Ensino Fundamental ⇒ Quadrado Tópico resolvido

[Flavio2020]

ABCD é um quadrado de centro O. Se BG=2(GC), calcule a razão de áreas das regiões ABCD é EFGD.

IMG-20220526-WA0006.jpg (73.45 KiB) Exibido 434 vezes

a)[tex3]\frac{8}{5}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{8}{5}[/tex3]

b)[tex3]\frac{6}{5}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{6}{5}[/tex3]

c)[tex3]\frac{10}{7}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{10}{7}[/tex3]

d)[tex3]\frac{3}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{3}{2}[/tex3]

e)[tex3]\frac{9}{5}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{9}{5}[/tex3]

Resposta

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e

[geobson]

.........U P ...................
Serve de inspiração.....

[petras]

Flavio2020,

[tex3]\mathsf{BG=2a \implies CG = a\\
\angle EDO =\angle OBG=45^o\\
\angle EOD = \angle BOG (o.p.v)\\
BO=DO\\
\therefore \triangle BOG \cong DOE(A.L.A) \implies BG=ED =2a\\
CG=AE = a\\
S_{ABCD}=(3a)^2 = 9a^2\\
S_ {\triangle{BGF}} = \frac{(2a)^2}{2} =\frac{4a^2}{2}(I)\\
S_{\triangle{GCD}}=\frac{a.3a}{2}=\frac{3a^2}{2}(II)\\
S_{\triangle AFE} = \frac{a.a}{2} = \frac{a^2}{2}(III)\\
S_{EFGD} = S_{ABCD}-[(I)+(II)+(III)] = 9a^2 - 4a^2 = 5a^2\\
\therefore \frac{S_{ABCD}}{S_{EFGD}}=\frac{9a^2}{5a^2}=\boxed{\frac{9}{5}}\color{green} \checkmark




}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{BG=2a \implies CG = a\\
\angle EDO =\angle OBG=45^o\\
\angle EOD = \angle BOG (o.p.v)\\
BO=DO\\
\therefore \triangle BOG \cong DOE(A.L.A) \implies BG=ED =2a\\
CG=AE = a\\
S_{ABCD}=(3a)^2 = 9a^2\\
S_ {\triangle{BGF}} = \frac{(2a)^2}{2} =\frac{4a^2}{2}(I)\\
S_{\triangle{GCD}}=\frac{a.3a}{2}=\frac{3a^2}{2}(II)\\
S_{\triangle AFE} = \frac{a.a}{2} = \frac{a^2}{2}(III)\\
S_{EFGD} = S_{ABCD}-[(I)+(II)+(III)] = 9a^2 - 4a^2 = 5a^2\\
\therefore \frac{S_{ABCD}}{S_{EFGD}}=\frac{9a^2}{5a^2}=\boxed{\frac{9}{5}}\color{green} \checkmark




}[/tex3]