Question

calcule a medida da área da regiao pintada de azul,sabendo que o lado dos quadradinhos da malha mede 0,5 cm​

Answer 1

A região em azul pode ser determinada subtraindo-se, da área total (retângulo), a área de cada figura geométrica que aparece (branco) dentro do retângulo.

$\boxed{Area_{Azul}~=~\left(^{Area~Total~do}_{~~retangulo}\right)~-~\left(^{Soma~das~areas~das}_{~figuras~em~branco}\right)}$

Vamos então começar extraindo as medidas de cada figura e calculando suas áreas. Para facilitar, acompanhe com auxilio do desenho anexado a resolução.

$\underline{Retangulo}:\\\\\\\left\{\begin{array}{ccl}Base&=&12\cdot 0,5~~~~\Rightarrow~\boxed{Base=6\,cm}\\\\Altura&=&11\cdot 0,5~~~~\Rightarrow~\boxed{Altura=5,5\,cm}\end{array}\right.\\\\\\\\Area~=~Base\times Altura\\\\\\Area~=~6\times 5,5\\\\\\\boxed{Area~=~33~cm^2}$

$\underline{Losango~(Vermelho)}:\\\\\\\left\{\begin{array}{ccl}D&=&4\cdot 0,5~~~~\Rightarrow~\boxed{D=2\,cm}\\\\d&=&4\cdot 0,5~~~~\Rightarrow~\boxed{d=2\,cm}\end{array}\right.\\\\\\\\Area~=~\dfrac{D\times d}{2}\\\\\\Area~=~\dfrac{2\times 2}{2}\\\\\\Area~=~\dfrac{4}{2}\\\\\\\boxed{Area~=~2~cm^2}$

$\underline{Losango~(Azul)}:\\\\\\\left\{\begin{array}{ccl}D&=&4\cdot 0,5~~~~\Rightarrow~\boxed{D=2\,cm}\\\\d&=&2\cdot 0,5~~~~\Rightarrow~\boxed{d=1\,cm}\end{array}\right.\\\\\\\\Area~=~\dfrac{D\times d}{2}\\\\\\Area~=~\dfrac{2\times 1}{2}\\\\\\Area~=~\dfrac{2}{2}\\\\\\\boxed{Area~=~1~cm^2}$

$\underline{Triangulo}:\\\\\\\left\{\begin{array}{ccl}base&=&5\cdot 0,5~~~~\Rightarrow~\boxed{base=2,5\,cm}\\\\altura&=&4\cdot 0,5~~~~\Rightarrow~\boxed{altura=2\,cm}\end{array}\right.\\\\\\\\Area~=~\dfrac{base\times altura}{2}\\\\\\Area~=~\dfrac{2,5\times 2}{2}\\\\\\Area~=~\dfrac{5}{2}\\\\\\\boxed{Area~=~2,5~cm^2}$

$\underline{Trapezio}:\\\\\\\left\{\begin{array}{ccl}base_{maior}&=&4\cdot 0,5~~~~\Rightarrow~\boxed{base_{maior}=2\,cm}\\\\base_{menor}&=&2\cdot0,5~~~~\Rightarrow~\boxed{base_{menor}=1\,cm}\\\\altura&=&2\cdot 0,5~~~~\Rightarrow~\boxed{altura=1\,cm}\end{array}\right.\\\\\\\\Area~=~\dfrac{(base_{maior}+base_{menor})\times altura}{2}\\\\\\Area~=~\dfrac{(2+1)\times 1}{2}\\\\\\Area~=~\dfrac{3\times 1}{2}\\\\\\\boxed{Area~=~1,5~cm^2}$

$\underline{Paralelogramo}:\\\\\\\left\{\begin{array}{ccl}base&=&3\cdot 0,5~~~~\Rightarrow~\boxed{base=1,5\,cm}\\\\altura&=&3\cdot 0,5~~~~\Rightarrow~\boxed{altura=1,5\,cm}\end{array}\right.\\\\\\\\Area~=~base\times altura\\\\\\Area~=~1,5\times 1,5\\\\\\\boxed{Area~=~2,25~cm^2}$

Podemos agora calcular a soma das áreas das figuras em branco como proposto no inicio da resolução:

$^{Soma~das~areas~das}_{~figuras~em~branco}~=~2~+~1~+~2,5~+~1,5~+~2,25\\\\\\\boxed{^{Soma~das~areas~das}_{~figuras~em~branco}~=~9,25~cm^2}$

Por fim, subtraindo esta área da área do retângulo, podemos determinar a área em azul:

$\boxed{Area_{Azul}~=~\left(^{Area~Total~do}_{~~retangulo}\right)~-~\left(^{Soma~das~areas~das}_{~figuras~em~branco}\right)}\\\\\\Area_{Azul}~=~33~-~9,25\\\\\\\boxed{Area_{Azul}~=~23,75~cm^2}~~~\Rightarrow~Resposta\\\\\\\\\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$

Answer 2

Primeira figura=

=22/2

=2cm2

-Segunda figura=

=2.5.2/2

=2,5cm2

-Terceira figura

=21/2

=1cm2

Quarta figura

= (1+2)1/2

=3/2

=1,5cm2

Quinta figura

=1,5.1,5

=2,25cm2

Azul=33-9,25

Azul=23,75cm2

*resposta:

A area pintada de Azul mede 23,75cm2.