SPM Mathematics 2019, Paper 2 (Question 11)

Question 11:
Diagram 6 shows a semicircle landscape garden built by Kamal at his house. The garden consists of a herbal garden and a quadrant fish pound. He plans to fence the herbal garden.

Diagram 6

POKQ is a straight line and K is the midpoint of OQ. It is given that PQ = 14 m and OP : OK = 2:1. Calculate
(a) the total length, in m, of the fence for the herbal garden, [3 marks]
(b) the area, in m², of the herbal garden. [3 marks]

Solution:
(a)

$\begin{array}{l} Length of arc P Q = \frac{180^{o}}{360^{o}} \times 2 π r \\ = \frac{180^{o}}{360^{o}} \times 2 \times \frac{22}{7} \times 7 \\ = 22 m \\ Length of arc L K = \frac{90^{o}}{360^{o}} \times 2 \times \frac{22}{7} \times 3.5 \\ = 5.5 m \\ Total length of fence = (22 + 3.5 + 5.5 + 3.5 + 7) m \\ = 41.5 m \end{array}$

(b)

$\begin{array}{l} Total area of semicircle = \frac{1}{2} π r^{2} \\ = \frac{1}{2} \times \frac{22}{7} \times 7^{2} \\ = 77 m^{2} \\ Total area of quadrant of fish pond \\ = \frac{1}{4} π r^{2} \\ = \frac{1}{4} \times \frac{22}{7} \times {3.5}^{2} \\ = 9.625 m^{2} \\ Area of herbal garden = 77 m^{2} - 9.625 m^{2} \\ = 67.375 m^{2} \end{array}$

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