Question

En la figura, $\angle EAB$ y $\angle ABC$ son ángulos rectos, $AB = 4,$ $BC = 6,$ $AE = 8,$ y $\overline{AC}$ y $\overline{BE}$ se cortan en $D.$ ¿Cuál es la diferencia entre las áreas de $\triangle ADE$ y $\triangle BDC$ ?

In the figure, $\angle EAB$ and $\angle ABC$ are right angles, $AB = 4,$ $BC = 6,$ $AE = 8,$ and $\overline{AC}$ and $\overline{BE}$ intersect at $D.$ What is the difference between the areas of $\triangle ADE$ and $\triangle BDC?$

$2$

$4$

$5$

$8$

$9$

Solución:

Sea $[ABD]$ el área compartida por ambos triángulos grandes. Entonces $[ABE] = [ADE] + [ABD]$ y $[ABC] = [BDC] + [ABD].$

Restando, $\begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC]. \end{aligned}$ Como $\angle EAB$ y $\angle ABC$ son ángulos rectos, $\begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12. \end{aligned}$

La diferencia es $16 - 12 = 4.$

Por lo tanto, la respuesta correcta es B.

Let $[ABD]$ be the area shared by both large triangles. Then $[ABE] = [ADE] + [ABD]$ and $[ABC] = [BDC] + [ABD].$

Subtracting, $\begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC]. \end{aligned}$ Since $\angle EAB$ and $\angle ABC$ are right angles, $\begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12. \end{aligned}$

The difference is $16 - 12 = 4.$

Thus, the correct answer is B.

2004 AMC 10A Problema 9

Solución:

El Problema 9 en otros años