Categoria: Módulo inicial

0

Uma caixa com latas de refrigerante

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 39 Ex. 31

Enunciado

Base da caixa

Imagine que alguém pensou acondicionar latas de $75$ cl de refrigerante numa caixa prismática cuja base é um paralelogramo obliquângulo, como mostra a figura.

  1. Se o raio da base de cada lata medir $4$ cm, qual é a área da base da caixa?
    Sugestão: No esquema, marcaram-se vários raios de circunferências.
0

Dois quadrados sobrepostos

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 39 Ex. 30

Enunciado

Dois quadrados congruentes de $6$ cm de lado estão sobrepostos como mostra a figura.

O vértice de um dos quadrados está no centro do outro quadrado.

Qual é a área da parte sobreposta?

Resolução >> Resolução

Explore a animação seguinte:

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"}; // is3D=is 3D applet using 3D view, AV=Algebra View, SV=Spreadsheet View, CV=CAS View, EV2=Graphics View 2, CP=Construction Protocol, PC=Probability Calculator, DA=Data Analysis, FI=Function Inspector, PV=Python, macro=Macro View var views = {'is3D': 0,'AV': 0,'SV': 0,'CV': 0,'EV2': 0,'CP': 0,'PC': 0,'DA': 0,'FI': 0,'PV': 0,'macro': 0}; var applet = new GGBApplet(parameters, '5.0', views); window.onload = function() {applet.inject('ggbApplet')};

 

Qual é a justificação para o resultado obtido?…

1

A pirâmide de Quéops

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 39 Ex. 30

Enunciado

Conta-se que Thales de Mileto (séc. VI a.C.), considerado por alguns autores como um dos sete sábios da Antiguidade, se ofereceu para determinar a altura da pirâmide de Quéops, sem escalar o monumento.

Segundo a lenda, a prova ter-se-á realizado na presença do faraó Amasis. Thales espetou perpendiculamente ao chão a sua bengala e mediu as sombras da bengala e da pirâmide.…

0

Em volta de um retângulo

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 38 Ex. 29

Enunciado

Observe a figura ao lado, onde [ABCD] é um retângulo.

  1. Se o segmento de reta [AM] é perpendicular a BD, demonstre que os triângulos [MAD] e [ABD] são semelhantes.
     
  2. Se AM e PC são paralelas e AM e BD são perpendiculares, demostre que os triângulos [MAD] e [PBC] são semelhantes e conclua que $\frac{{\overline {AD} }}{{\overline {PC} }} = \frac{{\overline {MD} }}{{\overline {BP} }}$.
Dois triângulos semelhantes 1

Dois triângulos semelhantes

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 38 Ex. 28

Enunciado

Determine a medida da hipotenusa de um triângulo retângulo de perímetro igual a $24$ cm que é semelhante a outro cujos catetos medem $3$ cm e $4$ cm.

Resolução >> Resolução

De acordo com o Teorema de Pitágoras, a hipotenusa do triângulo retângulo cujas medidas dos catetos são conhecidas tem $5$ cm de comprimento.…

Duas circunferências concêntricas 0

Duas circunferências concêntricas

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 38 Ex. 27

Enunciado

Os raios de duas circunferências concêntricas medem $4$ cm e $5$ cm.

Determine o comprimento da corda da circunferência maior que é tangente à circunferência menor.

Resolução >> Resolução

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Consideremos a construção acima, que ilustra a situação descrita.…

Resolva as equações 0

Resolva as equações

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 38 Ex. 26

Enunciado

Resolva, em $\mathbb{R}$, as seguintes equações:

  1. $$\frac{{2\left( {x + 1} \right)}}{3} + 5\left( {x + 2} \right) = 8 – 3x$$
     
  2. $$3\left( {\frac{{x + 1}}{2} + \frac{{x – 1}}{3}} \right) = 5x – 2$$
     
  3. $$5 – \frac{{2\left( {x + 1} \right)}}{4} = \frac{{3x – 1}}{7}$$
     
  4. $$\frac{{x + 4}}{6} – \frac{{2\left( {x + 1} \right)}}{9} = \frac{{x – 2}}{6} + \frac{{11 – 2x}}{{18}}$$
     
