| Date | May 2010 | Marks available | 3 | Reference code | 10M.2.sl.TZ2.4 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate | Question number | 4 | Adapted from | N/A |
Question
A gardener has to pave a rectangular area 15.4 metres long and 5.5 metres wide using rectangular bricks. The bricks are 22 cm long and 11 cm wide.
The gardener decides to have a triangular lawn ABC, instead of paving, in the middle of the rectangular area, as shown in the diagram below.
The distance AB is 4 metres, AC is 6 metres and angle BAC is 40°.
In another garden, twelve of the same rectangular bricks are to be used to make an edge around a small garden bed as shown in the diagrams below. FH is the length of a brick and C is the centre of the garden bed. M and N are the midpoints of the long edges of the bricks on opposite sides of the garden bed.
The garden bed has an area of 5419 cm2. It is covered with soil to a depth of 2.5 cm.
It is estimated that 1 kilogram of soil occupies 514 cm3.
Calculate the total area to be paved. Give your answer in cm2.
Write down the area of each brick.
Find how many bricks are required to pave the total area.
Find the length of BC.
Hence write down the perimeter of the triangular lawn.
Calculate the area of the lawn.
Find the percentage of the rectangular area which is to be lawn.
Find the angle FCH.
Calculate the distance MN from one side of the garden bed to the other, passing through C.
Find the volume of soil used.
Find the number of kilograms of soil required for this garden bed.
Markscheme
15.4 × 5.5 (M1)
84.7 m2 (A1)
= 847000 cm2 (A1)(G3)
Note: Award (G2) if 84.7 m2 seen with no working.
OR
1540 × 550 (A1)(M1)
= 847000 cm2 (A1)(ft)(G3)
Note: Award (A1) for both dimensions converted correctly to cm, (M1) for multiplication of both dimensions. (A1)(ft) for correct product of their sides in cm.
[3 marks]
242 cm2 (0.0242 m2) (A1)
[1 marks}
\(\frac {15.4}{0.22} = 70\) (M1)
\(\frac{5.5}{0.11} = 50\)
\(70 \times 50 = 3500\) (A1)(G2)
OR
\(\frac {847000}{242} = 3500\) (M1)(A1)(ft)(G2)
Note: Follow through from parts (a) (i) and (ii).
[2 marks]
\({\text{B}}{{\text{C}}^2} = {4^2} + {6^2}-2 \times 4 \times 6 \times \cos 40^\circ \) (M1)(A1)
\({\text{BC}} = 3.90{\text{ m}}\) (A1)(G2)
Note: Award (M1) for correct substituted formula, (A1) for correct substitutions, (A1) for correct answer.
[3 marks]
perimeter = 13.9 m (A1)(ft)(G1)
Notes: Follow through from part (b) (i).
[1 mark]
\({\text{Area}} = \frac{1}{2} \times 4 \times 6 \times \sin 40^\circ \) (M1)
= 7.71 m2 (A1)(ft)(G2)
Notes: Award (M1) for correct formula and correct substitution, (A1)(ft) for correct answer.
[2 marks]
\(\frac{{7.713}}{{84.7}} \times 100{\text{ }}\% = 9.11{\text{ }}\% \) (A1)(M1)(A1)(ft)(G2)
Notes: Accept 9.10 %.
Award (A1) for both measurements correctly written in the same unit, (M1) for correct method, (A1)(ft) for correct answer.
Follow through from (b) (iii) and from consistent error in conversion of units throughout the question.
[3 marks]
\(\frac{{360^\circ }}{{12}}\) (M1)
\( = 30^\circ\) (A1)(G2)
[2 marks]
\(\text{MN} = 2 \times \frac{11}{\tan 15} \) (A1)(ft)(M1)
OR
\(\text{MN} = 2 \times 11 \tan 75^\circ \)
\({\text{MN}} = 82.1{\text{ cm}}\) (A1)(ft)(G2)
Notes: Award (A1) for 11 and 2 seen (implied by 22 seen), (M1) for dividing by tan15 (or multiplying by tan 75).
Follow through from their angle in part (c) (i).
[3 marks]
volume = 5419 × 2.5 (M1)
= 13500 cm3 (A1)(G2)
[2 marks]
\(\frac{{13547.34 \ldots }}{{514}} = 26.4\) (M1)(A1)(ft)(G2)
Note: Award (M1) for dividing their part (d) by 514.
Accept 26.3.
[2 marks]
Examiners report
Part (a) was well done except for the fact that very few students were able to convert correctly from m2 to cm2 and this was very disappointing.
Part (a) was well done except for the fact that very few students were able to convert correctly from m2 to cm2 and this was very disappointing.
Part (a) was well done except for the fact that very few students were able to convert correctly from m2 to m2 and this was very disappointing.
In part (b) the cosine rule and the area of a triangle were well done. In some cases units were missing and therefore a unit penalty was applied.
In part (b) the cosine rule and the area of a triangle were well done. In some cases units were missing and therefore a unit penalty was applied.
In part (b) the cosine rule and the area of a triangle were well done. In some cases units were missing and therefore a unit penalty was applied.
In part (b) the cosine rule and the area of a triangle were well done. In some cases units were missing and therefore a unit penalty was applied.
Part (c) was clearly the most difficult one for the students. The general impression was that they did not read the diagram in detail. A number of candidates could not distinguish the circle from the triangle and hence used an incorrect method to find the radius.
Part (c) was clearly the most difficult one for the students. The general impression was that they did not read the diagram in detail. A number of candidates could not distinguish the circle from the triangle and hence used an incorrect method to find the radius.
It was pleasing to see candidates recovering well to get full marks for the last two parts.
It was pleasing to see candidates recovering well to get full marks for the last two parts.
