Calculate the volume of the pyramid.

The glass weighs 9.3 grams per cm³. Calculate the weight of the pyramid.

Show that the length of the sloping edge VC of the pyramid is 3.6 cm.

Calculate the angle at the vertex, \({\text{B}}{\operatorname {\hat V}}{\text{C}}\).

Calculate the total surface area of the pyramid.

Unit penalty (UP) is applicable in question parts (a), (b) and (e) only.

\({\text{V}} = \frac{1}{3} \times {3.2^2} \times 2.8\)     (M1)

(M1) for substituting in correct formula

(UP) = 9.56 cm³     (A1)(G2)

[2 marks]

Unit penalty (UP) is applicable in question parts (a), (b) and (e) only.

\(9.56 \times 9.3\)     (M1)

(UP) = 88.9 grams     (A1)(ft)(G2)

[2 marks]

\(\frac{1}{2} {\text{base}} = 1.6 {\text{ seen}}\)     (M1)

award (M1) for halving base

\({\text{OC}}^2 = 1.6^2 + 1.6^2 = 5.12\)     (A1)

award (A1) for one correct use of Pythagoras

\(5.12 + 2.8^2 = 12.96 = {\text{VC}}^2\)     (M1)

award (M1) for using Pythagoras again to find VC²

VC = 3.6 AG

award (A1) for 3.6 obtained from 12.96 only (not 12.95…)     (A1)

OR

\({\text{AC}}^2 = 3.2^2 + 3.2^2 = 20.48\)     (A1)

award (A1) for one correct use of Pythagoras

({\text{OC}} = \frac{1}{2} \sqrt{20.48}\) ( = 2.26...)     (M1)

award (M1) for halving AC

\(2.8^2 + (2.26...)^2 = {\text{VC}}^2 = 12.96\)     (M1)

award (M1) for using Pythagoras again to find VC²

VC = 3.6 AG     (A1)

award (A1) for 3.6 obtained from 12.96 only (not 12.95…)

[4 marks]

\(3.2^2 = 3.6^2 + 3.6^2 - 2 \times (3.6) (3.6) \cos\) \({\text{B}}{\operatorname {\hat V}}{\text{C}}\)     (M1)(A1)

\({\text{B}}{\operatorname {\hat V}}{\text{C}}\) \( = {52.8^\circ }\) (no (ft) here)     (A1)(G2)

award (M1) for substituting in correct formula, (A1) for correct substitution

OR

\(\sin\) \({\text{B}}{\operatorname {\hat V}}{\text{M}}\) \( = \frac{{1.6}}{{3.6}}\) where M is the midpoint of BC     (M1)(A1)

\({\text{B}}{\operatorname {\hat V}}{\text{C}}\) \( = {52.8^\circ}\) (no (ft) here)     (A1)

[3 marks]

Unit penalty (UP) is applicable in question parts (a), (b) and (e) only.

\(4 \times \frac{1}{2}{(3.6)^2} \times \sin (52.8^\circ ) + {(3.2)^2}\)     (M1)(M1)(M1)

award (M1) for \( \times 4\), (M1) for substitution in relevant triangle area, (\(\frac{1}{2}(3.2)(2.8)\) gets (M0))

(M1) for \(+ {(3.2)^2}\)

(UP) = 30.9 cm² ((ft) from their (d))     (A1)(ft)(G2)

[4 marks]

The volume of the pyramid and the weight were well done. Many candidates lost their unit penalty here. They had trouble showing that the sloping edge was 3.6 cm. The angle BVC was done well but not the total surface area. They knew that they needed four sides and the base, but finding the area of the triangle proved difficult for the less able candidates.

The volume of the pyramid and the weight were well done. Many candidates lost their unit penalty here. They had trouble showing that the sloping edge was 3.6 cm. The angle BVC was done well but not the total surface area. They knew that they needed four sides and the base, but finding the area of the triangle proved difficult for the less able candidates.

The volume of the pyramid and the weight were well done. Many candidates lost their unit penalty here. They had trouble showing that the sloping edge was 3.6 cm. The angle BVC was done well but not the total surface area. They knew that they needed four sides and the base, but finding the area of the triangle proved difficult for the less able candidates.

The volume of the pyramid and the weight were well done. Many candidates lost their unit penalty here. They had trouble showing that the sloping edge was 3.6 cm. The angle BVC was done well but not the total surface area. They knew that they needed four sides and the base, but finding the area of the triangle proved difficult for the less able candidates.

The volume of the pyramid and the weight were well done. Many candidates lost their unit penalty here. They had trouble showing that the sloping edge was 3.6 cm. The angle BVC was done well but not the total surface area. They knew that they needed four sides and the base, but finding the area of the triangle proved difficult for the less able candidates.