AMC12 2004 A

AMC12 2004 A · Q8

AMC12 2004 A · Q8. It mainly tests Area & perimeter, Coordinate geometry.

In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$, $\angle{EAB}$ 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 $\triangle{BDC}$?

在重叠的三角形 $\triangle{ABC}$ 和 $\triangle{ABE}$ 中，它们共有边 $AB$，$\angle{EAB}$ 和 $\angle{ABC}$ 是直角，$AB=4$，$BC=6$，$AE=8$，且 $\overline{AC}$ 与 $\overline{BE}$ 相交于 $D$。$\triangle{ADE}$ 与 $\triangle{BDC}$ 的面积之差是多少？

(A) 2 2

(B) 4 4

(D) 8 8

(E) 9 9

Answer

Correct choice: (B)

正确答案：（B）

Solution

Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ - $[\triangle CDB]$. So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$. Therefore, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{\mathrm{(B)}\ 4}$

观察可得，$[\triangle EBA]$ 的面积为 16，$[\triangle ABC]$ 的面积为 12。设 $[\triangle ADB]$ 的面积为 x。我们要求 $[\triangle ADE]-[\triangle CDB]$。于是 $[\triangle ADE]=[\triangle EBA]-[\triangle ADB]=16-x$，$[\triangle DBC]=[\triangle ABC]-[\triangle ADB]=12-x$。因此，$[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{\mathrm{(B)}\ 4}$

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