Question

CALCULE A ÁREA E O PERIMETRO E URGENTE!!!!

Answer 1

Primeira figura:  Como não há qualquer outra informação, vamos calcular separadamente as áreas e os perímetros do retângulo e triangulo da figura.

$Area_{retangulo}~=~Comprimento\times Largura\\\\\\Area_{retangulo}~=~4m\times 6m\\\\\\\boxed{Area_{retangulo}~=~24m^2}\\\\\\\\Perimetro_{retangulo}~=~2\times Comprimento~+~2\times Largura\\\\\\Perimetro_{retangulo}~=~2\times 6m~+~2\times 4m\\\\\\Perimetro_{retangulo}~=~12m~+~8m\\\\\\\boxed{Perimetro_{retangulo}~=~20m}$

$Area_{triangulo}~=~\dfrac{Base\times Altura}{2}\\\\\\Area_{triangulo}~=~\dfrac{6m\times 4m}{2}\\\\\\Area_{triangulo}~=~\dfrac{24m^2}{2}\\\\\\\boxed{Area_{triangulo}~=~12m^2}\\\\\\\\Perimetro_{triangulo}~=~Lado_1+Lado_2+Lado_3\\\\\\Perimetro_{triangulo}~=~6m+3m+3m\\\\\\\boxed{Perimetro_{triangulo}~=~12m}$

Segunda Figura: Como não há qualquer outra informação, vamos considerar que o topo do desenho seja um semicirculo (metade de circulo) de raio 6cm.

$Area~=~Area_{retangulo}+Area_{semicirculo}\\\\\\Area~=~(Comprimento\times Largura)~+~\left(\dfrac{\pi\,.\,Raio^2}{2}\right)\\\\\\Area~=~(8cm\times 6cm)~+~\left(\dfrac{\pi\,.\,(6cm)^2}{2}\right)\\\\\\Area~=~(48cm^2)~+~\left(\dfrac{\pi\,.\,36cm^2}{2}\right)\\\\\\Area~=~48cm^2~+~\pi\,.\,18cm^2\\\\\\\boxed{Area~=~(48+18\pi)\,cm^2}$

$Perimetro~=~(2\times Comprimento_{ret}+Largura_{ret})~+~(Comprimento_{semicirculo})\\\\\\Perimetro~=~(2\times 8cm+6cm)~+~\dfrac{2\pi~.~Raio}{2}\\\\\\Perimetro~=~(16cm+6cm)~+~\dfrac{2\pi~.~6cm}{2}\\\\\\Perimetro~=~(22cm)~+~6\pi\\\\\\\boxed{Perimetro~=~(22~+~6\pi)\,cm}$