The answer
(a) \(AE = \sqrt{512.5} = 22.6\) cm (shown)
(b) \(70.7^{\circ}\)
(c) \(\approx 866 \text{ cm}^2\)
(d) \(144 \text{ cm}^2\)
O-Level E-Math 2018 Paper 2 Question 8 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 8 of the O-Level E-Math 2018 Paper 2. It tests slant edge via 3d pythagoras, in the Mensuration (pyramids, similar solids) area. It is worth 11 marks: (a) 3, (b) 3, (c) 3, (d) 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.
(a) The centre of the base is the midpoint of each diagonal; the diagonal of the square is \(15\sqrt2\) cm, so the distance from the centre to a corner is \(\dfrac{15\sqrt2}{2} = 10.607\) cm. The slant edge \(AE\) is the hypotenuse of the right triangle with legs (height) 20 and (centre-to-corner) \(\dfrac{15\sqrt2}{2}\): \[AE = \sqrt{20^2 + \left(\tfrac{15\sqrt2}{2}\right)^2} = \sqrt{400 + 112.5} = \sqrt{512.5} = 22.638 = 22.6 \text{ cm (3 s.f.).}\]
(b) Angle \(BAE\) is the angle at \(A\) in triangle \(ABE\). By symmetry \(AE = BE = 22.638\), so the triangle is isosceles; dropping a perpendicular from \(E\) to the midpoint of \(AB\) (half of \(AB\) = 7.5) gives \(\cos BAE = \dfrac{7.5}{22.638} = 0.3313\), so \(BAE = 70.7^{\circ}\) (1 d.p.).
(c) Total surface area \(=\) square base \(+\) four triangular faces. Base \(= 15^2 = 225 \text{ cm}^2\). The slant height of a face (from \(E\) to the midpoint of a base edge) is \(l = \sqrt{20^2 + 7.5^2} = \sqrt{456.25} = 21.36\) cm. Each face area \(= \dfrac12 \times 15 \times 21.36 = 160.2 \text{ cm}^2\), so four faces \(= 640.8 \text{ cm}^2\). Total \(= 225 + 640.8 = 866 \text{ cm}^2\) (3 s.f.).
(d) The frustum is 4 cm tall, so the removed top pyramid has height \(20 - 4 = 16\) cm. It is similar to the original (height 20, base side 15), with linear scale factor \(\dfrac{16}{20} = \dfrac45\). Its base, which is the top \(PQRS\) of the frustum, has side \(15 \times \dfrac45 = 12\) cm, so area \(PQRS = 12^2 = 144 \text{ cm}^2\).
Answer: (a) \(AE = \sqrt{512.5} = 22.6\) cm (shown)
(b) \(70.7^{\circ}\)
(c) \(\approx 866 \text{ cm}^2\)
(d) \(144 \text{ cm}^2\)
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 slant edge via 3d pythagoras question from Mensuration (pyramids, similar solids), worth 11 marks: (a) 3, (b) 3, (c) 3, (d) 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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