The answer
(a)(i) congruent by SAS
(a)(ii)(a) \(\angle OBC = (90 - x)^{\circ}\)
(a)(ii)(b) \(\angle DPE = 2x^{\circ}\)
(b)(i) \(\approx 53.8\) cm
(b)(ii) \(\approx 13.1\%\)
O-Level E-Math 2020 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 2020 Paper 2. It tests congruent triangles (sas), in the Circle geometry & radian measure area. It is worth 11 marks: (a)(i) 3, (a)(ii)(a) 1, (a)(ii)(b) 1, (b)(i) 2, (b)(ii) 4. 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)(i) \(FB = FC\) (tangents from \(F\) to circle \(O\) are equal); \(FD = FE\) (tangents from \(F\) to circle \(P\) are equal); \(\angle BFD = \angle CFE\) (vertically opposite angles). Hence \(\triangle BDF \cong \triangle CEF\) (SAS).
(ii)(a) \(OA = OB\) (radii), so \(\angle OBA = \angle OAB = x^{\circ}\) and \(\angle AOB = 180^{\circ} - 2x\). As \(AOC\) is a straight line, \(\angle BOC = 2x\). Triangle \(OBC\) is isosceles (\(OB = OC\)), so \(\angle OBC = \tfrac{180^{\circ} - 2x}{2} = (90 - x)^{\circ}\).
(ii)(b) In kite \(OBFC\) the radii meet the tangents at right angles, so \(\angle BFC = 180^{\circ} - \angle BOC = 180^{\circ} - 2x\). The vertically opposite angle \(\angle DFE = 180^{\circ} - 2x\), and in kite \(PDFE\), \(\angle DPE = 180^{\circ} - \angle DFE = 2x^{\circ}\). (Equivalently, the two central angles subtended by the tangent points are equal: \(\angle DPE = \angle BOC = 2x\).) *(The publisher key prints \(180 - 2x\); a concrete two-circle construction confirms \(\angle DPE = \angle BOC = 2x\).)*
(b)(i) Major arc \(= r(2\pi - 1.8) = 12(2\pi - 1.8) = 53.8\) cm.
(ii) Shaded minor segment \(= \tfrac12 r^2(\theta - \sin\theta) = \tfrac12 (12^2)(1.8 - \sin 1.8) = 59.42\) cm². Circle area \(= \pi(12^2) = 452.39\) cm². Percentage \(= \dfrac{59.42}{452.39} \times 100 = 13.1\%\).
Answer: (a)(i) congruent by SAS
(a)(ii)(a) \(\angle OBC = (90 - x)^{\circ}\)
(a)(ii)(b) \(\angle DPE = 2x^{\circ}\)
(b)(i) \(\approx 53.8\) cm
(b)(ii) \(\approx 13.1\%\)
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 congruent triangles (sas) question from Circle geometry & radian measure, worth 11 marks: (a)(i) 3, (a)(ii)(a) 1, (a)(ii)(b) 1, (b)(i) 2, (b)(ii) 4.
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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