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Date November 2007 Marks available 2 Reference code 07N.2.sl.TZ0.2
Level SL only Paper 2 Time zone TZ0
Command term Calculate Question number 2 Adapted from N/A

Question

ABCDV is a solid glass pyramid. The base of the pyramid is a square of side 3.2 cm. The vertical height is 2.8 cm. The vertex V is directly above the centre O of the base.

Calculate the volume of the pyramid.

[2]
a.

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

[2]
b.

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

[4]
c.

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

[3]
d.

Calculate the total surface area of the pyramid.

[4]
e.

Markscheme

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]

a.

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]

b.

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

c.

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

d.

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]

e.

Examiners report

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.

a.

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.

b.

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.

c.

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.

d.

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.

e.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.5 » Volume and surface areas of the three-dimensional solids defined in 5.4.
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