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.