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Date	November 2016	Marks available	2	Reference code	16N.2.SL.TZ0.T_5
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Calculate	Question number	T_5	Adapted from	N/A

Question

A farmer owns a plot of land in the shape of a quadrilateral ABCD.

${\text{AB}} = 105{\text{ m, BC}} = 95{\text{ m, CD}} = 40{\text{ m, DA}} = 70{\text{ m}}$ and angle ${\text{DCB}} = 90^\circ$ .

N16/5/MATSD/SP2/ENG/TZ0/05

The farmer wants to divide the land into two equal areas. He builds a fence in a straight line from point B to point P on AD, so that the area of PAB is equal to the area of PBCD.

Calculate

the length of BD;

[2]

the size of angle DAB;

[3]

the area of triangle ABD;

[3]

the area of quadrilateral ABCD;

[2]

the length of AP;

[3]

the length of the fence, BP.

[3]

Markscheme

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

$({\text{BD}} = ){\text{ }}\sqrt {{{95}^2} + {{40}^2}}$ (M1)

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

$= 103{\text{ }}({\text{m}}){\text{ }}\left( {103.077 \ldots ,{\text{ }}25\sqrt {17} } \right)$ (A1)(G2)

[2 marks]

$\cos {\rm{B\hat AD}} = \frac{{{{105}^2} + {{70}^2} - {{(103.077 \ldots )}^2}}}{{2 \times 105 \times 70}}$ (M1)(A1)(ft)

Note: Award (M1) for substitution into cosine rule, (A1)(ft) for their correct substitutions. Follow through from part (a).

$({\rm{B\hat AD}}) = 68.9^\circ {\text{ }}(68.8663 \ldots )$ (A1)(ft)(G2)

Note: If their 103 used, the answer is $68.7995 \ldots$

[3 marks]

$({\text{Area of ABD}} = )\frac{1}{2} \times 105 \times 70 \times \sin (68.8663 \ldots )$ (M1)(A1)(ft)

Notes: Award (M1) for substitution into the trig form of the area of a triangle formula.

Award (A1)(ft) for their correct substitutions.

Follow through from part (b).

If 68.8° is used the area $= 3426.28 \ldots {\text{ }}{{\text{m}}^2}$ .

$= 3430{\text{ }}{{\text{m}}^2}{\text{ }}(3427.82 \ldots )$ (A1)(ft)(G2)

[3 marks]

${\text{area of ABCD}} = \frac{1}{2} \times 40 \times 95 + 3427.82 \ldots$ (M1)

Note: Award (M1) for correctly substituted area of triangle formula added to their answer to part (c).

$= 5330{\text{ }}{{\text{m}}^2}{\text{ }}(5327.83 \ldots )$ (A1)(ft)(G2)

[2 marks]

$\frac{1}{2} \times 105 \times {\text{AP}} \times \sin (68.8663 \ldots ) = 0.5 \times 5327.82 \ldots$ (M1)(M1)

Notes: Award (M1) for the correct substitution into triangle formula.

Award (M1) for equating their triangle area to half their part (d).

$({\text{AP}} = ){\text{ }}54.4{\text{ }}({\text{m}}){\text{ }}(54.4000 \ldots )$ (A1)(ft)(G2)

Notes: Follow through from parts (b) and (d).

[3 marks]

${\text{B}}{{\text{P}}^2} = {105^2} + {(54.4000 \ldots )^2} - 2 \times 105 \times (54.4000 \ldots ) \times \cos (68.8663 \ldots )$ (M1)(A1)(ft)

Notes: Award (M1) for substituted cosine rule formula.

Award (A1)(ft) for their correct substitutions. Accept the exact fraction $\frac{{53}}{{147}}$ in place of $\cos (68.8663 \ldots )$ .

Follow through from parts (b) and (e).

$({\text{BP}} = ){\text{ }}99.3{\text{ }}({\text{m}}){\text{ }}(99.3252 \ldots )$ (A1)(ft)(G2)

Notes: If 54.4 and $\cos (68.9)$ are used the answer is $99.3567 \ldots$

[3 marks]

Examiners report

[N/A]

Syllabus sections

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

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Topic 3—Geometry and trigonometry

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