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Date	May 2019	Marks available	2	Reference code	19M.2.SL.TZ1.T_2
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Justify and State	Question number	T_2	Adapted from	N/A

Question

Olivia’s house consists of four vertical walls and a sloping roof made from two rectangles. The height, ${\text{CD}}$ , from the ground to the base of the roof is 4.5 m.

The base angles of the roof are ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}} = 27^\circ$ and ${\text{A}}\mathop {\text{C}}\limits^ \wedge {\text{B}} = 26^\circ$ .

The house is 10 m long and 5 m wide.

The length ${\text{AC}}$ is approximately 2.84 m.

Olivia decides to put solar panels on the roof. The solar panels are fitted to both sides of the roof.

Each panel is 1.6 m long and 0.95 m wide. All the panels must be arranged in uniform rows, with the shorter edge of each panel parallel to ${\text{AB}}$ or ${\text{AC}}$ . Each panel must be at least 0.3 m from the edge of the roof and the top of the roof, ${\text{AF}}$ .

Olivia estimates that the solar panels will cover an area of 29 m².

Find the length ${\text{AB}}$ , giving your answer to four significant figures.

[5]

Find the total area of the two rectangles that make up the roof.

[3]

Find the maximum number of complete panels that can be fitted to the whole roof.

[3]

Find the percentage error in her estimate.

[3]

Olivia investigates arranging the panels, such that the longer edge of each panel is parallel to ${\text{AB}}$ or ${\text{AC}}$ .

State whether this new arrangement will allow Olivia to fit more solar panels to the roof. Justify your answer.

[2]

Markscheme

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

180° − 27° − 26° (M1)

Note: Award (M1) for correct working to find angle ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ or 127 seen.

$\frac{{{\text{AB}}}}{{{\text{sin}}\,26^\circ }} = \frac{5}{{{\text{sin}}\,127^\circ }}$ (M1)(A1)

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

2.74450 (m) (A1)

( ${\text{AB}}$ =) 2.745 (m) (A1)(ft)(G4)

Note: The final (A1)(ft) is for correctly rounding their unrounded ${\text{AB}}$ to 4 sf. If 2.745 is given as the final answer, the unrounded answer need not be seen, award (M1)(M1)(A1)(A2). For all other answers, the unrounded answer must be seen to an accuracy greater than 4 sf.

Award (G3) for a final answer of 2.74450…(m) with no working. If radians are used then award at most (M1)(M1)(A1)(A0)(A1)(ft) for an answer of 3.920 (m).

[5 marks]

Units are required in this question part.

10 × 2.84 + 10 × 2.74450… (M1)(M1)

Note: Award (M1) for finding their area of each rectangle and (M1) for adding their areas.

10 × (2.84 + 2.74450…) (M1)(M1)

Note: Award (M1) for adding ${\text{AC}}$ and their ${\text{AB}}$ . Award (M1) for multiplying their total area by 10.

55.8 (55.8450…) m² (A1)(ft)(G3)

Note: Follow through from their ${\text{AB}}$ in part (a).

[3 marks]

$\frac{{10 - 2\left( {0.3} \right)}}{{1.6}}$ (M1)

Note: Award (M1) for correct calculation of the number of panels on the long side.

$\frac{{2.745 - 2\left( {0.3} \right)}}{{0.95}}$ OR $\frac{{2.84 - 2\left( {0.3} \right)}}{{0.95}}$ (M1)

Note: Award (M1) for correct calculation of the number of panels on either short side with no further incorrect working.

20 (A1)(ft)(G2)

Note: Follow through from part (a). Do not award (M0)(M1)(A1)(ft).

[3 marks]

20 × 1.6 × 0.95 (= 30.4) (M1)

Note: Award (M1) for their 20 × 1.6 × 0.95 or 30.4 seen. Follow through from their 20 in part (c). Award (M0) if their 20 is not an integer.

$\left| {\frac{{29 - 30.4}}{{30.4}}} \right| \times 100{\text{% }}$ (M1)

Note: Award (M1) for correct substitution of their 30.4 into the percentage error formula. Their 30.4 must be exact.

found. Accept a method in two steps where “×100” is implicit from their answer.

The second (M1) is contingent on the first (M1) being awarded, eg do not award (M0)(M1)(A0).

4.61 (%) (4.60526 (%)) (A1)(ft)(G3)

Note: Follow through from their answer to part (c). Percentage sign is not required.

Award (G2) for an unsupported final answer of 4.61.

[3 marks]

1 × 9 (array) OR 18 (total panels) (R1)(ft)

Note: Award (R1) for one correct array seen (1 × 9) or total number of panels (18). Working is not required, but award (R0) for incorrect working seen. Correct working is as follows. $\left( {\frac{{10 - 0.6}}{{0.95}}{\text{,}}\,\,\frac{{2.84 - 0.6}}{{1.6}}{\text{,}}\,\,\frac{{2.745 - 0.6}}{{1.6}}} \right)$
Reasoning may compare both sides of the roof or just one side; accept correct comparisons with part (c) values. Follow through from their treatment of tolerances in part (c) and maximum number of panels.
Award (R0) for any approach with no clearance or for any method which includes further incorrect working.

No (new arrangement will mean fewer solar panels) (A1)(ft)

Note: Follow through from their maximum number of panels in part (c). Do not award (R0)(A1)(ft).

[2 marks]

Examiners report

[N/A]

Syllabus sections

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

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Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area

Topic 3— Geometry and trigonometry

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