Question

q) Calcule a área do triângulo ABC, sabendo que h= $\sqrt{2}$ cm, α = 30° e β = 45° .


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Answer 1

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$\sf tg(30^\circ)=\dfrac{AH}{\sqrt{2}}\\\sf AH=\sqrt{2}\cdot\dfrac{\sqrt{3}}{3}\\\sf AH=\dfrac{\sqrt{6}}{3}\\\sf tg(45^\circ)=\dfrac{HB}{\sqrt{2}}\\\sf HB=\sqrt{2}\cdot1=\sqrt{2}\\\sf A_{\triangle}=\dfrac{1}{2}\cdot(AH+BH)\cdot h\\\sf A_{\triangle}=\dfrac{1}{2}\cdot\left(\dfrac{\sqrt{6}}{3}+\sqrt{2}\right)\cdot\sqrt{2}\\\sf A_{\triangle}=\dfrac{1}{2}\cdot\left(\dfrac{\sqrt{12}}{3}+\sqrt{4}\right)\\\sf A_{\triangle}=\dfrac{1}{2}\cdot\left(\dfrac{2\sqrt{3}}{3}+2\right)$

$\huge\boxed{\boxed{\boxed{\boxed{\sf A_{\triangle}=\dfrac{\sqrt{3}}{3}+1\checkmark}}}}$