The answer
(a) \(2100\) m³
(b) \(\approx 420\) m²
(c) \(\approx 13.6^{\circ}\)
O-Level E-Math 2019 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 2019 Paper 2. It tests volume of a prism with a composite cross-section, in the Mensuration & 3D trigonometry area. It is worth 11 marks: 3 + 4 + 4. 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.
Set up the cross-section (the front pentagon). The base \(BD = BC + CD = 6 + 10 = 16\) m. The walls \(AB\) and \(ED\) are 4 m, so the eaves \(A\) and \(E\) are 4 m above the base. The ridge \(F\) is directly above \(C\) at height \(CG + FG = 5 + 1.5 = 6.5\) m, so \(F\) is \(6.5 - 4 = 2.5\) m above the eaves.
(a) Cross-sectional area = rectangle (walls) + roof triangle. - Rectangle \(ABDE\): \(16 \times 4 = 64\) m². - Roof triangle on top of \(AE\) (base 16 m, height 2.5 m above the eaves): \(\tfrac12 (16)(2.5) = 20\) m².
Cross-section \(= 64 + 20 = 84\) m². Volume \(= 84 \times 25 = 2100\) m³.
(b) Each sloping roof is a rectangle (slant length \(\times\) barn length 25). - Left slope \(AF\): horizontal run \(BC = 6\) m, rise \(2.5\) m, so \(AF = \sqrt{6^2 + 2.5^2} = \sqrt{42.25} = 6.5\) m. - Right slope \(FE\): horizontal run \(CD = 10\) m, rise \(2.5\) m, so \(FE = \sqrt{10^2 + 2.5^2} = \sqrt{106.25} = 10.308\) m.
Total roof area \(= (6.5 + 10.308) \times 25 = 16.808 \times 25 = 420.2 \approx 420\) m².
(c) \(P\) is the ridge apex at the back of the barn, 6.5 m above the ground and directly above the back-end position of \(C\). \(D\) is a front base corner. The horizontal distance from \(D\) to the point on the ground below \(P\) has two parts: along the length of the barn (25 m) and across the cross-section from \(D\) to \(C\) (\(CD = 10\) m). So horizontal distance \(= \sqrt{25^2 + 10^2} = \sqrt{725} = 26.926\) m. The vertical height of \(P\) above \(D\) is \(6.5\) m. Hence the angle of elevation is \[\tan^{-1}\!\left(\frac{6.5}{26.926}\right) = \tan^{-1}(0.2414) = 13.57° \approx 13.6°.\]
Answer: (a) \(2100\) m³
(b) \(\approx 420\) m²
(c) \(\approx 13.6^{\circ}\)
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 volume of a prism with a composite cross-section question from Mensuration & 3D trigonometry, worth 11 marks: 3 + 4 + 4.
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