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⠀⠀☞ Por um método geométrico encontramos que a área do quadrilátero dados seus vértices é de 25 u.a., o que nos leva à opção e). ✅
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- ⠀⠀ Para resolver por um método mais rápido e algébrico (matriz) confira o link ao final da explicação.
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⠀⠀Vamos inicialmente visualizar os pontos e a figura formada pelo sistema cartesiano:
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[tex]\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\vector(1,0){5}}\put(0,0){\vector(0,1){5}}\put(0,0){\vector(-1,0){5}}\put(0,0){\vector(0,-1){5}}\put(4.8,0.2){x}\put(0.2,4.8){y}\put(-1.50,0.50){\circle*{0.13}}\put(-1.25,0.75){$\sf P_{A}$}\put(0.0,2.0){\circle*{0.13}}\put(0.25,2.25){$\sf P_{B}$}\bezier{15}(-1.50,0.50)(-0.75,0.50)(0,0.50)\bezier{5}(-1.50,0.50)(-1.50,0.25)(-1.50,0)\bezier(-1.50,0.50)(-0.75,1.25)(0.0,2.0)\put(2.50,2.0){\circle*{0.13}}\put(2.75,2.25){$\sf P_{C}$}\put(-1.0,-1.50){\circle*{0.13}}\put(-0.75,-1.25){$\sf P_{D}$}\bezier{25}(2.50,2.0)(1.25,2.0)(0,2.0)\bezier{10}(-1.0,-1.50)(-0.50,-1.50)(0,-1.50)\bezier{20}(2.50,2.0)(2.50,1.0)(2.50,0)\bezier{15}(-1.0,-1.50)(-1.0,-0.75)(-1.0,0)\bezier(2.50,2.0)(0.75,0.25)(-1.0,-1.50)\bezier(-1.50,0.50)(-1.25,-0.50)(-1.0,-1.50)\bezier(0.0,2.0)(1.25,2.0)(2.50,2.0)\end{picture}[/tex]
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[tex]\footnotesize\text{$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$}[/tex] ☹ )
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⠀⠀Uma técnica eficiente na geometria é ver o que não está ali. Para aplicarmos este método, completaremos um grande retângulo e em seguida poderemos subtrair os 3 triângulos que foram adicionados em excesso:
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[tex]\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\vector(1,0){5}}\put(0,0){\vector(0,1){5}}\put(0,0){\vector(-1,0){5}}\put(0,0){\vector(0,-1){5}}\put(4.8,0.2){x}\put(0.2,4.8){y}\put(-1.50,0.50){\circle*{0.13}}\put(-1.25,0.75){$\sf P_{A}$}\put(0.0,2.0){\circle*{0.13}}\put(0.25,2.25){$\sf P_{B}$}\bezier{15}(-1.50,0.50)(-0.75,0.50)(0,0.50)\bezier{5}(-1.50,0.50)(-1.50,0.25)(-1.50,0)\bezier(-1.50,0.50)(-0.75,1.25)(0.0,2.0)\put(2.50,2.0){\circle*{0.13}}\put(2.75,2.25){$\sf P_{C}$}\put(-1.0,-1.50){\circle*{0.13}}\put(-0.75,-1.25){$\sf P_{D}$}\bezier{25}(2.50,2.0)(1.25,2.0)(0,2.0)\bezier{10}(-1.0,-1.50)(-0.50,-1.50)(0,-1.50)\bezier{20}(2.50,2.0)(2.50,1.0)(2.50,0)\bezier{15}(-1.0,-1.50)(-1.0,-0.75)(-1.0,0)\bezier(2.50,2.0)(0.75,0.25)(-1.0,-1.50)\bezier(0.0,2.0)(1.25,2.0)(2.50,2.0)\bezier{30}(2.50,2.0)(0.50,2.0)(-1.50,2.0)\bezier{30}(-1.50,2.0)(-1.50,0.25)(-1.50,-1.50)\bezier{30}(-1.50,-1.50)(0.50,-1.50)(2.50,-1.50)\bezier{30}(2.50,-1.50)(2.50,0.25)(2.50,2.0)\bezier(-1.50,0.50)(-1.25,-0.50)(-1.0,-1.50)\put(-1.2,-1){\circle{0.5}}\put(-1.3,-1.1){1}\put(-1.2,1.5){\circle{0.5}}\put(-1.3,1.4){2}\put(2,-1.2){\circle{0.5}}\put(1.9,-1.3){3}\end{picture}[/tex]
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[tex]\footnotesize\text{$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$}[/tex] ☹ )
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⠀⠀Temos que a área total do nosso retângulo será o produto da base |xc - xa| pela altura |yc - yd|, que é igual a:
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[tex]\LARGE\blue{\text{$\sf |5 - (-3)| \cdot |4 - (-3)|$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = 8 \cdot 7$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = 56~u.a.$}}[/tex]
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⠀⠀A área total do nosso triângulo 1 será metade do produto da base |xd - xa| pela altura |yd - ya|:
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[tex]\LARGE\blue{\text{$\sf \dfrac{|-2 - (-3)| \cdot |-3 - 1|}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = \dfrac{1 \cdot 4}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = \dfrac{4}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = 2~u.a.$}}[/tex]
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⠀⠀A área do nosso triângulo 2 será metade do produto da base |xa| pela altura |yb - ya|:
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[tex]\LARGE\blue{\text{$\sf \dfrac{|-3| \cdot |4 - 1|}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = \dfrac{3 \cdot 3}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = \dfrac{9}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = 4,5~u.a.$}}[/tex]
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⠀⠀A área do nosso triângulo 3 será metade do produto da base |xc - xd| pela altura |yc - yd|:
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[tex]\LARGE\blue{\text{$\sf \dfrac{|5 - (-2)| \cdot |-3 - 4|}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = \dfrac{7 \cdot 7}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = \dfrac{49}{2}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf = 24,5~u.a.$}}[/tex]
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⠀⠀Temos, por fim, que a área delimitada pelos pontos A, B, C e D é de:
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[tex]\LARGE\blue{\text{$\sf A_T = 56 - 2 - 4,5 - 24,5$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf A_T = 25~u.a.$}}[/tex]
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[tex]\huge\green{\boxed{\rm~~~\red{e)}~\blue{ 25 }~~~}}[/tex] ✅
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[tex]\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]
⠀⠀☀️ Outro método (pseudo-determinante de matriz) para encontrar a área de um polígono dados seus vértices:
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✈ https://brainly.com.br/tarefa/4717722
[tex]\bf\large\red{\underline{\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}[/tex]✍
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_______________________________☁
⠀⠀⠀⠀☕ Bons estudos.
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(Dúvidas nos comentários) ☄
__________________________[tex]\LaTeX[/tex]✍
❄☃ [tex]\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})[/tex] ☘☀
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Absque sudore et labore nullum opus perfectum est.
