IB DP Mathematical Studies 5.2 Paper 2

 

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Question

Jenny has a circular cylinder with a lid. The cylinder has height 39 cm and diameter 65 mm.

An old tower (BT) leans at 10° away from the vertical (represented by line TG).

The base of the tower is at B so that \({\text{M}}\hat {\rm B}{\text{T}} = 100^\circ \).

Leonardo stands at L on flat ground 120 m away from B in the direction of the lean.

He measures the angle between the ground and the top of the tower T to be \({\text{B}}\hat {\rm L}{\text{T}} = 26.5^\circ \).

Calculate the volume of the cylinder in cm3. Give your answer correct to two decimal places.[3]

i.a.

The cylinder is used for storing tennis balls. Each ball has a radius of 3.25 cm.

Calculate how many balls Jenny can fit in the cylinder if it is filled to the top.[1]

i.b.

(i) Jenny fills the cylinder with the number of balls found in part (b) and puts the lid on. Calculate the volume of air inside the cylinder in the spaces between the tennis balls.

(ii) Convert your answer to (c) (i) into cubic metres.[4]

i.c.

(i) Find the value of angle \({\text{B}}\hat {\rm T}{\text{L}}\).

(ii) Use triangle BTL to calculate the sloping distance BT from the base, B to the top, T of the tower.[5]

ii.a.

Calculate the vertical height TG of the top of the tower.[2]

ii.b.

Leonardo now walks to point M, a distance 200 m from B on the opposite side of the tower. Calculate the distance from M to the top of the tower at T.[3]

ii.c.
Answer/Explanation

Markscheme

\(\pi \times 3.25^2 \times 39\)     (M1)(A1)

(= 1294.1398)

Answer 1294.14 (cm3)(2dp)     (A1)(ft)(G2)

(UP) not applicable in this part due to wording of question. (M1) is for substituting appropriate numbers from the problem into the correct formula, even if the units are mixed up. (A1) is for correct substitutions or correct answer with more than 2dp in cubic centimetres seen. Award (G1) for answer to > 2dp with no working and no attempt to correct to 2dp. Award (M1)(A0)(A1)(ft) for \(\pi  \times {32.5^2} \times 39{\text{ c}}{{\text{m}}^3}\) (= 129413.9824) = 129413.98

Use of \(\pi = \frac{22}{7}\) or 3.142 etc is premature rounding and is awarded at most (M1)(A1)(A0) or (M1)(A0)(A1)(ft) depending on whether the intermediate value is seen or not. For all other incorrect substitutions, award (M1)(A0) and only follow through the 2 dp correction if the intermediate answer to more decimal places is seen. Answer given as a multiple of \(\pi\) is awarded at most (M1)(A1)(A0). As usual, an unsubstituted formula followed by correct answer only receives the G marks.[3 marks]

i.a.

39/6.5 = 6     (A1)[1 mark]

i.b.

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

(UP) (i) Volume of one ball is \(\frac{4}{3} \pi \times 3.25^3 {\text{ cm}}^3\)     (M1)

\({\text{Volume of air}} = \pi  \times {3.25^2} \times 39 – 6 \times \frac{4}{3}\pi  \times {3.25^3} = 431{\text{ c}}{{\text{m}}^3}\)     (M1)(A1)(ft)(G2)

Award first (M1) for substituted volume of sphere formula or for numerical value of sphere volume seen (143.79… or 45.77… \( \times \pi\)). Award second (M1) for subtracting candidate’s sphere volume multiplied by their answer to (b). Follow through from parts (a) and (b) only, but negative or zero answer is always awarded (A0)(ft)

(UP) (ii) 0.000431m3 or  4.31×10−4 m3     (A1)(ft) [4 marks]

i.c.

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

(i) \({\text{Angle B}}\widehat {\text{T}}{\text{L}} = 180 – 80 – 26.5\) or \(180 – 90 – 26.5 – 10\)     (M1)

\(= 73.5^\circ\)     (A1)(G2) 

(ii) \(\frac{{BT}}{{\sin (26.5^\circ )}} = \frac{{120}}{{\sin (73.5^\circ )}}\)     (M1)(A1)(ft)

(UP) BT = 55.8 m (3sf)     (A1)(ft)[5 marks]


If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but

negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

ii.a.

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

TG = 55.8sin(80°) or 55.8cos(10°)     (M1)

(UP) = 55.0 m (3sf)     (A1)(ft)(G2)

Apply (AP) if 0 missing[2 marks]


If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but

negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

ii.b.

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

\({\text{MT}}^2 = 200^2 + 55.8^2 – 2 \times 200 \times 55.8 \times \cos(100^\circ)\)     (M1)(A1)(ft)

(UP) MT = 217 m  (3sf)     (A1)(ft)

Follow through only from part (ii)(a)(ii). Award marks at discretion for any valid alternative method.[3 marks]


If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but

negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

ii.c.

