Write down the length of ${\text{AO}}$.

Find the size of the angle that the sloping edge ${\text{VA}}$ makes with the base of the pyramid.

Hence, or otherwise, find the area of the triangle ${\text{CAV}}$.

${\text{AO}} = 4{\text{ (cm)}}$     (A1)     (C1)

$\cos {\rm{O\hat AV}} = \frac{4}{{10}}$     (M1)

Note: Award (M1) for their correct trigonometric ratio.

OR

$\cos {\rm{O\hat AV}} = \frac{{{{10}^2} + {8^2} - {{10}^2}}}{{2 \times 10 \times 8}}\;\;\;$OR$\;\;\;\frac{{{{10}^2} + {4^2} - {{(9.16515 \ldots )}^2}}}{{2 \times 10 \times 4}}$     (M1)

Note: Award (M1) for correct substitution into the cosine rule formula.

${\rm{O\hat AV}} = 66.4^\circ \;\;\;(66.4218 \ldots )$     (A1)(ft)     (C2)

Notes: Follow through from their answer to part (a).

${\text{area}} = \frac{{8 \times 10 \times \sin (66.4218 \ldots ^\circ)}}{2}\;\;\;$OR$\;\;\;\frac{1}{2} \times 8 \times \sqrt {{{10}^2} - {4^{\text{2}}}}$

OR$\;\;\;\frac{1}{2} \times 10 \times 10 \times \sin (47.1563 \ldots ^\circ )$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitutions. Follow through from their answer to part (b) and/or part (a).

${\text{area}} = 36.7{\text{ c}}{{\text{m}}^2}\;\;\;(36.6606 \ldots {\text{ c}}{{\text{m}}^2})$     (A1)(ft)     (C3)

Notes: Accept an answer of $8\sqrt {21} {\text{ c}}{{\text{m}}^2}$ which is the exact answer.