如图所示，矩形 ABCD 中，AD = 15 cm。 E 在 DC 的延长线上，使得 DE = 8 cm。 阴影部分 X 的面积比阴影部分 Y 的面积大 12 $\text{cm}^2$。 求 AB 的长度。 OAB 是一个三角形，C 是 AB 上的一点，使得 AC : CB = 2 : 1。 边 OB 延长至点 D，使得 OB : BD = 3 : 2。 $\overrightarrow{OA} = a$ 且 $\overrightarrow{OB} = b$。 尽可能简捷地用 $a$ 和 $b$ 表示 $\overrightarrow{OC}$。 尽可能简捷地用 $a$ 和 $b$ 表示 $\overrightarrow{CD}$。 答案 ... E 是 OA 上的一点，使得 $\overrightarrow{OE} = \tfrac{5}{9}a$。 证明 D、C、E 三点共线。 答案 写出以下比值： 三角形 OEC 的面积 三角形 OCD 的面积 答案 ... 写出以下比值： 三角形 EAC 的面积 三角形 OAB 的面积 答案 ...

1. Let the unknown length be represented algebraically and find the relevant whole areas.

Let $AB = x$ cm.

Since $ABCD$ is a rectangle, $DC = AB = x$ cm.

Area of rectangle $ABCD$:
$15x\text{ cm}^2$

Area of triangle $ADE$:
$\frac12 \times DE \times AD = \frac12 \times 8 \times 15 = 60\text{ cm}^2$

2. Relate the given shaded-area difference to the difference of the two whole areas.

The rectangle $ABCD$ consists of the unshaded part inside the rectangle and shaded region $X$.

Triangle $ADE$ consists of the same unshaded part and shaded region $Y$.

Hence,
$\text{Area of rectangle }ABCD - \text{Area of triangle }ADE = X - Y$

Given that $X$ is $12\text{ cm}^2$ more than $Y$,
$15x - 60 = 12$

3. Solve the resulting linear equation.

$15x - 60 = 12$
$15x = 72$
$x = 4.8$

Therefore,
$AB = 4.8\text{ cm}$

1. Express the vector from $A$ to $C$ using the given ratio on $AB$.

Since $AC:CB=2:1$, the whole segment $AB$ has $3$ equal parts and
$\overrightarrow{AC}=\frac{2}{3}\overrightarrow{AB}.$

2. Write $\overrightarrow{OC}$ in terms of $a$ and $b$ and simplify.

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=b-a.$
Hence
$\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}$
$=a+\frac{2}{3}(b-a)$
$=\frac{1}{3}a+\frac{2}{3}b.$

1. Find the position vector of $D$ from the ratio $OB:BD=3:2$.

Since $OB:BD=3:2$,
$\overrightarrow{BD}=\frac{2}{3}\overrightarrow{OB}=\frac{2}{3}b.$
Hence
$\overrightarrow{OD}=\overrightarrow{OB}+\overrightarrow{BD}=b+\frac{2}{3}b=\frac{5}{3}b.$

2. Use $\overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}$ and simplify.

From part (i),
$\overrightarrow{OC}=\frac{1}{3}a+\frac{2}{3}b.$
Therefore
$\overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}$
$=\frac{5}{3}b-\left(\frac{1}{3}a+\frac{2}{3}b\right)$
$=\frac{5}{3}b-\frac{1}{3}a-\frac{2}{3}b$
$=b-\frac{1}{3}a.$

1. Write the position vectors of $C$ and $E$.

From part (i),
$\overrightarrow{OC}=\frac{1}{3}a+\frac{2}{3}b.$
Given
$\overrightarrow{OE}=\frac{5}{9}a.$

2. Find $\overrightarrow{CE}$ in terms of $a$ and $b$.

$\overrightarrow{CE}=\overrightarrow{OE}-\overrightarrow{OC}$
$=\frac{5}{9}a-\left(\frac{1}{3}a+\frac{2}{3}b\right)$
$=\frac{2}{9}a-\frac{2}{3}b.$

3. Compare $\overrightarrow{CE}$ with $\overrightarrow{CD}$ to prove collinearity.

From part (ii),
$\overrightarrow{CD}=b-\frac{1}{3}a.$
Now
$\overrightarrow{CE}=\frac{2}{9}a-\frac{2}{3}b$
$=-\frac{2}{3}\left(b-\frac{1}{3}a\right)$
$=-\frac{2}{3}\overrightarrow{CD}.$
Therefore $\overrightarrow{CE}$ is a scalar multiple of $\overrightarrow{CD}$, so $D$, $C$ and $E$ lie on a straight line.

1. Use collinearity to compare the areas by their bases.

Since $D$, $C$ and $E$ lie on a straight line and $\overrightarrow{CE}=-\dfrac{2}{3}\overrightarrow{CD}$,
$EC:CD=2:3.$
Triangles $OEC$ and $OCD$ have the same height from $O$ to the straight line $ECD$.

Therefore,
$\text{area of }\triangle OEC : \text{area of }\triangle OCD = EC:CD=2:3.$

1. Compare the two triangles using products of corresponding side ratios from the common angle at $A$.

Since $\overrightarrow{OE}=\dfrac{5}{9}a$, $OE:OA=5:9$.

Hence
$AE:AO=(9-5):9=4:9.$

Also, $AC:CB=2:1$, so
$AC:AB=2:3.$

Triangles $EAC$ and $OAB$ have the included angle at $A$ common, so