Question

Mal is shopping for a school trip. He buys \(50\) tins of beans and \(20\) packets of cereal. The total cost is \(260\) Australian dollars (\({\text{AUD}}\)).

The triangular faces of a square based pyramid, \({\text{ABCDE}}\), are all inclined at \({70^ \circ }\) to the base. The edges of the base \({\text{ABCD}}\) are all \(10{\text{ cm}}\) and \({\text{M}}\) is the centre. \({\text{G}}\) is the mid-point of \({\text{CD}}\).

Write down an equation showing this information, taking \(b\) to be the cost of one tin of beans and \(c\) to be the cost of one packet of cereal in \({\text{AUD}}\).[1]

i.a.

Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).

Write down another equation to represent this information.[1]

i.b.

Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).

Find the cost of one tin of beans.[2]

i.c.

(i)     Sketch the graphs of the two equations from parts (a) and (b).

(ii)    Write down the coordinates of the point of intersection of the two graphs.[4]

i.d.

Using the letters on the diagram draw a triangle showing the position of a \({70^ \circ }\) angle.[1]

ii.a.

Show that the height of the pyramid is \(13.7{\text{ cm}}\), to 3 significant figures.[2]

ii.b.

Calculate

(i)     the length of \({\text{EG}}\);

(ii)    the size of angle \({\text{DEC}}\).[4]

ii.c.

Find the total surface area of the pyramid.[2]

ii.d.

Find the volume of the pyramid.[2]

ii.e.
Answer/Explanation

Markscheme

\(50b + 20c = 260\)     (A1)[1 mark]

i.a.

\(12b + 6c = 66\)     (A1)[1 mark]

i.b.

Solve to get \(b = 4\)     (M1)(A1)(ft)(G2)

Note: (M1) for attempting to solve the equations simultaneously.[2 marks]

i.c.

(i)

     (A1)(A1)(A1)


Notes: Award (A1) for labels and some idea of scale, (A1)(ft)(A1)(ft) for each line.
The axis can be reversed.

(ii)    \((4,3)\) or \((3,4)\)     (A1)(ft)

Note: Accept \(b = 4\), \(c = 3\)[4 marks]

i.d.

     (A1)

[1 mark]

ii.a.

\(\tan 70 = \frac{h}{5}\)     (M1)

\(h = 5\tan 70 = 13.74\)     (A1)

\(h = 13.7{\text{ cm}}\)     (AG)[2 marks]

ii.b.

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

(i)     \({\text{E}}{{\text{G}}^2} = {5^2} + {13.7^2}\) OR \({5^2} + {(5\tan 70)^2}\)     (M1)

(UP)     \({\text{EG}} = 14.6{\text{ cm}}\)     (A1)(G2)

(ii)    \({\text{DEC}} = 2 \times {\tan ^{ – 1}}\left( {\frac{5}{{14.6}}} \right)\)     (M1)

\( = {37.8^ \circ }\)     (A1)(ft)(G2)[4 marks]

ii.c.

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

\({\text{Area}} = 10 \times 10 + 4 \times 0.5 \times 10 \times 14.619\)     (M1)

(UP)     \( = 392{\text{ c}}{{\text{m}}^2}\)     (A1)(ft)(G2)[2 marks]

ii.d.

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

\({\text{Volume}} = \frac{1}{3} \times 10 \times 10 \times 13.7\)     (M1)

(UP)     \( = 457{\text{ c}}{{\text{m}}^3}\) (\(458{\text{ c}}{{\text{m}}^3}\))     (A1)(G2)[2 marks]

ii.e.

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.
Answer/Explanation

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.

Question

In the diagram below A, B and C represent three villages and the line segments AB, BC and CA represent the roads joining them. The lengths of AC and CB are 10 km and 8 km respectively and the size of the angle between them is 150°.

Find the length of the road AB.[3]

a.

Find the size of the angle CAB.[3]

b.

Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the point where this road cuts AC. This information is shown in the diagram below.

Write down the distance from A to D.[1]

c.

Show that the distance from D to T is 2.06 km correct to three significant figures.[2]

d.

A bus starts and ends its journey at A taking the route AD to DT to TA.

Find the total distance for this journey.[3]

e.

The average speed of the bus while it is moving on the road is 70 km h–1. The bus stops for 5 minutes at each of D and T .

Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.[4]

f.
Answer/Explanation

Markscheme

AB2 = 102 + 82 – 2 × 10 × 8 × cos150°     (M1)(A1)

AB = 17.4 km     (A1)(G2)

Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer.[3 marks]

a.

\(\frac{8}{{\sin {\text{C}}\hat {\rm A}{\text{B}}}} = \frac{{17.4}}{{\sin 150^\circ }}\)     (M1)(A1)

\({\text{C}}\hat {\rm A}{\text{B}} = 13.3^\circ \)     (A1)(ft)(G2)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer. Follow through from their answer to part (a).[3 marks]

b.

AD = 8.70 km (8.7 km)     (A1)(ft)

Note: Follow through from their answer to part (a).[1 mark]

c.

