The answer
(a) \(20\pi \approx 62.8\) cm
(b) \(\approx 198 \text{ cm}^2\)
O-Level E-Math 2016 Paper 1 Question 16 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 16 of the O-Level E-Math 2016 Paper 1. It tests pythagoras for the rectangle diagonal, in the Pythagoras & circle properties area. It is worth 4 marks: 2 + 2. 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.
Each rhombus vertex is the midpoint of a side of the rectangle, so a side of the rhombus joins the midpoints of two adjacent sides. If the rectangle is \(w\) by \(h\), a rhombus side \(= \tfrac12\sqrt{w^2 + h^2}\). With rhombus side \(10\): \[\tfrac12\sqrt{w^2 + h^2} = 10 \;\Rightarrow\; \sqrt{w^2 + h^2} = 20 \;\Rightarrow\; w^2 + h^2 = 400.\] The quantity \(\sqrt{w^2 + h^2} = 20\) is exactly the rectangle's diagonal, which (as the rectangle is inscribed in the circle) is the diameter of the circle.
(a) Diameter \(= 20\) cm, so circumference \(= \pi \times 20 = 20\pi = 62.83\ldots \approx 62.8\) cm.
(b) The shorter side is \(13\), so the longer side is \(\sqrt{400 - 13^2} = \sqrt{400 - 169} = \sqrt{231} = 15.198\ldots\) cm. Area of rectangle \(= 13 \times \sqrt{231} = 197.58\ldots \approx 198 \text{ cm}^2\) (3 s.f.).
Answer: (a) \(20\pi \approx 62.8\) cm
(b) \(\approx 198 \text{ cm}^2\)
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 pythagoras for the rectangle diagonal question from Pythagoras & circle properties, worth 4 marks: 2 + 2.
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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