The answer
(a) shown
(b) \(\approx 310 \text{ cm}^2\)
(c)(i) \(\approx 46\,500 \text{ cm}^3\)
(ii) \(\approx 170\) litres
O-Level E-Math 2017 Paper 2 Question 8 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 8 of the O-Level E-Math 2017 Paper 2. It tests angle at centre via right-angled trig, in the Mensuration (circle segment, prism volume) area. It is worth 11 marks: 2 + 4 + 2 + 3. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
(a) Since \(M\) is the midpoint of chord \(AB\), \(OM \perp AB\), so triangle \(OMA\) is right-angled at \(M\) with \(OA = 30\), \(OM = 20\). Then \(\cos \angle AOM = \dfrac{OM}{OA} = \dfrac{20}{30} = \dfrac{2}{3}\), so \(\angle AOM = 48.19^{\circ}\). By symmetry \(\angle AOB = 2 \times 48.19^{\circ} = 96.38^{\circ} = 96.4^{\circ}\) (3 s.f.). (shown)
(b) The shaded region is the minor segment cut off by chord \(AB\): segment \(=\) sector \(AOB\) \(-\) triangle \(AOB\). Sector area \(= \dfrac{96.38^{\circ}}{360^{\circ}} \times \pi (30)^2 = \dfrac{96.38}{360} \times 900\pi = 757.0 \text{ cm}^2\). Triangle area \(= \dfrac{1}{2}(30)(30)\sin 96.38^{\circ} = 450 \times 0.99381 = 447.2 \text{ cm}^2\). Shaded (segment) area \(= 757.0 - 447.2 = 309.8 \approx 310 \text{ cm}^2\) (3 s.f.).
(c)(i) Length of trough \(= 1.5\) m \(= 150\) cm. Volume of water \(= \) shaded area \(\times\) length \(= 309.8 \times 150 = 46\,463 \approx 46\,500 \text{ cm}^3\) (3 s.f.).
(ii) The full trough (the whole semicircle as cross-section) has volume \(= \dfrac{1}{2}\pi (30)^2 \times 150 = 450\pi \times 150 = 67\,500\pi = 212\,058 \text{ cm}^3\). Water still needed \(= 212\,058 - 46\,463 = 165\,595 \text{ cm}^3 = 165.6\) litres. To the nearest 10 litres, about \(170\) litres must be added.
Answer: (a) shown
(b) \(\approx 310 \text{ cm}^2\)
(c)(i) \(\approx 46\,500 \text{ cm}^3\)
(ii) \(\approx 170\) litres
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a angle at centre via right-angled trig question from Mensuration (circle segment, prism volume), worth 11 marks: 2 + 4 + 2 + 3.
Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.
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