$OABC$ is a trapezium such that $OA$ is parallel to $CB$, and $CB$ : $OA$ = $k:1$ ,

Question

$OABC$ is a trapezium such that $OA$ is parallel to $CB$, and $CB$ : $OA$ = $k:1$ , where $k$ is a positive constant, and $k\ne 1$. Given that $\overrightarrow{OA}$ = $\mathbf{a}$, $\overrightarrow{OB}$ = $\mathbf{b}$, and $X$ and $Y$ are the midpoints of $OB$ and $AC$ respectively, find the following vectors in terms of $k$, $\mathbf{a}$ and $\mathbf{b}$. $\overrightarrow{OC},$ [1] $\overrightarrow{OY}.$[2] Hence show that $XY$ is parallel to $OA$.[2] It is given that $OB$ and $AC$ intersect at the point $D$. Find the ratio, in terms of $k$, between the area of the triangle $XYD$ and the area of the triangle $BCD$.[2]

Accepted Answer

(i) $\overrightarrow{OC}=\mathbf b-k\mathbf a$. (ii) $\overrightarrow{OY}=\frac12\mathbf b+\frac{1-k}{2}\mathbf a$. Also, $\overrightarrow{XY}=\frac{1-k}{2}\mathbf a$, so $XY\parallel OA$; $\overrightarrow{OD}=\frac1{1+k}\mathbf b$; and $\text{Area}(XYD):\text{Area}(BCD)=(1-k)^2:4k^2$.

2013 RI P1 Q5

2013 RI P1 Q5

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