DT = tan (13.29…°) × 8.697… = 2.0550…     (M1)(A1)

= 2.06     (AG)

Notes: Award (M1) for correct substitution in the correct formula, award (A1) for the unrounded answer seen. If 2.06 not seen award at most (M1)(AO).[2 marks]

d.

\(\sqrt {{{8.70}^2} + {{2.06}^2}}  + 8.70 + 2.06\)     (A1)(M1)

= 19.7 km     (A1)(ft)(G2)

Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT, (A1)(ft) for answer. Follow through from their answer to part (c).[3 marks]

e.

\(\frac{{19.7}}{{70}} \times 60 + 10\)     (M1)(M1)

= 26.9     (A1)(ft)

Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for unrounded answer. Follow through from their answer to (e).

= 27  (nearest minute)     (A1)(ft)(G3)

Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.[4 marks]

f.

Question

The diagram represents a small, triangular field, ABC , with \({\text{BC}} = 25{\text{ m}}\) , \({\text{angle BAC}} = {55^ \circ }\) and \({\text{angle ACB}} = {75^ \circ }\) .

Write down the size of angle ABC.[1]

a.

Calculate the length of AC.[3]

b.

Calculate the area of the field ABC.[3]

c.

N is the point on AB such that CN is perpendicular to AB. M is the midpoint of CN.

Calculate the length of NM.[3]

d.

A goat is attached to one end of a rope of length 7 m. The other end of the rope is attached to the point M.

Decide whether the goat can reach point P, the midpoint of CB. Justify your answer.[5]

e.
Answer/Explanation

Markscheme

\({\text{Angle ABC}} = {50^ \circ }\)     (A1)[1 mark]

a.

\(\frac{{{\text{AC}}}}{{\sin {{50}^ \circ }}} = \frac{{25}}{{\sin {{55}^ \circ }}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their angle ABC.

\({\text{AC}} = 23.4{\text{ m}}\)     (A1)(ft)(G2)[3 marks]

b.

\({\text{Area of }}\Delta {\text{ ABC}} = \frac{1}{2} \times 23.379 \ldots  \times 25 \times \sin {75^ \circ }\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their AC.

OR

\({\text{Area of triangle ABC}} = \frac{{29.479 \ldots  \times 19.151 \ldots }}{2}\)     (A1)(ft)(M1)

Note: (A1)(ft) for correct values of AB (29.479…) and CN (19.151…). Follow through from their (a) and /or (b). Award (M1) for substitution of their values of AB and CN into the correct formula.

\({\text{Area of }}\Delta {\text{ ABC}} = 282{\text{ }}{{\text{m}}^2}\)     (A1)(ft)(G2)

Note: Accept \(283{\text{ }}{{\text{m}}^2}\) if \(23.4\) is used.[3 marks]

c.

\({\text{NM}} = \frac{{25 \times \sin {{50}^ \circ }}}{2}\)     (M1)(M1)

Note: Award (M1) for \({25 \times \sin {{50}^ \circ }}\) or equivalent for the length of CN. (M1) for dividing their CN by \(2\).

\({\text{NM}} = 9.58{\text{ m}}\)     (A1)(ft)(G2)

Note: Follow through from their angle ABC.

Notes: Premature rounding of CN leads to the answers \(9.60\) or \(9.6\). Award at most (M1)(M1)(A0) if working seen. Do not penalize with (AP). CN may be found in (c).

Note: The working for this part of the question may be in part (b).[3 marks]

d.

\({\text{Angle NCB}} = {40^ \circ }\) seen     (A1)(ft)

Note: Follow through from their (a).

From triangle MCP:

\({\text{M}}{{\text{P}}^2} = {(9.5756 \ldots )^2} + {12.5^2} – 2 \times 9.5756 \ldots  \times 12.5 \times \cos ({40^ \circ })\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(ft)(G3)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{CP}} = 12.5{\text{ m}}\) seen     (A1)

\({\text{M}}{{\text{P}}^2} = {(12.5)^2} – {(9.575 \ldots )^2}\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{Angle MCP}} = {40^ \circ }\) seen     (A1)(ft)

\(\frac{{{\text{MP}}}}{{12.5}} = \sin ({40^ \circ })\) or equivalent     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (a). Award (G3) for correct value of MP seen without working.

The goat cannot reach point P as \({\text{MP}} > 7{\text{ m}}\) .     (A1)(ft)

Note: Award (A1)(ft) only if their value of MP is compared to \(7{\text{ m}}\), and conclusion is stated.[5 marks]

e.

Question

A solid metal cylinder has a base radius of 4 cm and a height of 8 cm.

Find the area of the base of the cylinder.[2]

a.

Show that the volume of the metal used in the cylinder is 402 cm3, given correct to three significant figures.[2]

b.

