gouy,
- fig1.jpg (25.47 KiB) Exibido 1116 vezes
[tex3]\mathtt{
Base_p:\triangle GIJ\\
\triangle GHJ: GJ^2=1^2+2^2\implies GJ= \sqrt5\\
\triangle GIH: GI^2=1^2+2^2 \implies GI=\sqrt5\\
\triangle HIJ: IJ^2=1^2+1^2 \implies IJ = \sqrt2\\
h_{\triangle {GIJ}}: h^2= \sqrt5^2-(\frac{\sqrt2}{2})^2\implies h = \frac{3\sqrt2}{2} \\
\therefore S_{\triangle {GIJ}}=\frac{1}{2}.\frac{3\sqrt2}{2}.\sqrt2=\underline {\frac{3}{2}}\\
H_p=KM\\
\triangle AKD:KD = \sqrt{2^2+1^2} = \sqrt5 \\
\triangle KDJ: KJ=\sqrt{1^2+\sqrt5^2}=\sqrt6\\
\triangle KHD: KH^2= \sqrt{2^2+\sqrt5^2}=3\\
\triangle JMH: Sendo~MH=x\implies JM^2 = x^2-1^2\\
\triangle JKM: JM^2 = \sqrt6^2-(3-x)^2 \\
Igualando: x^2-1^2 = 6-9+6x-x^2 \implies x = \frac{2}{3}\\
\therefore KM = 3-\frac{2}{3} =\underline{ \frac{7}{3}}\\
\therefore V_p = \frac{1}{3}.\frac{3}{2}.\frac{7}{3}=\boxed{\frac{7}{6}}\color{green}\checkmark
}[/tex3]