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Date	November 2017	Marks available	3	Reference code	17N.2.SL.TZ0.T_3
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Calculate	Question number	T_3	Adapted from	N/A

Question

Abdallah owns a plot of land, near the river Nile, in the form of a quadrilateral ABCD.

The lengths of the sides are ${\text{AB}} = {\text{40 m, BC}} = {\text{115 m, CD}} = {\text{60 m, AD}} = {\text{84 m}}$ and angle ${\rm{B\hat AD}} = 90^\circ$ .

This information is shown on the diagram.

N17/5/MATSD/SP2/ENG/TZ0/03

The formula that the ancient Egyptians used to estimate the area of a quadrilateral ABCD is

${\text{area}} = \frac{{({\text{AB}} + {\text{CD}})({\text{AD}} + {\text{BC}})}}{4}$ .

Abdallah uses this formula to estimate the area of his plot of land.

Show that ${\text{BD}} = 93{\text{ m}}$ correct to the nearest metre.

[2]

a.

Calculate angle ${\rm{B\hat CD}}$ .

[3]

b.

Find the area of ABCD.

[4]

c.

Calculate Abdallah’s estimate for the area.

[2]

d.i.

Find the percentage error in Abdallah’s estimate.

[2]

d.ii.

Markscheme

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

${\text{B}}{{\text{D}}^2} = {40^2} + {84^2}$ (M1)

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

Accept correct substitution into cosine rule.

${\text{BD}} = 93.0376 \ldots$ (A1)

$= 93$ (AG)

Note: Both the rounded and unrounded value must be seen for the (A1) to be awarded.

[2 marks]

a.

$\cos C = \frac{{{{115}^2} + {{60}^2} - {{93}^2}}}{{2 \times 115 \times 60}}{\text{ }}({93^2} = {115^2} + {60^2} - 2 \times 115 \times 60 \times \cos C)$ (M1)(A1)

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

$= 53.7^\circ {\text{ }}(53.6679 \ldots ^\circ )$ (A1)(G2)

[3 marks]

b.

$\frac{1}{2}(40)(84) + \frac{1}{2}(115)(60)\sin (53.6679 \ldots )$ (M1)(M1)(A1)(ft)

Note: Award (M1) for correct substitution into right-angle triangle area. Award (M1) for substitution into area of triangle formula and (A1)(ft) for correct substitution.

$= 4460{\text{ }}{{\text{m}}^2}{\text{ }}(4459.30 \ldots {\text{ }}{{\text{m}}^2})$ (A1)(ft)(G3)

Notes: Follow through from part (b).

[4 marks]

c.

$\frac{{(40 + 60)(84 + 115)}}{4}$ (M1)

Note: Award (M1) for correct substitution in the area formula used by ‘Ancient Egyptians’.

$= 4980{\text{ }}{{\text{m}}^2}{\text{ }}(4975{\text{ }}{{\text{m}}^2})$ (A1)(G2)

[2 marks]

d.i.

$\left| {\frac{{4975 - 4459.30 \ldots }}{{4459.30 \ldots }}} \right| \times 100$ (M1)

Notes: Award (M1) for correct substitution into percentage error formula.

$= 11.6{\text{ }}(\% ){\text{ }}(11.5645 \ldots )$ (A1)(ft)(G2)

Notes: Follow through from parts (c) and (d)(i).

[2 marks]

d.ii.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.i.

[N/A]

d.ii.

Syllabus sections

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

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