Find the total surface area of the cylinder.[3]

c.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the height, OC, of the cone.[3]

d.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the size of angle BCO.[2]

e.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the slant height, CB.[2]

f.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the total surface area of the cone.[4]

g.
Answer/Explanation

Markscheme

\( \pi \times 4^2\)     (M1)

= 50.3 (16\(\pi\)) cm(50.2654…)     (A1)(G2)

Note: Award (M1) for correct substitution in area formula. The answer is 50.3 cm2, the units are required.[2 marks]

a.

50.265…× 8     (M1)

Note: Award (M1) for correct substitution in the volume formula.

= 402.123…     (A1)
= 402 (cm3)     (AG)

Note: Both the unrounded and the rounded answer must be seen for the (A1) to be awarded. The units are not required[2 marks]

b.

\(2 \times \pi \times 4 \times 8 + 2 \times \pi \times 4^2\)     (M1)(M1)

Note: Award (M1) for correct substitution in the curved surface area formula, (M1) for adding the area of their two bases.

= 302 cm2 (96π cm2) (301.592…)     (A1)(ft)(G2)

Notes: The answer is 302 cm2, the units are required. Do not penalise for missing or incorrect units if penalised in part (a). Follow through from their answer to part (a).[3 marks]

c.

\(\frac{1}{3} \pi \times 6^2 \times \text{OC} = 402\)     (M1)(M1)

Note: Award (M1) for correctly substituted volume formula, (M1) for equating to 402 (402.123…).

\({\text{OC}} = 10.7{\text{ (cm)}}\left( {{\text{10}}\frac{2}{3},{\text{ }}10.6666…} \right)\)     (A1)(G2)[3 marks]

d.

\(\tan \text{BCO} = \frac{6}{10.66…}\)     (M1)

Note: Award (M1) for use of correct tangent ratio.

\({\text{B}}{\operatorname{\hat C}}{\text{O}} = 29.4^\circ \) (29.3577…)     (A1)(ft)(G2)

Notes: Accept 29.3° (29.2814…) if 10.7 is used. An acceptable alternative method is to calculate CB first and then angle BCO. Allow follow through from parts (d) and (f). Answers range from 29.2° to 29.5°.[2 marks]

e.

\(\text{CB} = \sqrt{{6^2} + {(10.66…)^2}}\)     (M1)

OR

\(\sin 29.35…^\circ = \frac{6}{\text{CB}}\)     (M1)

OR

\(\cos 29.35…^\circ = \frac{10.66…}{\text{CB}}\)     (M1)

CB = 12.2 (cm) (12.2383…)     (A1)(ft)(G2)

Note: Accept 12.3 (12.2674…) if 10.7 (and/or 29.3) used. Follow through from part (d) or part (e) as appropriate.[2 marks]

f.

\(\pi \times 6 \times 12.2383… + \pi \times 6^2\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution in curved surface area formula, (M1) for correct substitution in area of circle formula, (M1) for addition of the two areas.

= 344 cm2 (343.785…)     (A1)(ft)(G3)

Note: The answer is 344 cm2, the units are required. Do not penalise for missing or incorrect units if already penalised in either part (a) or (c). Accept 345 cm2 if 12.3 is used and 343 cm2 if 12.2 is used. Follow through from their part (f).[4 marks]

g.

Question

The Great Pyramid of Cheops in Egypt is a square based pyramid. The base of the pyramid is a square of side length 230.4 m and the vertical height is 146.5 m. The Great Pyramid is represented in the diagram below as ABCDV . The vertex V is directly above the centre O of the base. M is the midpoint of BC.

(i) Write down the length of OM .

(ii) Find the length of VM .[3]

a.

Find the area of triangle VBC .[2]

b.

Calculate the volume of the pyramid.[2]

c.

Show that the angle between the line VM and the base of the pyramid is 52° correct to 2 significant figures.[2]

d.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Write down the size of angle VMP .[1]

e.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Using your value of VM from part (a)(ii), find the value of x.[4]

f.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Ahmed walks 50 m from P to Q.

Find the length of QV, the distance from Ahmed to the vertex of the pyramid.[4]

g.
Answer/Explanation

Markscheme

(i) 115.2 (m)     (A1)

Note: Accept 115 (m)

(ii) \(\sqrt{(146.5^2 + 115.2^2)}\)     (M1)

Note: Award (M1) for correct substitution.

186 (m) (186.368…)     (A1)(ft)(G2)

Note: Follow through from part (a)(i).[3 marks]

a.

\(\frac{1}{2} \times 230.4 \times 186.368…\)     (M1)

Note: Award (M1) for correct substitution in area of the triangle formula.

21500 m2 (21469.6…m2)     (A1)(ft)(G2)

Notes: The final answer is 21500 m2; units are required. Accept 21400 m2 for use of 186 m and/or 115 m.[2 marks]

b.

\(\frac{1}{3} \times 230.4^2 \times 146.5\)     (M1)

Note: Award (M1) for correct substitution in volume formula.

2590000 m3 (2592276.48 m3)     (A1)(G2)

Note: The final answer is 2590000 m3; units are required but do not penalise missing or incorrect units if this has already been penalised in part (b). [2 marks]

c.

