The answer
(a)(i) angle \(ABD = 67^{\circ}\)
(ii) angle \(EAD = 25^{\circ}\)
(b)(i) angle \(POQ = 1.8\) rad
(ii) shaded area \(\approx 59.2\) cm²
O-Level E-Math 2019 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 2019 Paper 2. It tests isosceles triangle from radii, in the Circle properties & mensuration area. It is worth 11 marks: 3 + 3 + 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)(i) \(OA = OD\) (radii), so triangle \(OAD\) is isosceles and \(\angle ODA = \angle OAD = 23^{\circ}\). Hence the angle at the centre \(\angle AOD = 180^{\circ} - 23^{\circ} - 23^{\circ} = 134^{\circ}\) (angle sum of a triangle). Angle \(ABD\) stands on the same arc \(AD\) as the central angle \(AOD\), so by the angle-at-the-centre property, \(\angle ABD = \dfrac{1}{2}\angle AOD = \dfrac{1}{2}(134^{\circ}) = 67^{\circ}\).
(ii) The inscribed angle \(\angle ADE = 42^{\circ}\) stands on arc \(AE\), so the central angle for arc \(AE\) is \(\angle AOE = 2(42^{\circ}) = 84^{\circ}\). From part (i), arc \(AD\) has central angle \(134^{\circ}\), and \(E\) lies on arc \(AD\) (between \(A\) and \(D\)), so the central angle for arc \(ED\) is \(134^{\circ} - 84^{\circ} = 50^{\circ}\). Angle \(EAD\) is the inscribed angle standing on arc \(ED\), so \(\angle EAD = \dfrac{1}{2}(50^{\circ}) = 25^{\circ}\).
(b)(i) The minor sector \(OPQ\) has radius \(OP = OQ = 4\) cm and angle \(\theta = \angle POQ\). Its perimeter is the two radii plus the arc: \(4 + 4 + 4\theta = 15.2\). So \(8 + 4\theta = 15.2 \Rightarrow 4\theta = 7.2 \Rightarrow \theta = 1.8\) radians.
(ii) The shaded region is the whole ring (annulus) between the two circles, minus the annular slice lying between the lines \(OR\) and \(OS\), plus the inner sector \(OPQ\). - Ring (annulus) area \(= \pi(6^2 - 4^2) = \pi(36 - 16) = 20\pi = 62.83\) cm². - Annular slice between \(OR\) and \(OS\) (angle 1.8 rad) \(= \tfrac12(6^2)(1.8) - \tfrac12(4^2)(1.8) = 32.4 - 14.4 = 18\) cm². - Inner sector \(OPQ\) \(= \tfrac12(4^2)(1.8) = 14.4\) cm².
Total shaded \(= 62.83 - 18 + 14.4 = 59.23 \approx 59.2\) cm².
Answer: (a)(i) angle \(ABD = 67^{\circ}\)
(ii) angle \(EAD = 25^{\circ}\)
(b)(i) angle \(POQ = 1.8\) rad
(ii) shaded area \(\approx 59.2\) cm²
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 isosceles triangle from radii question from Circle properties & mensuration, worth 11 marks: 3 + 3 + 2 + 3.
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