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Date May 2008 Marks available 2 Reference code 08M.2.sl.TZ2.3
Level SL only Paper 2 Time zone TZ2
Command term Show that Question number 3 Adapted from N/A

Question

Amir needs to construct an isosceles triangle \({\text{ABC}}\) whose area is \(100{\text{ cm}}^2\). The equal sides, \({\text{AB}}\) and \({\text{BC}}\), are \(20{\text{ cm}}\) long.

Sylvia is making a square-based pyramid. Each triangle has a base of length \(12{\text{ cm}}\) and a height of \(10{\text{ cm}}\).

Angle \({\text{ABC}}\) is acute. Show that the angle \({\text{ABC}}\) is \({30^ \circ }\).

[2]
i.a.

Find the length of \({\text{AC}}\).

[3]
i.b.

Show that the height of the pyramid is \(8{\text{ cm}}\).

[2]
ii.a.

\({\text{M}}\) is the midpoint of the base of one of the triangles and \({\text{O}}\) is the apex of the pyramid.

Find the angle that the line \({\text{MO}}\) makes with the base of the pyramid.

[3]
ii.b.

Calculate the volume of the pyramid.

[2]
ii.c.

Daniel wants to make a rectangular prism with the same volume as that of Sylvia’s pyramid. The base of his prism is to be a square of side \(10{\text{ cm}}\). Calculate the height of the prism.

[2]
ii.d.

Markscheme

\(\frac{1}{2}{20^2}\sin B = 100\)     (M1)(A1)

\(B = {30^ \circ }\)    (AG)

Note: (M1) for correct substituted formula and (A1) for correct substitution. \(B = {30^ \circ }\) must be seen or previous (A1) mark is lost.

[2 marks]

i.a.

Unit penalty (UP) is applicable where indicated in the left hand column.

\({\overline {{\text{AC}}} ^2} = 2 \times {20^2} - 2 \times {20^2} \times \cos {30^ \circ }\)     (M1)(A1)

(UP)     \(\overline {{\text{AC}}}  = 10.4{\text{ cm}}\)     (A1)(G2)

Note: (M1) for using cosine rule, (A1) for correct substitution. Last (A1) is for the correct answer. Accept use of sine rule or any correct method e.g. \({\text{AC}} = 2 \times 20\sin {15^ \circ }\) .

[3 marks]

i.b.

\({x^2} + {6^2} = {10^2}\)     (A1)(M1)

\(x = 8{\text{ cm}}\)     (AG)

Note: (A1) for \(6\) (or \(36\)) seen and (M1) for using Pythagoras with correct substitution. \(x = 8\) must be seen or previous (M1) mark is lost.

[2 marks]

ii.a.

\(\cos \beta  = \frac{6}{{10}}\)     (M1)(A1)

\(\beta  = {53.1^ \circ }\)     (A1)(G2)

OR equivalent

Note: (M1) for use of trigonometric ratio with numbers from question. (A1) for use of correct numbers, and (A1) for correct answer.

[3 marks]

ii.b.

Unit penalty (UP) is applicable where indicated in the left hand column.

\(vol = \frac{{{{12}^2} \times 8}}{3}\)     (M1)

(UP)     \( = 384{\text{ c}}{{\text{m}}^3}\)     (A1)(G2)

Note: (M1) for correct formula and correct substitution, (A1) for correct answer.

[2 marks]

ii.c.

Unit penalty (UP) is applicable where indicated in the left hand column.

Let h be the height

\({10^2}h = 384\)     (M1)

(UP)     \( = 3.84{\text{ cm}}\)     (A1)(ft)(G2)

Note: (M1) for correct formula and correct substitution, (A1) for correct answer. (ft) from answer to part (c).

[2 marks]

ii.d.

Examiners report

Many students did not write the units in their answers and were penalized with the UP in this question.

Part (a) was not very well answered. It looked as if the candidates did not understand the question. Many candidates did not draw a sketch of the triangle; this would have helped them to solve the question. Many candidates simply calculated the remaining angles of the triangle and showed that the sum was \({180^ \circ }\) . This was a clear example of the misunderstanding of the term "show that". Part (b) was well done though some candidates lost a mark for not giving the answer to the correct accuracy.

i.a.

Many students did not write the units in their answers and were penalized with the UP in this question.

Part (a) was not very well answered. It looked as if the candidates did not understand the question. Many candidates did not draw a sketch of the triangle; this would have helped them to solve the question. Many candidates simply calculated the remaining angles of the triangle and showed that the sum was \({180^ \circ }\) . This was a clear example of the misunderstanding of the term "show that". Part (b) was well done though some candidates lost a mark for not giving the answer to the correct accuracy.

i.b.

Many students did not write the units in their answers and were penalized with the UP in this question.

The weaker candidates spent a lot of time in (a) using the wrong triangle to find half of the diagonal of the base. Finally they used Pythagoras theorem with the wrong numbers. Part (b) was well answered by most of the students. For the volume of the pyramid in (c) they used the correct formula though not always with the correct substitutions. To find the height of the prism in (d) the most common error was multiplying the volume of the prism by \(\frac{1}{3}\). It seemed that many did not know the term 'prism'.

ii.a.

Many students did not write the units in their answers and were penalized with the UP in this question.

The weaker candidates spent a lot of time in (a) using the wrong triangle to find half of the diagonal of the base. Finally they used Pythagoras theorem with the wrong numbers. Part (b) was well answered by most of the students. For the volume of the pyramid in (c) they used the correct formula though not always with the correct substitutions. To find the height of the prism in (d) the most common error was multiplying the volume of the prism by \(\frac{1}{3}\). It seemed that many did not know the term 'prism'.

ii.b.

Many students did not write the units in their answers and were penalized with the UP in this question.

The weaker candidates spent a lot of time in (a) using the wrong triangle to find half of the diagonal of the base. Finally they used Pythagoras theorem with the wrong numbers. Part (b) was well answered by most of the students. For the volume of the pyramid in (c) they used the correct formula though not always with the correct substitutions. To find the height of the prism in (d) the most common error was multiplying the volume of the prism by \(\frac{1}{3}\). It seemed that many did not know the term 'prism'.

ii.c.

Many students did not write the units in their answers and were penalized with the UP in this question.

The weaker candidates spent a lot of time in (a) using the wrong triangle to find half of the diagonal of the base. Finally they used Pythagoras theorem with the wrong numbers. Part (b) was well answered by most of the students. For the volume of the pyramid in (c) they used the correct formula though not always with the correct substitutions. To find the height of the prism in (d) the most common error was multiplying the volume of the prism by \(\frac{1}{3}\). It seemed that many did not know the term 'prism'.

ii.d.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.3 » Use of the area of a triangle \( = \frac{1}{2}ab\sin C\).
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