Ensino Médio ⇒ Áreas triangulares. Tópico resolvido

[geobson]

Na figura , calcular S1 +S2:

Resposta

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[tex3]\frac{15\sqrt{7}}{4}u^{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{15\sqrt{7}}{4}u^{2}[/tex3]

[petras]

geobson,

[tex3]\mathsf{\\
S_{ABC}=\sqrt{(\frac{12+8+10}{2}(15-12)(15-8)(15-10)}=\sqrt{1575}=15\sqrt7\\
S_{ABC}=S_{ACD}+S_{DCB}\implies 15\sqrt7 = \frac{8.h}{2}+\frac{12.h}{2}=15\sqrt7\therefore \underline{h = \frac{3\sqrt7}{2}}\\
S_{ACD} = \frac{1}{2} . (8.\frac{3\sqrt7}{2}) = 6\sqrt7\\
S_{BDC} = \frac{1}{2}.(\frac{12.3\sqrt7}{2}) =9\sqrt7\\
\frac{S_{ACD}}{AD}=\frac{S_{DCB}}{DB} \implies \frac{6\sqrt7}{AD}=\frac{9\sqrt7}{DB} \therefore DB = \frac{3AD}{2}\\
DB+AD = 10 \implies \frac{3AD}{2}+AD = 10 \therefore \underline{AD = 4:BD = 6}\\
\triangle ADF: AF^2 = 4^2 - (\frac{3\sqrt7}{2})^2 \implies AF =\frac{1}{2}\\
\triangle DGB: BG^2 = 6^2 - (\frac{3\sqrt7}{2})^2 \implies BG = \frac{9}{2}\\
S_{ADF} = \frac{1}{2}.{\frac{1}2.\frac{3\sqrt7}{2}}=\frac{3\sqrt7}{8}\\
S_{DGB}:=\frac{1}{2}.\frac{9}{2}.\frac{3\sqrt7}{2}=\frac{27\sqrt7}{8}\\
S_{ADF}+S_{DGB}=\frac{3\sqrt7}{8}+\frac{27\sqrt3}{8}=\frac{30\sqrt7}{8}=\boxed{\frac{15\sqrt7}{4}}\color{green} \checkmark



}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{\\
S_{ABC}=\sqrt{(\frac{12+8+10}{2}(15-12)(15-8)(15-10)}=\sqrt{1575}=15\sqrt7\\
S_{ABC}=S_{ACD}+S_{DCB}\implies 15\sqrt7 = \frac{8.h}{2}+\frac{12.h}{2}=15\sqrt7\therefore \underline{h = \frac{3\sqrt7}{2}}\\
S_{ACD} = \frac{1}{2} . (8.\frac{3\sqrt7}{2}) = 6\sqrt7\\
S_{BDC} = \frac{1}{2}.(\frac{12.3\sqrt7}{2}) =9\sqrt7\\
\frac{S_{ACD}}{AD}=\frac{S_{DCB}}{DB} \implies \frac{6\sqrt7}{AD}=\frac{9\sqrt7}{DB} \therefore DB = \frac{3AD}{2}\\
DB+AD = 10 \implies \frac{3AD}{2}+AD = 10 \therefore \underline{AD = 4:BD = 6}\\
\triangle ADF: AF^2 = 4^2 - (\frac{3\sqrt7}{2})^2 \implies AF =\frac{1}{2}\\
\triangle DGB: BG^2 = 6^2 - (\frac{3\sqrt7}{2})^2 \implies BG = \frac{9}{2}\\
S_{ADF} = \frac{1}{2}.{\frac{1}2.\frac{3\sqrt7}{2}}=\frac{3\sqrt7}{8}\\
S_{DGB}:=\frac{1}{2}.\frac{9}{2}.\frac{3\sqrt7}{2}=\frac{27\sqrt7}{8}\\
S_{ADF}+S_{DGB}=\frac{3\sqrt7}{8}+\frac{27\sqrt3}{8}=\frac{30\sqrt7}{8}=\boxed{\frac{15\sqrt7}{4}}\color{green} \checkmark



}[/tex3]

[geobson]

petras, obrigado !