Question

4 - A área do triângulo ABC, de altura h = v2 , sendo X = 30° e Y = 45° é igual a:

Answer 1

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$\boxed{\begin{array}{l}\sf tg(30^\circ)=\dfrac{AH}{h}\implies AH=h\cdot tg(30^\circ)\\\sf AH=\sqrt{2}\cdot\dfrac{\sqrt{3}}{3}=\dfrac{\sqrt{6}}{3}\\\sf tg(45^\circ)=\dfrac{HB}{h}\implies HB=h\cdot tg(45^\circ)\\\sf HB=\sqrt{2}\cdot1=\sqrt{2}\\\sf AB=AH+HB\\\sf AB=\dfrac{\sqrt{6}}{3}+\sqrt{2}\\\sf A_{\triangle}=\dfrac{1}{2}\cdot AB\cdot h\\\sf A_{\triangle}=\dfrac{1}{2}\cdot\bigg(\dfrac{\sqrt{6}}{3}+\sqrt{2}\bigg)\cdot\sqrt{2}\\\sf \end{array}}$

$\large\boxed{\begin{array}{l}\sf A_{\triangle}=\dfrac{1}{2}\cdot\bigg(\dfrac{\sqrt{12}}{3}+\sqrt{4}\bigg)\\\sf A_{\triangle}=\dfrac{1}{2}\cdot\bigg(\dfrac{2\sqrt{3}}{3}+2\bigg)\\\sf A_{\triangle}=\dfrac{1}{2}\cdot\bigg(\dfrac{2\sqrt{3}+6}{3}\bigg)\\\sf A_{\triangle}=\dfrac{1}{2}\bigg(\dfrac{1}{3}\cdot\bigg[2\sqrt{3}+6\bigg]\bigg)\\\sf A_{\triangle}=\dfrac{1}{6}\bigg(2\sqrt{3}+6\bigg)\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~b}}}}\end{array}}$