AMC10 2006 B

AMC10 2006 B · Q23

AMC10 2006 B · Q23. It mainly tests Linear equations, Area & perimeter.

A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?

一个三角形通过从顶点向对边画两条线，被分割成三个三角形和一个四边形。三个三角形的面积分别为3、7和7，如图所示。阴影四边形的面积是多少？

(A) 15 15

(B) 17 17

(D) 18 18

(E) $\frac{55}{3}$ $\frac{55}{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

gles be $R$ and $S$ as shown. Then the required area is $T = R + S$. Let $a$ and $b$, respectively, be the bases of the triangles with areas $R$ and $3$, as indicated. If two triangles have the same altitude, then the ratio of their areas is the same as the ratio of their bases. Thus $\dfrac{a}{b}=\dfrac{R}{3}=\dfrac{R+S+7}{3+7},\ \text{so}\ \dfrac{R}{3}=\dfrac{T+7}{10}.$ Similarly, $\dfrac{S}{7}=\dfrac{S+R+3}{7+7},\ \text{so}\ \dfrac{S}{7}=\dfrac{T+3}{14}.$ Thus $T=R+S=3\left(\dfrac{T+7}{10}\right)+7\left(\dfrac{T+3}{14}\right).$ From this we obtain $10T=3(T+7)+5(T+3)=8T+36,$ and it follows that $T=18$.

如图令两块面积分别为 $R$ 和 $S$，则所求面积为 $T=R+S$。分别令 $a$ 和 $b$ 为图示中面积为 $R$ 与 $3$ 的两个三角形的底边长度。若两个三角形高相同，则它们面积之比等于底边之比。因此 $\dfrac{a}{b}=\dfrac{R}{3}=\dfrac{R+S+7}{3+7},\ \text{所以}\ \dfrac{R}{3}=\dfrac{T+7}{10}.$ 同理， $\dfrac{S}{7}=\dfrac{S+R+3}{7+7},\ \text{所以}\ \dfrac{S}{7}=\dfrac{T+3}{14}.$ 因此 $T=R+S=3\left(\dfrac{T+7}{10}\right)+7\left(\dfrac{T+3}{14}\right).$ 由此得到 $10T=3(T+7)+5(T+3)=8T+36,$ 从而 $T=18$。

Topics