\(\tan^{-1}\left( {\frac{{146.5}}{{115.2}}} \right)\)     (M1)

Notes: Award (M1) for correct substituted trig ratio. Accept alternate correct trig ratios.

= 51.8203…= 52°     (A1)(AG)

Notes: Both the unrounded answer and the final answer must be seen for the (A1) to be awarded. Accept 51.96° = 52°, 51.9° = 52° or 51.7° = 52°

d.

128°     (A1)[1 mark]

e.

\(\frac{{186.368}}{{\sin27}} = \frac{{x}}{{\sin25}}\)     (A1)(M1)(A1)(ft)

Notes: Award (A1)(ft) for their angle MVP seen, follow through from their part (e). Award (M1) for substitution into sine formula, (A1) for correct substitutions. Follow through from their VM and their angle VMP.

x = 173 (m) (173.490…)     (A1)(ft)(G3)

Note: Accept 174 from use of 186.4.[4 marks]

f.

VQ2 = (186.368…)2 + (123.490…)2 − 2 × (186.368…) × (123.490…) × cos128     (A1)(ft)(M1)(A1)(ft)

Notes: Award (A1)(ft) for 123.490…(123) seen, follow through from their x (PM) in part (f), (M1) for substitution into cosine formula, (A1)(ft) for correct substitutions. Follow through from their VM and their angle VMP.

OR

173.490… − 50 = 123.490… (123)     (A1)(ft)

115.2 + 123.490… = 238.690…     (A1)(ft)

\(\text{VQ} = \sqrt{(146.5^2 + 238.690…^2)}\)     (M1)

VQ = 280 (m) (280.062…)     (A1)(ft)(G3)

Note: Accept 279 (m) from use of 3 significant figure answers.[4 marks]

g.

Question

A tent is in the shape of a triangular right prism as shown in the diagram below.

The tent has a rectangular base PQRS .

PTS and QVR are isosceles triangles such that PT = TS and QV = VR .

PS is 3.2 m , SR is 4.7 m and the angle TSP is 35°.

Show that the length of side ST is 1.95 m, correct to 3 significant figures.[3]

a.

Calculate the area of the triangle PTS.[3]

b.

Write down the area of the rectangle STVR.[1]

c.

Calculate the total surface area of the tent, including the base.[3]

d.

Calculate the volume of the tent.[2]

e.

A pole is placed from V to M, the midpoint of PS.

Find in metres,

(i) the height of the tent, TM;

(ii) the length of the pole, VM.[4]

f.

Calculate the angle between VM and the base of the tent.[2]

g.
Answer/Explanation

Markscheme

\({\text{ST}} = \frac{{1.6}}{{\cos 35^\circ }}\)     (M1)(A1)

Note: Award (M1) for correctly substituted trig equation, (A1) for 1.6 seen.

OR

\(\frac{{{\text{ST}}}}{{\sin 35^\circ }} = \frac{{3.2}}{{\sin 110^\circ }}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule equation, (A1) for correct substitutions.

ST = 1.95323…     (A1)

= 1.95 (m)     (AG)

Notes: Both unrounded and rounded answer must be seen for final (A1) to be awarded.

a.

\(\frac{1}{2} \times 3.2 \times 1.95323… \times \sin 35^\circ \) or \(\frac{1}{2} \times 1.95323… \times 1.95323… \times \sin 110^\circ \)     (M1)(A1)

Note: Award (M1) for substituted area formula, (A1) for correct substitutions. Do not award follow through marks.

= 1.79 m2 (1.79253…m2)     (A1)(G2)

Notes: The answer is 1.79 m2, units are required. Accept 1.78955… from using 1.95.

OR

\(\frac{1}{2} \times 3.2 \times 1.12033…\)     (A1) (M1)

Note: Award (A1) for the correct value for TM (1.12033…) OR correct expression for TM (i.e. 1.6tan35°, \(\sqrt {{{(1.95323…)}^2} – {{1.6}^2}} \)), (M1) for correctly substituted formula for triangle area.

= 1.79 m2 (1.79253…m2)     (A1)(G2)

Notes: The answer is 1.79 m2, units are required. Accept 1.78 m2 from using 1.95.

b.

9.18 m2 (9.18022 m2)     (A1)(G1)

Notes: The answer is 9.18 m2, units are required. Do not penalize if lack of units was already penalized in (b). Do not award follow through marks here. Accept 9.17 m2 (9.165 m2) from using 1.95.

c.

\(2 \times 1.79253… + 2 \times 9.18022… + 4.7 \times 3.2\)     (M1)(A1)(ft)

Note: Award (M1) for addition of three products, (A1)(ft) for three correct products.

= 37.0 m2 (36.9855…m2)     (A1)(ft)(G2)

Notes: The answer is 37.0 m2, units are required. Accept 36.98 m2 from using 3sf answers. Follow through from their answers to (b) and (c). Do not penalize if lack of units was penalized earlier in the question.

d.

\(1.79253… \times 4.7\)     (M1)

Note: Award (M1) for their correctly substituted volume formula.

