Long Question (Question 9)

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