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Date	May 2018	Marks available	2	Reference code	18M.2.SL.TZ1.T_1
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Calculate	Question number	T_1	Adapted from	N/A

Question

Farmer Brown has built a new barn, on horizontal ground, on his farm. The barn has a cuboid base and a triangular prism roof, as shown in the diagram.

The cuboid has a width of 10 m, a length of 16 m and a height of 5 m.
The roof has two sloping faces and two vertical and identical sides, ADE and GLF.
The face DEFL slopes at an angle of 15° to the horizontal and ED = 7 m .

The roof was built using metal supports. Each support is made from five lengths of metal AE, ED, AD, EM and MN, and the design is shown in the following diagram.

ED = 7 m , AD = 10 m and angle ADE = 15° .
M is the midpoint of AD.
N is the point on ED such that MN is at right angles to ED.

Farmer Brown believes that N is the midpoint of ED.

Calculate the area of triangle EAD.

[3]

Calculate the total volume of the barn.

[3]

Calculate the length of MN.

[2]

Calculate the length of AE.

[3]

Show that Farmer Brown is incorrect.

[3]

Calculate the total length of metal required for one support.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(Area of EAD =) $\frac{1}{2} \times 10 \times 7 \times {\text{sin}}15$ (M1)(A1)

Note: Award (M1) for substitution into area of a triangle formula, (A1) for correct substitution. Award (M0)(A0)(A0) if EAD or AED is considered to be a right-angled triangle.

= 9.06 m² (9.05866… m²) (A1) (G3)

[3 marks]

(10 × 5 × 16) + (9.05866… × 16) (M1)(M1)

Note: Award (M1) for correct substitution into volume of a cuboid, (M1) for adding the correctly substituted volume of their triangular prism.

= 945 m³ (944.938… m³) (A1)(ft) (G3)

Note: Follow through from part (a).

[3 marks]

$\frac{{{\text{MN}}}}{5} = \,\,\,{\text{sin}}15$ (M1)

Note: Award (M1) for correct substitution into trigonometric equation.

(MN =) 1.29(m) (1.29409… (m)) (A1) (G2)

[2 marks]

(AE² =) 10² + 7² − 2 × 10 × 7 × cos 15 (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, and (A1) for correct substitution.

(AE =) 3.71(m) (3.71084… (m)) (A1) (G2)

[3 marks]

ND² = 5² − (1.29409…)² (M1)

Note: Award (M1) for correct substitution into Pythagoras theorem.

(ND =) 4.83 (4.82962…) (A1)(ft)

Note: Follow through from part (c).

$\frac{{1.29409 \ldots }}{{{\text{ND}}}} = {\text{tan}}\,15^\circ$ (M1)

Note: Award (M1) for correct substitution into tangent.

(ND =) 4.83 (4.82962…) (A1)(ft)

Note: Follow through from part (c).

$\frac{{{\text{ND}}}}{5} = {\text{cos }}15^\circ$ (M1)

Note: Award (M1) for correct substitution into cosine.

(ND =) 4.83 (4.82962…) (A1)(ft)

Note: Follow through from part (c).

ND² = 1.29409…² + 5² − 2 × 1.29409… × 5 × cos 75° (M1)

Note: Award (M1) for correct substitution into cosine rule.

(ND =) 4.83 (4.82962…) (A1)(ft)

Note: Follow through from part (c).

4.82962… ≠ 3.5 (ND ≠ 3.5) (R1)(ft)

4.82962… ≠ 2.17038… (ND ≠ NE) (R1)(ft)

(hence Farmer Brown is incorrect)

Note: Do not award (M0)(A0)(R1)(ft). Award (M0)(A0)(R0) for a correct conclusion without any working seen.

[3 marks]

(EM² =) 1.29409…² + (7 − 4.82962…)² (M1)

Note: Award (M1) for their correct substitution into Pythagoras theorem.

(EM² =) 5² + 7² − 2 × 5 × 7 × cos 15 (M1)

Note: Award (M1) for correct substitution into cosine rule formula.

(EM =) 2.53(m) (2.52689...(m)) (A1)(ft) (G2)(ft)

Note: Follow through from parts (c), (d) and (e).

(Total length =) 2.52689… + 3.71084… + 1.29409… +10 + 7 (M1)

Note: Award (M1) for adding their EM, their parts (c) and (d), and 10 and 7.

= 24.5 (m) (24.5318… (m)) (A1)(ft) (G4)

Note: Follow through from parts (c) and (d).

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.1—3d space, volume, angles, midpoints

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Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig

Topic 3—Geometry and trigonometry

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