= 8.42 m3 (8.42489…m3)     (A1)(ft)(G2)

Notes: The answer is 8.42 m3, units are required. Accept 8.41 m3 from use of 1.79. An answer of 8.35, from use of TM = 1.11, will receive follow-through marks if working is shown. Follow through from their answer to part (b). Do not penalize if lack of units was penalized earlier in the question.

e.

(i) \({\text{TM}} = 1.6\tan {35^\circ }\)     (M1)

Notes: Award (M1) for their correct substitution in trig ratio.

OR

\({\text{TM}} = \sqrt {{{(1.95323…)}^2} – {{1.6}^2}} \)     (M1)

Note: Award (M1) for correct substitution in Pythagoras’ theorem.

OR

\(\frac{{3.2 \times {\text{TM}}}}{2} = 1.79253…\)     (M1)

Note: Award (M1) for their correct substitution in area of triangle formula.

= 1.12 (m) (1.12033…)     (A1)(ft)(G2)

Notes: Follow through from their answer to (b) if area of triangle is used. Accept 1.11 (1.11467) from use of ST = 1.95.

(ii) \({\text{VM}} = \sqrt {{{1.12033…}^2} + {{4.7}^2}} \)     (M1)

Note: Award (M1) for their correct substitution in Pythagoras’ theorem.

= 4.83 (m) (4.83168 )     (A1)(ft)(G2)

Notes: Follow through from (f)(i).

f.

\({\sin ^{ – 1}}\left( {\frac{{1.12033…}}{{4.83168…}}} \right)\)     (M1)

OR

\({\cos^{ – 1}}\left( {\frac{{4.7}}{{4.83168…}}} \right)\)     (M1)

OR

\({\tan^{ – 1}}\left( {\frac{{1.12033…}}{{4.7}}} \right)\)     (M1)

Note: Award (M1) for correctly substituted trig equation.

OR

\({\cos ^{ – 1}}\left( {\frac{{{{4.7}^2} + {{(4.83168…)}^2} – {{(1.12033…)}^2}}}{{2 \times 4.7 \times 4.83168…}}} \right)\)     (M1)

Note: Award (M1) for correctly substituted cosine formula.

= 13.4° (13.4073…)     (A1)(ft)(G2)

Notes: Accept 13.3°. Follow through from part (f).

g.

Question

ABC is a triangular field on horizontal ground. The lengths of AB and AC are 70 m and 50 m respectively. The size of angle BCA is 78°.

Find the size of angle \(ABC\).[3]

a.

Find the area of the triangular field.[4]

b.

\({\text{M}}\) is the midpoint of \({\text{AC}}\).

Find the length of \({\text{BM}}\).[3]

c.

A vertical mobile phone mast, \({\text{TB}}\), is built next to the field with its base at \({\text{B}}\). The angle of elevation of \({\text{T}}\) from \({\text{M}}\) is \(63.4^\circ \). \({\text{N}}\) is the midpoint of the mast.

Calculate the angle of elevation of \({\text{N}}\) from \({\text{M}}\).[5]

d.
Answer/Explanation

Markscheme

\(\frac{{70}}{{\sin 78}} = \frac{{50}}{{\sin {\rm{A\hat BC}}}}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule, (A1) for correct substitution.

\({\rm{A\hat BC}} = 44.3^\circ \) (\(44.3209…\))     (A1)(G3)

Note: If radians are used the answer is \(0.375918…\), award at most (M1)(A1)(A0).[3 marks]

a.

\({\text{area }}\Delta {\text{ABC}} = \frac{1}{2} \times 70 \times 50 \times \sin (57.6790 \ldots )\)     (A1)(M1)(A1)(ft)
 

Notes: Award (A1)(ft) for their \(57.6790…\) seen, (M1) for substituted area formula, (A1)(ft) for correct substitution.

     Follow through from part (a).

\( = 1480{\text{ }}{{\text{m}}^2}\)  \((1478.86 \ldots )\)     (A1)(ft)(G3)

Notes: The answer is \(1480{\text{ }}{{\text{m}}^2}\), units are required. \(1479.20…\) if 3 sf used.

     If radians are used the answer is \(1554.11 \ldots {{\text{m}}^2}\), award (A1)(ft)(M1)(A1)(ft)(A1)(ft)(G3).[4 marks]

b.

\({\text{B}}{{\text{M}}^2} = {70^2} + {25^2} – 2 \times 70 \times 25 \times \cos (57.6790 \ldots )\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted cosine rule, (A1)(ft) for correct substitution. Follow through from their angle in part (b).

\({\text{BM}} = 60.4{\text{ (m)}}\)   \((60.4457 \ldots )\)     (A1)(ft)(G2)

Notes: If the 3 sf answer is used the answer is \(60.5{\text{ }}({\text{m}})\).

     If radians are used the answer is \(62.5757… {\text{ }} ({\text{m}})\), award (M1)(A1)(ft)(A1)(ft)(G2).[3 marks]

c.

