Question
Quadrilateral OABC is shown on the following set of axes.
$OABC\text{ is symmetrical about }[OB].$
$A\text{ has coordinates }(6,0)\text{ and }C\text{ has coordinates }(3,3\sqrt{3}).$
$(a)\quad(i)\text{ Write down the coordinates of the midpoint of }[AC].$
$\quad\quad(ii)\text{ Hence or otherwise, find the equation of the line passing through the points }O\text{ and }B.$
$(b)\text{ Given that }[OA]\text{ is perpendicular to }[AB],\text{ find the area of the quadrilateral }OABC.$
▶️Answer/Explanation
Solution:-
(b) substituting $x=6$ into their equation
so at B $y=2\sqrt{3}$
area of triangle $OAB=\frac{1}{2}\times6\times2\sqrt{3}=6\sqrt{3}$
area of quadrilateral $OABC=12\sqrt{3}$
Solution
(a) (i)The mid point of AC given
$A\text{ has coordinates }(6,0)\text{ and }C\text{ has coordinates }(3,3\sqrt{3})$, let the mid point be M
Therefore, the mid point M is $\frac{6+3}{2}, \frac{0+3\sqrt{3}}{2}$
Coordinates of M = $\frac{9}{2}, \frac{3\sqrt{3}}{2}$
(ii) Point M lies on OB, the slope of line OB can be found by Coordinates of point O(0,0) and coordinates of M = $\frac{9}{2}, \frac{3\sqrt{3}}{2}$
Slope $ tan\theta = m = \frac{\frac{3\sqrt{3}}{2}-0}{\frac{9}{2}-0}$ = $\frac{1}{\sqrt{3}}$
Therefore, equation of line OB passing through origin, y=$\frac{1}{\sqrt{3}}$x
(b) Since $OABC\text{ is symmetrical about }[OB]$
Therefore area of the quadrilateral }OABC = $(2\times area of triangle OAB)$
= $(2\times \frac{1}{2} OA. AB)$
$\text{ Given that }[OA]\text{ is perpendicular to }[AB]$, so $ tan\theta$ = $\frac{1}{\sqrt{3}}$ = $\frac{AB}{OA}$ = $\frac{AB}{6}$
Therefore AB =\(\frac{6}{\sqrt{3}}\),Therefore area of the quadrilateral OABC =\( 2 \times\frac{6}{\sqrt{3}}\)
area of quadrilateral OABC=\(12\sqrt{3}\)
Question
ABCD is a quadrilateral where \({\text{AB}} = 6.5,{\text{ BC}} = 9.1,{\text{ CD}} = 10.4,{\text{ DA}} = 7.8\) and \({\rm{C\hat DA}} = 90^\circ \). Find \({\rm{A\hat BC}}\), giving your answer correct to the nearest degree.
▶️Answer/Explanation
Markscheme
\({\text{A}}{{\text{C}}^2} = {7.8^2} + {10.4^2}\) (M1)
\({\text{AC}} = 13\) (A1)
use of cosine rule eg, \(\cos ({\rm{A\hat BC}}) = \frac{{{{6.5}^2} + {{9.1}^2} – {{13}^2}}}{{2(6.5)(9.1)}}\) M1
\({\rm{A\hat BC}} = 111.804 \ldots ^\circ {\text{ }}( = 1.95134 \ldots )\) (A1)
\( = 112^\circ \) A1
[5 marks]
Examiners report
Well done by most candidates. A small number of candidates did not express the required angle correct to the nearest degree.