{2})^2\rightarrow (\frac{5a}{3})^2 +25 =\frac{25a^2}{4}\rightarrow \\
\boxed{a=\frac{6\sqrt{5}}{5}}\rightarrow \boxed{h = \frac{5}{3} . \frac{6\sqrt{5}}{5}=2\sqrt{5}}\\
\triangle EDA:h^2 = x.5\rightarrow (2\sqrt{5})^2 = x.5 \therefore \boxed{\color{red}x = 4}} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{\triangle FCB \sim \triangle ECA\rightarrow \frac{a}{h}=\frac{3}{5}\rightarrow \boxed{h = \frac{5a}{3}}\\
\frac{b}{3}=\frac{b+a}{5}\therefore\boxed{b = \frac{3a}{2}}\\
\boxed{EC = a + \frac{3a}{2}=\frac{5a}{2}}\\
Pit.~em ~ \triangle AEC: h^2+5^2 = (\frac{5a}{2})^2\rightarrow (\frac{5a}{3})^2 +25 =\frac{25a^2}{4}\rightarrow \\
\boxed{a=\frac{6\sqrt{5}}{5}}\rightarrow \boxed{h = \frac{5}{3} . \frac{6\sqrt{5}}{5}=2\sqrt{5}}\\
\triangle EDA:h^2 = x.5\rightarrow (2\sqrt{5})^2 = x.5 \therefore \boxed{\color{red}x = 4}} [/tex3]

[geobson]

petras, obrigado, meu amigo.