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

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

Find the area of ABCD.

Calculate Abdallah’s estimate for the area.

Find the percentage error in Abdallah’s estimate.

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

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

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

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

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