AMC12 2000 A

AMC12 2000 A · Q19

AMC12 2000 A · Q19. It mainly tests Triangles (properties), Area & perimeter.

In triangle $ABC$, $AB = 13$, $BC = 14$, $AC = 15$. Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$. Which of the following is closest to the area of the triangle $ADE$?

在三角形$ABC$中，$AB = 13$，$BC = 14$，$AC = 15$。设$D$为$\overline{BC}$的中点，$E$为$\overline{BC}$与$\angle BAC$的角平分线的交点。以下哪一项最接近三角形$ADE$的面积？

(A) 2 2

(B) 2.5 2.5

(D) 3.5 3.5

(E) 4 4

Answer

Correct choice: (B)

正确答案：（B）

Solution

The answer is exactly $3$, choice $\mathrm{(C)}$. We can find the area of triangle $ADE$ by using the simple formula $\frac{bh}{2}$. Dropping an altitude from $A$, we see that it has length $12$ (we can split the large triangle into a $9-12-15$ and a $5-12-13$ triangle). Then we can apply the Angle Bisector Theorem on triangle $ABC$ to solve for $BE$. Solving $\frac{13}{BE}=\frac{15}{14-BE}$, we get that $BE=\frac{13}{2}$. $D$ is the midpoint of $BC$ so $BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\frac{1}{2}$ units long. Applying the formula $\frac{bh}{2}$, we get $\frac{12*\frac{1}{2}}{2}=3$.

答案恰为$3$，选项$\mathrm{(C)}$。我们可以用简单的公式$\frac{bh}{2}$求三角形$ADE$的面积。从$A$作高，可见其长度为$12$（可将大三角形分成一个$9-12-15$三角形和一个$5-12-13$三角形）。然后在三角形$ABC$中应用角平分线定理求$BE$。解$\frac{13}{BE}=\frac{15}{14-BE}$，得$BE=\frac{13}{2}$。$D$是$BC$的中点，所以$BD=7$。因此三角形$ADE$的底$DE$为$\frac{1}{2}$。应用公式$\frac{bh}{2}$，得到$\frac{12*\frac{1}{2}}{2}=3$。

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