Question

in the diagram, or is parallel to pq and $\frac{pq}{or}=\frac{2}{3}$. op and rq produced meet at s. $overrightarrow{op}=p$ and $overrightarrow{pq}=q$. (i) express in terms of $p$ and/or $q$. (a) $overrightarrow{or}$, 1 (b) $overrightarrow{rq}$. 2 (ii) write down, in its lowest terms, the ratio (a) $\frac{qs}{rs}$, 1 (b) $\frac{\text{area of }\triangle pqs}{\text{area of trapezium opqr}}$. 2

Explanation:

Step1: Find $\overrightarrow{OR}$

Since $\frac{PQ}{OR}=\frac{2}{3}$, we can express $\overrightarrow{OR}$ in terms of $\overrightarrow{PQ}$. Given $\overrightarrow{PQ} = q$, then $\overrightarrow{OR}=\frac{3}{2}q$.

Step2: Find $\overrightarrow{RQ}$

We know that $\overrightarrow{OP} = p$ and $\overrightarrow{PQ}=q$. Consider the vectors in the trapezium. Since $\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}=p + q$. Also, $\overrightarrow{OR}=\frac{3}{2}q$. Then $\overrightarrow{RQ}=\overrightarrow{OQ}-\overrightarrow{OR}=(p + q)-\frac{3}{2}q=p-\frac{1}{2}q$.

Step3: Find the ratio $\frac{QS}{RS}$

Because $OR\parallel PQ$, we have similar - triangles $\triangle SOP\sim\triangle SOR$. The ratio of corresponding sides of similar - triangles is the same. Since $\frac{PQ}{OR}=\frac{2}{3}$, from the property of similar - triangles, $\frac{QS}{RS}=\frac{2}{3}$.

Step4: Find the ratio of areas

Let the height of $\triangle SOP$ and $\triangle SOR$ (with bases $OP$ and $OR$ respectively) be $h_1$ and the height of trapezium $OPQR$ be $h_2$. The height of $\triangle PQS$ and $\triangle SOR$ are in the ratio of the ratio of their parallel - sides. Let the height of $\triangle SOR$ be $h$. The height of $\triangle PQS$ is $\frac{2}{3}h$ and the height of trapezium $OPQR$ is $h-\frac{2}{3}h=\frac{1}{3}h$.
The area of $\triangle PQS=\frac{1}{2}\times base\times height$. Let the base of $\triangle PQS$ be considered with respect to the height from $S$ to $PQ$. The area of $\triangle PQS = \frac{1}{2}\times PQ\times\frac{2}{3}h$. The area of trapezium $OPQR=\frac{1}{2}(PQ + OR)\times\frac{1}{3}h$. Substituting $OR=\frac{3}{2}PQ$, we get:
\[

$$\begin{align*} \frac{\text{area of }\triangle PQS}{\text{area of trapezium }OPQR}&=\frac{\frac{1}{2}\times PQ\times\frac{2}{3}h}{\frac{1}{2}(PQ+\frac{3}{2}PQ)\times\frac{1}{3}h}\\ &=\frac{\frac{2}{3}PQ}{\frac{1}{3}( \frac{2PQ + 3PQ}{2})}\\ &=\frac{\frac{2}{3}PQ}{\frac{1}{3}\times\frac{5}{2}PQ}\\ &=\frac{4}{5} \end{align*}$$

\]

Answer:

(i) (a) $\overrightarrow{OR}=\frac{3}{2}q$
(b) $\overrightarrow{RQ}=p-\frac{1}{2}q$
(ii) (a) $\frac{QS}{RS}=\frac{2}{3}$
(b) $\frac{\text{area of }\triangle PQS}{\text{area of trapezium }OPQR}=\frac{4}{5}$