Question

A área do triângulo ABC, de altura h = , sendo X = 30° e Y = 45° é.

Answer 1

$\displaystyle \text S_{\text {ABC}} = \frac{\text{AB.h}}{2}$

Façamos :

*

$\displaystyle \text{Tg}(30^\circ) = \frac{\text{AH}}{\text h} \\\\\\ \frac{\sqrt{3}}{3} = \frac{\text{AH}}{\sqrt{2}} \\\\\\ \text{AH} = \frac{\sqrt{6}}{3}$

*

$\displaystyle \text{Tg}(45^\circ )=\frac{\text{BH}}{\text h} \\\\\\ 1 = \frac{\text{BH}}{\sqrt{2}} \\\\\\ \text{BH} = \sqrt{2}$

E temos que :

AB = AH + BH

$\displaystyle \text{AB} = \frac{\sqrt{6}}{3}+\sqrt{2} \\\\\\ \text{AB}= \frac{\sqrt{6}+3\sqrt{2}}{3} \\\\\\ \text{AB}=\frac{\sqrt{2}(\sqrt{3}+3)}{3}$

Substituindo na área :

$\displaystyle \text S_{\text {ABC}} = \frac{\text{AB.h}}{2} \\\\\\ \text S_{\text{ABC}} = \frac{\sqrt{2}\ [ \ \sqrt{2}(\sqrt{3}+3)\ ] }{3}.\frac{1}{2} \\\\\\ \huge\boxed{\text S_{\text {ABC}} = \frac{\sqrt{3}+3}{3}\ }\checkmark$