\(\tan 63.4^\circ  = \frac{{{\text{TB}}}}{{60.4457 \ldots }}\)     (M1)

Note: Award (M1) for their correctly substituted trig equation.

\({\text{TB}} = 120.707 \ldots \)     (A1)(ft)

Notes: Follow through from part (c). If 3 sf answers are used throughout, \({\text{TB}} = 120.815 \ldots \)

If \({\text{TB}} = 120.707 \ldots \) is seen without working, award (A2).

\(\tan {\rm{N\hat MB = }}\frac{{\left( {\frac{{120.707 \ldots }}{2}} \right)}}{{60.4457 \ldots }}\)     (A1)(ft)(M1)

Notes: Award (A1)(ft) for their \({\text{TB}}\) divided by \(2\) seen, (M1) for their correctly substituted trig equation.

     Follow through from part (c) and within part (d).

\({\rm{N\hat MB = 45.0^\circ }}\)   \({\text{(44.9563}} \ldots {\text{)}}\)     (A1)(ft)(G3)

Notes:     If 3 sf are used throughout, answer is \(45^\circ \).

     If radians are used the answer is \(0.308958…\), and if full working is shown, award at most (M1)(A1)(ft)(A1)(ft)(M1)(A0).

     If no working is shown for radians answer, award (G2).

OR

\(\tan {\rm{N\hat MB}} = \frac{{{\text{NB}}}}{{{\text{BM}}}}\)     (M1)

\(\tan 63.4^\circ  = \frac{{2 \times {\text{NB}}}}{{{\text{BM}}}}\)     (A1)(M1)

Note: Award (A1) for \(2 \times {\text{NB}}\) seen.

\(\tan {\rm{N\hat MB}} = \frac{1}{2}\tan 63.4^\circ \)     (M1)

\({\rm{N\hat MB}} = 45.0^\circ \)   \((44.9563 \ldots )\)     (A1)(G3)

Notes: If radians are used the answer is \(0.308958…\), and if full working is shown, award at most (M1)(A1)(M1)(M1)(A0). If no working is shown for radians answer, award (G2).[5 marks]

d.

Question

A boat race takes place around a triangular course, \({\text{ABC}}\), with \({\text{AB}} = 700{\text{ m}}\), \({\text{BC}} = 900{\text{ m}}\) and angle \({\text{ABC}} = 110^\circ \). The race starts and finishes at point \({\text{A}}\).

Calculate the total length of the course.[4]

a.

It is estimated that the fastest boat in the race can travel at an average speed of \(1.5\;{\text{m}}\,{{\text{s}}^{ – 1}}\).

Calculate an estimate of the winning time of the race. Give your answer to the nearest minute.[3]

b.

It is estimated that the fastest boat in the race can travel at an average speed of \(1.5\;{\text{m}}\,{{\text{s}}^{ – 1}}\).

Find the size of angle \({\text{ACB}}\).[3]

c.

To comply with safety regulations, the area inside the triangular course must be kept clear of other boats, and the shortest distance from \({\text{B}}\) to \({\text{AC}}\) must be greater than \(375\) metres.

Calculate the area that must be kept clear of boats.[3]

d.

To comply with safety regulations, the area inside the triangular course must be kept clear of other boats, and the shortest distance from \({\text{B}}\) to \({\text{AC}}\) must be greater than \(375\) metres.

Determine, giving a reason, whether the course complies with the safety regulations.[3]

e.

The race is filmed from a helicopter, \({\text{H}}\), which is flying vertically above point \({\text{A}}\).

The angle of elevation of \({\text{H}}\) from \({\text{B}}\) is \(15^\circ\).

Calculate the vertical height, \({\text{AH}}\), of the helicopter above \({\text{A}}\).[2]

f.

The race is filmed from a helicopter, \({\text{H}}\), which is flying vertically above point \({\text{A}}\).

The angle of elevation of \({\text{H}}\) from \({\text{B}}\) is \(15^\circ\).

Calculate the maximum possible distance from the helicopter to a boat on the course.[3]

g.
Answer/Explanation

Markscheme

\({\text{A}}{{\text{C}}^2} = {700^2} + {900^2} – 2 \times 700 \times 900 \times \cos 110^\circ \)     (M1)(A1)

\({\text{AC}} = 1315.65 \ldots \)     (A1)(G2)

length of course \( = 2920{\text{ (m)}}\;\;\;(2915.65 \ldots {\text{ m)}}\)     (A1)

Notes: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution, (A1) for correct answer.

Award (G3) for \(2920\;\;\;(2915.65 \ldots )\) seen without working.

The final (A1) is awarded for adding \(900\) and \(700\) to their \({\text{AC}}\) irrespective of working seen.

a.

\(\frac{{2915.65}}{{1.5}}\)     (M1)

Note: Award (M1) for their length of course divided by \(1.5\).

Follow through from part (a).

