SPM Additional Mathematics 2018, Paper 2 (Question 3 & 4)

Question 3 (6 marks):
Solution by scale drawing is not accepted.
Diagram 1 shows a triangle OCD.
Diagram 1

(a) Given the area of triangle OCD is 30 units², find the value of h.

(b) Point Q (2, 4) lies on the straight line CD.
(i) Find CQ : QD.
(ii) Point P moves such that PD = 2 PQ.
  Find the equation of the locus P.


Solution:
(a)
$\begin{array}{l}\text{Given Area of }△OCD\text{ = }30\\ \\ \frac{1}{2}|\begin{array}{l}\text{0 }h-6\\ \text{0 2 8 }\end{array}\begin{array}{l}0\\ 0\end{array}|=30\\ \\ |\left(0\right)\left(2\right)+\left(h\right)\left(8\right)+\left(-6\right)\left(0\right)-\left(0\right)\left(h\right)-\left(2\right)\left(-6\right)-\left(8\right)\left(0\right)|=60\\ |0+8h+0-0+12-0|=60\\ |8h+12|=60\\ \\ 8h+12=60\\ 8h=48\\ h=6\\ \text{or }\\ 8h+12=-60\\ 8h=-72\\ h=-9\left(\text{ignore}\right)\end{array}$


(b)(i)

$\begin{array}{l}\left[\frac{6\left(m\right)+\left(-6\right)\left(n\right)}{m+n},\frac{2\left(m\right)+\left(8\right)\left(n\right)}{m+n}\right]=\left(2,4\right)\\ \frac{6m-6n}{m+n}=2\\ 6m-6n=2m+2n\\ 4m=8n\\ \frac{m}{n}=\frac{8}{4}\\ \frac{m}{n}=\frac{2}{1}\\ \\ \frac{2m+8n}{m+n}=4\\ 2m+8n=4m+4n\\ 2m=4n\\ \frac{m}{n}=\frac{4}{2}\\ \frac{m}{n}=\frac{2}{1}\\ \\ \text{Thus, }CQ=QD=2:1\end{array}$


(b)(ii)
$\begin{array}{l}PD=2PQ\\ \sqrt{{\left(x-6\right)}^2+{\left(y-2\right)}^2}=2\sqrt{{\left(x-2\right)}^2+{\left(y-4\right)}^2}\\ {\left(x-6\right)}^2+{\left(y-2\right)}^2=4\left[{\left(x-2\right)}^2+{\left(y-4\right)}^2\right]\\ x^2-12x+36+y^2-4y+4=4\left[x^2-4x+4+y^2-8y+16\right]\\ x^2-12x+36+y^2-4y+4=4x^2-16x+16+4y^2-32y+64\\ \text{The equation of locus }P:\\ 3x^2+3y^2-4x-28y+40=0\end{array}$

Question 4 (7 marks):
Diagram 2 shows the plan of a rectangular garden ABCD. The garden consists of a semicircular pond ATD and grassy area ABCDT.
It is given that DC = 6y metre and BC = 7x metre, x ≠ y. The area of the rectangular garden ABCD is 168 metre² and the perimeter of the grassy area is 60 metre. The pond with uniform depth contains 15.4 metre³ of water.
By using  $\pi =\frac{22}{7}$ , find the depth, in metre, of water in the pond.


Solution:
$\begin{array}{l}\text{Area of }ABCD=168\\ \left(6y\right)\left(7x\right)=168\\ 42xy=168\\ xy=\mathrm{4\dots \dots \dots \dots \dots \dots .}\left(1\right)\\ \\ \text{Perimeter of grassy area}=60\\ 6y+6y+7x+\left(\frac{1}{\cancel{2}}\times \cancel{2}\times \frac{{\cancel{22}}^{11}}{\cancel{7}}\times \frac{\cancel{7}x}{\cancel{2}}\right)=60\\ 12y+18x=60\\ 2y+3x=\mathrm{10\dots \dots \dots \dots \dots \dots .}\left(2\right)\\ \\ \text{From }\left(1\right):xy=4\\ x=\frac{4}{y}\mathrm{\dots \dots \dots \dots \dots \dots .}\left(3\right)\\ \\ \text{Substitute (3) into (2):}\\ 2y+3\left(\frac{4}{y}\right)=10\\ 2y^2+12=10y\\ y^2-5y+6=0\\ \left(y-2\right)\left(y-3\right)=0\\ y-2=0\\ y=2\\ \text{or }\\ y-3=0\\ y=3\\ \\ \text{Substitute the values of }y\text{ into (3):}\\ \text{When }y=2\\ x=\frac{4}{2}=2\left(\text{ignored, }x\ne y\right)\\ \text{When }y=3\\ x=\frac{4}{3}\\ \\ \text{Given volume of water}=15.4\\ \frac{1}{2}\left(\frac{22}{7}\right){\left(\frac{7x}{2}\right)}^{2}d=15.4\\ \frac{1}{2}\left(\frac{22}{7}\right){\left[\frac{7\left(\frac{4}{3}\right)}{2}\right]}^{2}d=15.4\\ \frac{11}{7}{\left(\frac{14}{3}\right)}^{2}d=15.4\\ \frac{308}{9}d=15.4\\ d=0.45\text{ m}\\ \\ \text{Thus, the depth of water in the pond is 0}\text{.45 m}\text{.}\end{array}$