  5. $$\left( {3x – \frac{2}{3}} \right)\left( {3x + \frac{2}{3}} \right) – 4 = {\left( {3x – 5} \right)^2} + \frac{5}{9}$$
     
  6. $$5{\left( {x – 2} \right)^2} – 500 = 0$$

 

Resolução >> Resolução

  1. $$\begin{array}{*{20}{l}}   {\mathop {\frac{{2\left( {x + 1} \right)}}{3}}\limits_{(1)}  + \mathop {5\left( {x + 2} \right)}\limits_{(3)}  = \mathop 8\limits_{(3)}  – \mathop {3x}\limits_{(3)} }& \Leftrightarrow &{2\left( {x + 1} \right) + 15\left( {x + 2} \right) = 24 – 9x} \\   {}& \Leftrightarrow &{2x + 2 + 15x + 30 = 24 – 9x} \\   {}& \Leftrightarrow &{26x =  – 8} \\   {}& \Leftrightarrow &{x =  – \frac{4}{{13}}} \end{array}$$
     
  2. $$\begin{array}{*{20}{l}}   {3\left( {\frac{{x + 1}}{2} + \frac{{x – 1}}{3}} \right) = 5x – 2}& \Leftrightarrow &{\frac{{3x + 3}}{2} + x – 1 = 5x – 2} \\   {}& \Leftrightarrow &{3x + 3 + 2x – 2 = 10x – 4} \\   {}& \Leftrightarrow &{ – 5x =  – 5} \\   {}& \Leftrightarrow &{x = 1} \end{array}$$
     
  3. $$\begin{array}{*{20}{l}}   {\mathop 5\limits_{(28)}  – \mathop {\frac{{2\left( {x + 1} \right)}}{4}}\limits_{(7)}  = \mathop {\frac{{3x – 1}}{7}}\limits_{(4)} }& \Leftrightarrow &{140 – 14\left( {x + 1} \right) = 12x – 4} \\   {}& \Leftrightarrow &{140 – 14x – 14 = 12x – 4} \\   {}& \Leftrightarrow &{ – 26x =  – 130} \\   {}& \Leftrightarrow &{x = 5} \end{array}$$
     
  4. $$\begin{array}{*{20}{l}}   {\mathop {\frac{{x + 4}}{6}}\limits_{(3)}  – \mathop {\frac{{2\left( {x + 1} \right)}}{9}}\limits_{(2)}  = \mathop {\frac{{x – 2}}{6}}\limits_{(3)}  + \mathop {\frac{{11 – 2x}}{{18}}}\limits_{(1)} }& \Leftrightarrow &{3\left( {x + 4} \right) – 4\left( {x + 1} \right) = 3\left( {x – 2} \right) + 11 – 2x} \\   {}& \Leftrightarrow &{3x + 12 – 4x – 4 = 3x – 6 + 11 – 2x} \\   {}& \Leftrightarrow &{ – 2x =  – 3} \\   {}& \Leftrightarrow &{x = \frac{3}{2}} \end{array}$$
     
  5. $$\begin{array}{*{20}{l}}   {\left( {3x – \frac{2}{3}} \right)\left( {3x + \frac{2}{3}} \right) – 4 = {{\left( {3x – 5} \right)}^2} + \frac{5}{9}}& \Leftrightarrow &{{{\left( {3x} \right)}^2} – {{\left( {\frac{2}{3}} \right)}^2} – 4 = 9{x^2} – 30x + 25 + \frac{5}{9}} \\   {}& \Leftrightarrow &{9{x^2} – \frac{4}{9} – 4 = 9{x^2} – 30x + 25 + \frac{5}{9}} \\   {}& \Leftrightarrow &{30x = 29 + 1} \\   {}& \Leftrightarrow &{x = 1} \end{array}$$
     
    RECORDE:
    $$\left( {A + B} \right)\left( {A – B} \right) = {A^2} – B{}^2$$
    $${\left( {A + B} \right)^2} = {A^2} + 2AB + {B^2}$$
     
