Question

3. E:m triângulo retângul onde b=12 e c=9, podemos concluir que o produto m.n vale:
b) 5,4
c) 9,6
d) 51,84​

Answer 1

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$\Huge\green{\boxed{\rm~~~\red{3.~d)}~~\blue{ 51,84 }~~~}}$

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍  

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☺lá, Giovana, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo. ✌  

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☔ Inicialmente vamos desenhar nosso triângulo

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,0){9.7}}\put(0,0){\line(2,3){3}}\put(3,4.5){\line(3,-2){6.7}}\bezier(0.45,0.65)(0.8,0.6)(0.9,0)\put(0.4,0.2){ \alpha }\bezier(8.7,0.65)(8.5,0.7)(8.45,0)\put(8.7,0.2){ \beta }\put(2.75,4.1){\line(3,-2){0.4}}\put(3.15,3.85){\line(2,3){0.25}}\put(3.5,4.5){A}\put(-0.6,0){B}\put(9.9,0){C}\put(3.1,4.15){\circle*{0.1}}\put(7,4){\dashbox{0.1}(3.5,1){\Large \sf \alpha + \beta = 90^{\circ} }}\put(0.8,2.5){\LARGE \sf 9 }\put(6.5,2.5){\LARGE \sf 12 }\end{picture}$  

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,0){9.7}}\put(0,0){\line(2,3){3}}\put(3,4.5){\line(3,-2){6.7}}\put(3,0){\line(0,1){4.5}}\bezier(0.45,0.65)(0.8,0.6)(0.9,0)\put(0.4,0.2){ \alpha }\bezier(8.7,0.65)(8.5,0.7)(8.45,0)\put(8.7,0.2){ \beta }\put(3.5,4.5){A}\put(-0.6,0){B}\put(9.9,0){C}\put(2.5,0.5){\line(1,0){1}}\put(2.5,0){\line(0,1){0.5}}\put(3.5,0){\line(0,1){0.5}}\put(3.7,0.2){D}\put(2.77,0.23){\circle*{0.1}}\put(3.27,0.23){\circle*{0.1}}\put(3.2,2){h}\put(1.7,-0.5){m}\put(5.7,-0.5){n}\put(7,4){\dashbox{0.1}(3.5,1){\Large \sf ABD \equiv ADC }}\end{picture}$  

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)(0,0)\thicklines\put(0,0){\line(1,0){7}}\put(7,0){\line(0,1){5.2}}\put(4.98,0.01){\line(0,1){3.7}}\put(0,0){\line(4,3){7}}\qbezier(0.7,0.5)(1,0.5)(1,0)\put(0.5,0.1){ \alpha }\put(4.58,0.4){\line(1,0){0.4}}\put(4.58,0){\line(0,1){0.4}}\put(6.6,0){\line(0,1){0.4}}\put(6.6,0.4){\line(1,0){0.4}}\put(4.8,0.2){\circle*{0.1}}\put(6.8,0.2){\circle*{0.1}}\put(2.7,-0.5){m}\put(5.7,-0.5){h - m}\put(5.2,1.7){h}\put(7.2,2.7){n}\put(5.2,5.1){\large \sf 12 - 9 }\put(2.2,2.3){\LARGE \sf 9 }\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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☔ Pela Função Seno temos que

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$\begin{cases}\blue{\sf~I)~sen(\alpha) = \dfrac{h}{9} = \dfrac{n}{12} \Longrightarrow \dfrac{9}{12} = \dfrac{h}{n}}\\\\\\ \blue{\sf~II)~cos(\alpha) = \dfrac{m}{9} = \dfrac{h}{12} \Longrightarrow \dfrac{9}{12} = \dfrac{m}{h} } \end{cases}$

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☔ Sendo 9/12 = 9/12 então temos que

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➡ $\large\blue{\sf \dfrac{h}{n} = \dfrac{m}{h} }$  

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$\Huge\gray{\boxed{\sf\blue{~~m \cdot n = h^2~~}}}$

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• "- Mas qual será a altura deste triângulo?"

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☔ Temos que a base deste triângulo, que é a sua hipotenusa, vale

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➡ $\large\blue{\sf a^2 = 9^2 + 12^2 }$  

➡ $\large\blue{\sf a^2 = 81 + 144 }$

➡ $\large\blue{\sf \sqrt{a^2} = \pm \sqrt{225} }$

➡ $\large\blue{\sf a = 15 }$

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☔ Sabemos portanto que m + n = 15, ou seja, n = 15 - m. Sabemos também que , pelo triângulo ABD

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➡ $\large\blue{\sf 9^2 = m^2 + h^2 }$

➡ $\large\blue{\sf 81 = m^2 + h^2 }$

➡ $\large\blue{\sf h^2 = 81 - m^2 }$

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e  pelo triângulo ADC

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➡ $\large\blue{\sf 12^2 = (15 - m)^2 + h^2 }$

➡ $\large\blue{\sf 144 = 225 - 30m + m^2 + h^2 }$

➡ $\large\blue{\sf h^2 = -m^2 + 30m - 81 }$

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☔ Sendo h² = h² então temos que

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➡ $\large\blue{\sf 81 - \diagup\!\!\!\!{m}^2 = -\diagup\!\!\!\!{m}^2 + 30m - 81 }$

➡ $\large\blue{\sf 81 = 30m - 81 }$

➡ $\large\blue{\sf 30m = 81 + 81 }$

➡ $\large\blue{\sf 30m = 162 }$

➡ $\large\blue{\sf m = \dfrac{162}{30} }$

➡ $\large\blue{\sf m = 5,4 }$

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☔ Com o valor de m temos portanto que

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➡ $\large\blue{\sf h^2 = 81 - (5,4)^2 }$

➡ $\large\blue{\sf h^2 = 81 - 29,16 }$

➡ $\large\blue{\sf h^2 = 51,84 }$

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$\Huge\green{\boxed{\rm~~~\red{d)}~~\blue{ 51,84 }~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁  

☕ $\bf\Large\blue{Bons\ estudos.}$  

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄  

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍  

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀  

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$