Vol. 09 - Geometria Plana FME 2004 ⇒ 296 - FME 09 -Teste de Vestibulares Tópico resolvido

[petrasMOD]

(COVEST-U.F.R.PE-91) A área do trapézio da figura é:
[tex3]a) x( { y + \frac{1}{2}\sqrt{z^2-x^2}})\\
b) x ( y - \frac{1}{2}\sqrt{z^2-x^2)}\\
c) \frac{1}{2}(z + x) y\\
d) \frac{1}{2} (x + y)z\\
e) xy + \frac{1}{2} xz[/tex3]

Resposta

---

Gabarito: a)

[petrasMOD]

[tex3]\triangle : a^2 = z^2 - x^2 \implies a = \sqrt{z^2-x^2}\\
S = \frac{y+y+\sqrt{z^2-x^2}}{2}.x = \frac{2y+\sqrt{z^2-x^2}}{2}.x\\
\therefore \boxed{S =x(y+\frac{\sqrt{z^2-x^2}}{2})} [/tex3]