2025 NASS P2 Q8

2025 NASS P2 Q8

Secondary 4
11 marks

The diagram shows the top view of a garden layout, where \(FC\) is a pathway that cuts across the garden. It is given that angle \(AFE=\theta\) and angle \(ABC=\text{angle}\hspace{0.5em}FAD=\text{angle}\hspace{0.5em}EDC=90^\circ\).

\(CF\) intersects \(AD\) at \(E\), with \(CE=4\text{ cm}\) and \(FE=6\text{ cm}\). \(ABCE\) is a raised garden bed in the shape of a trapezium and \(0^\circ<\theta<90^\circ\).

  1. Show that the perimeter of \(ABCE\), \(P\text{ cm}\), is given by \(P=4+16\sin\theta+4\cos\theta\).[2]
  2. Express \(P=4+16\sin\theta+4\cos\theta\) in the form \(c+R\sin(\theta+\alpha)\), where \(c\) is a constant, \(R\) is a positive constant and \(\alpha\) is acute.[3]
  3. Find the maximum exact value of \(P\) and the corresponding value of \(\theta\).[3]
  4. Find the value of \(\theta\) when the perimeter of \(ABCE\) is \(15\text{ cm}\).[2]
  5. Without calculating the exact value of \(\theta\), decide whether the perimeter of \(ABCE\) could be \(20\text{ cm}\). Explain your reasoning.[1]

Solution:

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Answer:(b) \(P=4+4\sqrt{17}\sin(\theta+\alpha)\), \(\alpha=\tan^{-1}\dfrac14\approx14.0^\circ\); (c) \(4+4\sqrt{17}\text{ cm}\) at \(76.0^\circ\); (d) \(27.8^\circ\); (e) yes.

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