The answer
(a) \(42^{\circ}\)
(b) AA similarity
(c)(i) \(AB \approx 2.90\) cm
(ii) arc \(\approx 3.27\) cm
O-Level E-Math 2022 Paper 2 Question 6 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 6 of the O-Level E-Math 2022 Paper 2. It tests angle in a semicircle, in the Circle properties / Pythagoras & trigonometry / Mensuration area. It is worth 12 marks: (a) 2, (b) 3, (c) 4 + 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) \(\angle DPE = \angle CPF = 96^{\circ}\) (vertically opposite angles, since \(DPF\) and \(CPE\) are straight lines through \(P\)). \(\triangle DPE\) is isosceles (\(PD = PE\), radii), so \(\angle DEP = \dfrac{180^{\circ} - 96^{\circ}}{2} = 42^{\circ}\).
(b) \(AC\) is a diameter of the small circle (\(A\), \(O\), \(C\) collinear), so \(\angle ABC = 90^{\circ}\) (angle in a semicircle). \(DF\) is a diameter of the large circle (\(D\), \(P\), \(F\) collinear), so \(\angle FCD = 90^{\circ}\) (angle in a semicircle). Hence \(\angle ABC = \angle FCD\). The circles touch at \(C\) and so share a common tangent there; by the alternate segment theorem applied to chord \(CB\) (small circle) and chord \(CD\) (large circle), and because \(B\), \(C\), \(D\) are collinear, \(\angle BAC = \angle DFC\). With two pairs of equal angles, \(\triangle ABC\) is similar to \(\triangle FCD\) (AA), with \(A \leftrightarrow F\), \(B \leftrightarrow C\), \(C \leftrightarrow D\).
(c)(i) In the large circle, chord \(CF\) subtends the same \(96^{\circ}\) central angle as \(DE\), so \(CF = DE = 7.21\) cm. In right-angled \(\triangle FCD\) (\(\angle FCD = 90^{\circ}\)): \(CD = \sqrt{DF^2 - CF^2} = \sqrt{9.70^2 - 7.21^2} = 6.49\) cm. Then \(BC = BD - CD = 9.10 - 6.49 = 2.61\) cm. By the similar triangles, \(\dfrac{AB}{FC} = \dfrac{BC}{CD}\), so \(AB = 7.21 \times \dfrac{2.61}{6.49} = 2.90\) cm.
(ii) \(AC = \sqrt{AB^2 + BC^2} = \sqrt{2.90^2 + 2.61^2} = 3.90\) cm, so the small circle's radius is \(1.95\) cm. The central angle \(\angle AOB = 2\angle ACB\) (angle at centre \(= 2 \times\) angle at circumference); \(\angle ACB = 48^{\circ}\), so \(\angle AOB = 96^{\circ}\). Minor arc \(AB = \dfrac{96}{360} \times 2\pi(1.95) = 3.27\) cm.
Answer: (a) \(42^{\circ}\)
(b) AA similarity
(c)(i) \(AB \approx 2.90\) cm
(ii) arc \(\approx 3.27\) cm
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 in a semicircle question from Circle properties / Pythagoras & trigonometry / Mensuration, worth 12 marks: (a) 2, (b) 3, (c) 4 + 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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