Question

No triângulo isósceles ABC, indicado na figura, AB = AC = 6, BC = 4 , DE = 2 e BC // DE. Sendo h a altura do trapézio DEBC, calcule 3h^2.

Answer 1

$\Delta \ ABC \\ AB=AC=6 \\ AD=AE \\ DE \ \Vert \ BC \ \ \ \ \ \Longrightarrow \ \Delta ADE \approx \Delta ABC \\\\\\ \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC} \\\\ \frac{AD}{6}=\frac{AE}{6}=\frac{2}{4} \\\\ \frac{AD}{6}=\frac{2}{4} \Longrightarrow AD=\frac{6*\not2}{\not4}= \frac{\not6}{\not2}=3 \\\\ AD=AE=3 \\\\ DB=EC \\\\ DB=AB-AD \\\\ DB=6-3 \\\\ DB=3$
 
$BDEC-trapezio \\ BD=EC \\Ser \ DM \perp BC \ \ \ \wedge \ \ \ EN \perp BC \\\\ DE=MN=2 \\\\ BM=NC=\frac{BC-MN}{2} = \frac{4-2}{2}=\frac{2}{2}=1 \\\\ BM=NC=1 \\\\\\ \Delta BDM; \ \angle M=90^0 \\\\ DM^2=BD^2-BM^2 \\\\ DM^2=3^2-1^2 \\\\ DM^2=9-1 \\\\ DM^2=8 \ \Longrightarrow DM=2\sqrt{2}$

$DM\to h \ trapezio \\\\ 3h^2= 3* (2\sqrt{2})^2= \\\\ =3*8= \\\\ = \boxed{24}$