The answer
\(\approx 716 \text{ cm}^2\)
O-Level E-Math 2015 Paper 1 Question 19 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 19 of the O-Level E-Math 2015 Paper 1. It tests perpendicular from centre to a chord, in the Mensuration (arc, sector, segment) area. It is worth 5 marks. 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.
\(D\) is the top of the circle and \(B\) lies on chord \(AC\), with \(OB\) perpendicular to \(AC\) (the perpendicular from the centre to a chord). \(D\), \(O\) and \(B\) are on the same vertical line, so \(BD = BO + OD\). Since \(OD\) is a radius, \(OD = 17\), hence \(OB = BD - OD = 25 - 17 = 8\) cm.
The perpendicular distance from \(O\) to the chord is \(8\), so the half-chord is \(AB = \sqrt{OA^2 - OB^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\) cm, and \(AC = 30\) cm.
Find the central angle \(\angle AOC\): in right-angled triangle \(OAB\), \(\cos(\angle AOB) = \dfrac{OB}{OA} = \dfrac{8}{17}\), so \(\angle AOB = 61.93^{\circ}\) and \(\angle AOC = 2 \times 61.93^{\circ} = 123.86^{\circ}\).
The major segment is the part of the circle on the side of chord \(AC\) that contains \(D\) (the larger part, containing \(O\)). Its area \(=\) (area of circle) \(-\) (area of the minor segment below \(AC\)): - Area of circle \(= \pi \times 17^2 = 908.04 \text{ cm}^2\). - Minor sector (angle \(123.86^{\circ}\)) \(= \dfrac{123.86}{360} \times 908.04 = 312.4 \text{ cm}^2\). - Triangle \(OAC\) \(= \dfrac{1}{2} \times AC \times OB = \dfrac{1}{2} \times 30 \times 8 = 120 \text{ cm}^2\). - Minor segment \(= 312.4 - 120 = 192.4 \text{ cm}^2\).
Major segment \(= 908.04 - 192.4 = 715.6 \approx 716 \text{ cm}^2\).
Answer: \(\approx 716 \text{ cm}^2\)
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 perpendicular from centre to a chord question from Mensuration (arc, sector, segment), worth 5 marks.
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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