  6. $$\begin{array}{*{20}{l}}   {5{{\left( {x – 2} \right)}^2} – 500 = 0}& \Leftrightarrow &{{{\left( {x – 2} \right)}^2} – 100 = 0} \\   {}& \Leftrightarrow &{{x^2} – 4x + 4 – 100 = 0} \\   {}& \Leftrightarrow &{x = \frac{{4 \pm \sqrt {16 – 4 \times 1 \times \left( { – 96} \right)} }}{2}} \\   {}& \Leftrightarrow &{x = \frac{{4 \pm \sqrt {400} }}{2}} \\   {}& \Leftrightarrow &{x = \frac{{4 \pm 20}}{2}} \\   {}& \Leftrightarrow &{\begin{array}{*{20}{c}}   {x =  – 8}& \vee &{x = 12} \end{array}} \end{array}$$
     
    ALTERNATIVA:
    $$\begin{array}{*{20}{l}}   {5{{\left( {x – 2} \right)}^2} – 500 = 0}& \Leftrightarrow &{{{\left( {x – 2} \right)}^2} = \frac{{500}}{5}} \\   {}& \Leftrightarrow &{{{\left( {x – 2} \right)}^2} = 100} \\   {}& \Leftrightarrow &{\begin{array}{*{20}{c}}   {x – 2 =  – 10}& \vee &{x – 2 =  + 10} \end{array}} \\   {}& \Leftrightarrow &{\begin{array}{*{20}{c}}   {x =  – 8}& \vee &{x = 12} \end{array}} \end{array}$$

 

<< Enunciado
0

Quanto mede o lado do quadrado?

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 38 Ex. 25

Enunciado

 Quanto mede, com aproximação às décimas, o lado do quadrado, sabendo que a área da parte mais clara é, aproximadamente, $4,35$ m2?

Resolução >> Resolução

Designado por $r$ o comprimento do lado do quadrado, em metros, temos:

$$\begin{array}{*{20}{l}} {{A_{Clara}} = 4,35}& \Leftrightarrow &{\begin{array}{*{20}{c}} {{r^2} – \frac{1}{4}\pi  \times {r^2} = 4,35}& \wedge &{r > 0} \end{array}}\\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {4{r^2} – \pi  \times {r^2} = 17,4}& \wedge &{r > 0} \end{array}}\\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {\left( {4 – \pi } \right){r^2} = 17,4}& \wedge &{r > 0} \end{array}}\\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {{r^2} = \frac{{17,4}}{{4 – \pi }}}& \wedge &{r > 0} \end{array}}\\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {r =  \pm \sqrt {\frac{{17,4}}{{4 – \pi }}} }& \wedge &{r > 0} \end{array}}\\ {}& \Leftrightarrow &{r = \sqrt {\frac{{17,4}}{{4 – \pi }}} } \end{array}$$

 

Portanto, $r \approx 4,5$ cm.…

0

Qual deve ser o valor de $y$?

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 37 Ex. 24

Enunciado

Qual deve ser o valor de $y$ para que a área da figura seja $845$ m2?

Resolução >> Resolução

O dodecágono pode ser decomposto em $5$ quadrados geometricamente iguais (de que forma?), cuja área individual pode ser expressa por ${y^2}$.

Equacionando o problema e resolvendo a equação, temos:

$$\begin{array}{*{20}{l}} {5{y^2} = 845}& \Leftrightarrow &{{y^2} = 169}\\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {y =  – \sqrt {169} }& \vee &{y =  + \sqrt {169} } \end{array}}\\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {y =  – 13}& \vee &{y = 13} \end{array}} \end{array}$$

Portanto, para que a área da figura seja $845$ m2, terá de ser ${y = 13}$ m.…

0

De quanto deve ser o aumento para que a área duplique?

Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 37 Ex. 23

Enunciado

Aumentou-se igualmente os lados do retângulo da figura ao lado.

De quanto deve ser o aumento para que a área duplique?

Resolução >> Resolução

O retângulo tem $2$ unidades de área.

A área do retângulo ampliado pode ser expressa por: $\left( {x + 1} \right)\left( {x + 2} \right)$.…