Estude! Com Prof. Caju

Resolva um triangulo conhecendo o perímetro [tex3]12 \sqrt{2}[/tex3] e a altura [tex3]12 \frac{\sqrt{2}}{5}[/tex3] relativa à hipotenusa.

@marcelojr,

Sejam a a hipotenusa, b e c os catetos.
[tex3]a + b + c = 12\sqrt{2}[/tex3]
Altura relativa à hipotenusa: [tex3]h = \frac{12\sqrt{2}}{5}[/tex3]
Relação de área:[tex3] bc = ah \implies bc = a \cdot \frac{12\sqrt{2}}{5}(II)[/tex3]
[tex3]b^2 + c^2 = a^2[/tex3]
[tex3](b+c)^2 = b^2 + c^2 + 2bc \implies (b+c)^2 = a^2 + 2ah[/tex3]
Do perímetro: [tex3]b + c = 12\sqrt{2} - a(I)[/tex3].
Substituindo na equação acima:[tex3](12\sqrt{2} - a)^2 = a^2 + 2a\left(\frac{12\sqrt{2}}{5}\right)[/tex3]
[tex3](12\sqrt{2})^2 - 24\sqrt{2}a + a^2 = a^2 + \frac{24\sqrt{2}}{5}a\\288 - 24\sqrt{2}a = \frac{24\sqrt{2}}{5} \implies 1440 - 120\sqrt{2}a = 24\sqrt{2}a\\1440 = 144\sqrt{2}a \implies a = \frac{1440}{144\sqrt{2}} = \frac{10}{\sqrt{2}}\\
\therefore \boxed{a = 5\sqrt{2}}[/tex3]
[tex3]De(I):b + c = 12\sqrt{2} - 5\sqrt{2} = 7\sqrt{2}(III)\\De(II):bc = ah = (5\sqrt{2}) \cdot \left(\frac{12\sqrt{2}}{5}\right) = 24(IV)[/tex3]
[tex3]De(III)e(IV): (b = 7\sqrt{2} - c) \implies (7\sqrt{2} - c) \cdot c = 24 \implies 7\sqrt{2}c - c^2 = 24 \implies c^2 - 7\sqrt{2}c + 24 = 0\\
Resolvendo:
c_1 = \frac{8\sqrt{2}}{2} = \boxed{4\sqrt{2}}:c_2 = \frac{6\sqrt{2}}{2} = \boxed{3\sqrt{2}}[/tex3]
[tex3]\boxed{Hipotenusa (a): 5\sqrt{2}\\
Cateto (b): 4\sqrt{2}\\
Cateto (c): 3\sqrt{2}}[/tex3]