AMC12 2009 B

AMC12 2009 B · Q9

AMC12 2009 B · Q9. It mainly tests Triangles (properties), Coordinate geometry.

Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$?

三角形 $ABC$ 的顶点为 $A = (3,0)$、$B = (0,3)$，以及 $C$，其中 $C$ 在直线 $x + y = 7$ 上。$\triangle ABC$ 的面积是多少？

(A) 6 6

(B) 8 8

(D) 12 12

(E) 14 14

Answer

Correct choice: (A)

正确答案：（A）

Solution

Because the line $x + y = 7$ is parallel to $\overline {AB}$, the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$. The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$. Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$, which is $\frac 4{\sqrt 2} = 2\sqrt 2$. Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$. The answer is $\mathrm{(A)}$. By Shoelace, our area is: \[\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.\] We know $x+y=7$ so we get: \[\frac {1}{2} \cdot |9-21|=\boxed 6\] WLOG, let the coordinates of $C$ be $(3,4)$ , or any coordinate, for that matter. Applying the shoelace formula, we get the area as $\boxed 6$.

因为直线 $x + y = 7$ 与 $\overline {AB}$ 平行，所以 $\triangle ABC$ 的面积与点 $C$ 在该直线上的位置无关。因此可假设 $C$ 为 $(7,0)$。此时三角形的底边 $AC = 4$，高为 $3$，所以面积为 $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$。三角形的底边为 $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$。其高为点 $A$ 到平行直线 $x + y = 7$ 的距离，即 $\frac 4{\sqrt 2} = 2\sqrt 2$。因此面积为 $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$。答案是 $\mathrm{(A)}$。用鞋带公式，面积为： \[\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.\] 已知 $x+y=7$，所以得到： \[\frac {1}{2} \cdot |9-21|=\boxed 6\] 不失一般性，令 $C$ 的坐标为 $(3,4)$（或任意坐标皆可）。应用鞋带公式，得到面积为 $\boxed 6$。

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