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Date	May 2022	Marks available	5	Reference code	22M.1.SL.TZ1.1
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	1	Adapted from	N/A

Question

The front view of a doghouse is made up of a square with an isosceles triangle on top.

The doghouse is $1.35 m$ high and $0.9 m$ wide, and sits on a square base.

The top of the rectangular surfaces of the roof of the doghouse are to be painted.

Find the area to be painted.

Markscheme

height of triangle at roof $= 1.35 - 0.9 = 0.45$ (A1)

Note: Award A1 for $0.45$ (height of triangle) seen on the diagram.

$slant height = \sqrt{0 . 45^{2} + 0 . 45^{2}}$ OR $\sin (45 °) = \frac{0.45}{slant height}$ (M1)

$= \sqrt{0.405} (0.636396 \dots, 0.45 \sqrt{2})$ A1

Note: If using $\sin (45 °) = \frac{0.45}{slant height}$ then (A1) for angle of $45 °$ , (M1) for a correct trig statement.

area of one rectangle on roof $= \sqrt{0.405} \times 0.9 (= 0.572756 \dots)$ M1

area painted $= (2 \times \sqrt{0.405} \times 0.9 = 2 \times 0.572756 \dots)$

$1.15 m^{2} (1.14551 \dots m^{2}, 0.81 \sqrt{2} m^{2})$ A1

[5 marks]

Examiners report

Although the first question on the paper, with appropriate low-level mathematics, the interpretation required seems to have been quite high, and many candidates found this challenging. Many candidates scored only the one mark for the height of the triangle. The most common wrong method seen was calculation of the area of the triangle and adding their result to the calculation 2 × 0.9 × 0.9. Stronger candidates lost the final mark either through premature rounding or incorrect units; both are aspects that can and do occur throughout candidate responses and hence clearly require focus in the classroom.

Syllabus sections

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

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