Question

Podem me ajudar???

Exercício sobre área de polígonos regulares.
***Com cálculo, pfvr!! ​

Answer 1

Resposta:

a)

Resolução:

$\sf \displaystyle A_{\triangle} = \dfrac{\sqrt{3} }{4} \: \mathit{l}^2$

$\sf \displaystyle A_{\triangle} = \dfrac{\sqrt{3} }{4} \: \mathit{(4\; cm)}^2$

$\sf \displaystyle A_{\triangle} = \dfrac{\sqrt{3} }{4} \: \mathit{16\; cm}^2$

$\boxed{ \boxed { \boldsymbol{ \sf \displaystyle A_{\triangle} = 4\:\sqrt{3} \: \mathit{ cm}^2 }}} \quad \gets \mathbf{ Resposta }$

b)

$\sf \displaystyle A_{\square} = \mathit{ l}^2$

$\sf \displaystyle A_{\square} = \mathit{(4\:cm)}^2$

$\boxed{ \boxed { \boldsymbol{ \sf \displaystyle A_{\square} = \mathit{16\:cm}^2 }}} \quad \gets \mathbf{ Resposta }$

c)

$\sf \displaystyle A = \dfrac{\mathit {n \cdot l}}{2} \cdot a$

$\sf \displaystyle A = \dfrac{\mathit {5 \cdot 4\:cm}}{2} \cdot \mathit { 2\:cm}$

$\sf \displaystyle A = \dfrac{\mathit {5 \cdot \diagup \!\!\! 4 \:cm}}{\diagup \!\!\! 2} \cdot \mathit { 2\:cm}$

$\sf \displaystyle A = \mathit {5 \cdot 2 \:cm} \cdot \mathit { 2\:cm}$

$\boxed{ \boxed { \boldsymbol{ \sf \displaystyle A = \mathit { 20\:cm^2} }}} \quad \gets \mathbf{ Resposta }$

d)

$\sf \displaystyle A = \dfrac{3\mathit{l}^2\sqrt{3} }{2}$

$\sf \displaystyle A = \dfrac{3 \cdot \mathit{(4\;cm)}^2\sqrt{3} }{2}$

$\sf \displaystyle A = \dfrac{3 \cdot \mathit{\diagup \!\!\! (16)\;cm}^2\sqrt{3} }{ \diagup \!\!\!2}$

$\sf \displaystyle A = 3 \cdot \mathit{8\;cm}^2\sqrt{3}$

$\boxed{ \boxed { \boldsymbol{ \sf \displaystyle A = \mathit{24 \: \sqrt{3} \;cm}^2 }}} \quad \gets \mathbf{ Resposta }$

Explicação passo-a-passo: