The figure shows a rectangle ABCD with AD = 15 cm. E is on DC produced such that DE = 8 cm. The area of shaded part X is 12 cm^2 more than the area of shaded part Y. Find the length of AB. OAB is a triangle and C is a point on AB such that AC : CB = 2 : 1. The side OB is produced to the point D such that OB : BD = 3 : 2. $\overrightarrow{OA} = a$ and $\overrightarrow{OB} = b$. Express $\overrightarrow{OC}$ in terms of $a$ and $b$, as simply as possible. Express $\overrightarrow{CD}$ in terms of $a$ and $b$, as simply as possible. Answer ... E is the point on OA such that $\overrightarrow{OE} = \tfrac{5}{9}a$. Show that D, C and E lie on a straight line. Answer Write down the ratio area of triangle OEC area of triangle OCD Answer ... Write down the ratio Area of triangle EAC Area of triangle OAB Answer ...

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