**Question 5**:

Lisa has a rectangular plot of land. She plants orchid and rears fish in the areas as shown on diagram above. The area used for planting orchid is 460 m

^{2}and the perimeter of the rectangular fish pond is 48 m. Find the value of*x*and*y*.

*Solution:*Area used for planting orchid = 460 m

^{2}10 (30 –

*y*) +*xy*= 460300 – 10

*y*+*xy*= 460*xy*– 10

*y*= 160

*y*(

*x*– 10) = 160

Perimeter of the rectangular fish pond = 48 m

2 (

*x*– 10) + 2 (30 –*y*) = 482

*x*– 20 + 60 – 2*y*= 482

*x*– 2*y*= 8*x*–*y*= 4*x*= 4 +

*y*—— (2)

Substitute (2) into (1):

$\begin{array}{l}y=\frac{160}{x-10}\\ y=\frac{160}{4+y-10}\\ y=\frac{160}{y-6}\\ {y}^{2}-6y-160=0\\ \left(y-16\right)\left(y+10\right)=0\\ y=16\text{or}y=-10\text{(notaccepted)}\end{array}$

$\begin{array}{l}y=\frac{160}{x-10}\\ y=\frac{160}{4+y-10}\\ y=\frac{160}{y-6}\\ {y}^{2}-6y-160=0\\ \left(y-16\right)\left(y+10\right)=0\\ y=16\text{or}y=-10\text{(notaccepted)}\end{array}$

From (2),

**When**

*y*= 16

**x****= 4 + 16 = 20**

**Question 6**:

In the diagram above,

*PQRS*is a rectangular piece of paper with an area of 112 cm

^{2}. A semicircle

*STR*is cut from this piece of paper. The perimeter of the remaining piece of paper is 52 cm. Using $\pi =\frac{22}{7}$ , find the integer value of

*x*and

*y*.

*Solution:*Area of the rectangle

*PQRS*= 112 cm^{2}Therefore, (14

*x*)(2*y*) = 112 28

*xy*= 112*xy*= 4 —— (1)

Perimeter of PSTRQ = 52 cm

*PS*+

*QR*+

*PQ*+ Length of arc

*STR*= 52

2

*y*+ 2*y*+ 14*x*+ ½ (2πr) = 524

*y*+ 14*x*+ $\left(\frac{22}{7}\right)\left(7x\right)$ = 524

*y*+ 14*x*+ 22*x*= 524

*y*+ 36*x*= 52*y*+ 9

*x*= 13 —— (2)

From equation (2) :

*y*= 13 – 9*x*—— (3)Substitute (3) into (1) :

*x*(13 – 9

*x*) = 4

13

*x*– 9*x*^{2}= 49

*x*^{2}– 13*x*+ 4 = 0(

*x*– 1)(9*x*– 4) = 0*x*= 1 or $\frac{4}{9}\text{(notaninterger,thereforenotaccepted)}$

From (3) :

When

**,***x*= 1

*y***= 13 – 9(1) = 4.**