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O-Level E-Math · 2016 · P1 Q24 Mensuration (areas of circle parts) · Semicircle area 5 marks · geometry & measurement (areas of circle parts in a square) difficulty 5 of 5

O-Level E-Math 2016 Paper 1, Question 24: Semicircle area

The answer

\(\dfrac{3}{4}\)

O-Level E-Math 2016 Paper 1 Question 24 · Verified worked solution by the Genius Plus Academy teaching team

What this question tests

This is Question 24 of the O-Level E-Math 2016 Paper 1. It tests semicircle area, in the Mensuration (areas of circle parts) 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.

Step-by-step solution

Take \(A(0,0)\), \(D(2r,0)\), \(C(2r,2r)\), \(B(0,2r)\), so the side is \(2r\) and the centre is \(O(r,r)\). The midpoints are \(E(r,0)\) (bottom), \(F(0,r)\) (left), \(G(r,2r)\) (top) and \(H(2r,r)\) (right). The chord \(EG\) is the vertical segment \(x = r\) joining \(E\) and \(G\).

**Semicircle \(EFG\), centre \(O\):** \(OE = OF = OG = r\), so this is a semicircle of radius \(r\) on diameter \(EG\), bulging left through \(F\). Its area is \(\tfrac12 \pi r^2\).

**Arc \(EG\), centre \(H\):** \(HE = HG = \sqrt{(2r-r)^2 + r^2} = \sqrt{2}\,r\) (given), and the angle \(EHG\) has \(\overrightarrow{HE} = (-r,-r)\), \(\overrightarrow{HG} = (-r, r)\), which are perpendicular, so angle \(EHG = 90^{\circ}\). The region between chord \(EG\) and this arc (the bulge to the left of \(EG\)) is a circular segment \(= \text{sector} - \text{triangle} = \tfrac14 \pi (\sqrt2 r)^2 - \tfrac12(\sqrt2 r)(\sqrt2 r) = \tfrac12\pi r^2 - r^2\).

Both bulges sit to the left of the chord \(EG\); the semicircle reaches further left (to \(F\)) than the \(H\)-arc, so the shaded crescent between them is \[\big(\tfrac12\pi r^2\big) - \big(\tfrac12\pi r^2 - r^2\big) = r^2.\] The square has area \((2r)^2 = 4r^2\), so the unshaded area is \(4r^2 - r^2 = 3r^2\), and the fraction not shaded is \(\dfrac{3r^2}{4r^2} = \dfrac{3}{4}\).

Answer: \(\dfrac{3}{4}\)

Same structure, different numbers

A question is hard because of its structure, not its surface.

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

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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.

Questions students ask

What does O-Level E-Math 2016 Paper 1 Question 24 test?

It is a semicircle area question from Mensuration (areas of circle parts), worth 5 marks.

Is this the same as IP Math?

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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