\( = 1943.76 \ldots {\text{ (seconds)}}\)     (A1)(ft)

\( = 32{\text{ (minutes)}}\)     (A1)(ft)(G2)

Notes: Award the final (A1) for correct conversion of their answer in seconds to minutes, correct to the nearest minute.

Follow through from part (a).

b.

\(\frac{{700}}{{\sin {\text{ACB}}}} = \frac{{1315.65 \ldots }}{{\sin 110^\circ }}\)     (M1)(A1)(ft)

OR

\(\cos {\text{ACB}} = \frac{{{{900}^2} + 1315.65{ \ldots ^2} – {{700}^2}}}{{2 \times 900 \times 1315.65 \ldots }}\)     (M1)(A1)(ft)

\({\text{ACB}} = 30.0^\circ \;\;\;(29.9979 \ldots ^\circ )\)     (A1)(ft)(G2)

Notes: Award (M1) for substitution into sine rule or cosine rule formula, (A1) for their correct substitution, (A1) for correct answer.

Accept \(29.9^\circ\) for sine rule and \(29.8^\circ\) for cosine rule from use of correct three significant figure values. Follow through from their answer to (a).

c.

\(\frac{1}{2} \times 700 \times 900 \times \sin 110^\circ \)     (M1)(A1)

Note: Accept \(\frac{1}{2} \times {\text{their AC}} \times {\text{900}} \times {\text{sin(their ACB)}}\). Follow through from parts (a) and (c).

\( = 296000{\text{ }}{{\text{m}}^2}\;\;\;(296003{\text{ }}{{\text{m}}^2})\)     (A1)(G2)

Notes: Award (M1) for substitution into area of triangle formula, (A1) for correct substitution, (A1) for correct answer.

Award (G1) if \(296000\) is seen without units or working.

d.

\(\sin 29.9979 \ldots  = \frac{{{\text{distance}}}}{{900}}\)     (M1)

\({\text{(distance}} = ){\text{ }}450{\text{ (m)}}\;\;\;{\text{(449.971}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Note: Follow through from part (c).

OR

\(\frac{1}{2} \times {\text{distance}} \times 1315.65 \ldots  = 296003\)     (M1)

\(({\text{distance}} = ){\text{ }}450{\text{ (m)}}\;\;\;{\text{(449.971}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Note: Follow through from part (a) and part (d).

\(450\) is greater than \(375\), thus the course complies with the safety regulations     (R1)

Notes:  A comparison of their area from (d) and the area resulting from the use of \(375\) as the perpendicular distance is a valid approach and should be given full credit. Similarly a comparison of angle \({\text{ACB}}\) and \({\sin ^{ – 1}}\left( {\frac{{375}}{{900}}} \right)\) should be given full credit.

Award (R0) for correct answer without any working seen. Award (R1)(ft) for a justified reason consistent with their working.

Do not award (M0)(A0)(R1).

e.

\(\tan 15^\circ  = \frac{{{\text{AH}}}}{{700}}\)     (M1)

Note: Award (M1) for correct substitution into trig formula.

\({\text{AH}} = 188{\text{ (m)}}\;\;\;(187.564 \ldots )\)     (A1)(ft)(G2)

f.

\({\text{H}}{{\text{C}}^2} = 187.564{ \ldots ^2} + 1315.65{ \ldots ^2}\)     (M1)(A1)

Note: Award (M1) for substitution into Pythagoras, (A1) for their \(1315.65{ \ldots}\) and their \(187.564{ \ldots}\) correctly substituted in formula.

\({\text{HC}} = 1330 \ldots {\text{ (m)}}\;\;\;(1328.95 \ldots )\)     (A1)(ft)(G2)

Note: Follow through from their answer to parts (a) and (f).

g.

Question

The Great Pyramid of Giza in Egypt is a right pyramid with a square base. The pyramid is made of solid stone. The sides of the base are \(230\,{\text{m}}\) long. The diagram below represents this pyramid, labelled \({\text{VABCD}}\).

\({\text{V}}\) is the vertex of the pyramid. \({\text{O}}\) is the centre of the base, \({\text{ABCD}}\) . \({\text{M}}\) is the midpoint of \({\text{AB}}\). Angle \({\text{ABV}} = 58.3^\circ \) .

Show that the length of \({\text{VM}}\) is \(186\) metres, correct to three significant figures.[3]

a.

Calculate the height of the pyramid, \({\text{VO}}\) .[2]

b.

Find the volume of the pyramid.[2]

c.

Write down your answer to part (c) in the form \(a \times {10^k}\)  where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\) .[2]

d.

Ahmad is a tour guide at the Great Pyramid of Giza. He claims that the amount of stone used to build the pyramid could build a wall \(5\) metres high and \(1\) metre wide stretching from Paris to Amsterdam, which are \(430\,{\text{km}}\) apart.

Determine whether Ahmad’s claim is correct. Give a reason.[4]

e.

Ahmad and his friends like to sit in the pyramid’s shadow, \({\text{ABW}}\), to cool down.
At mid-afternoon, \({\text{BW}} = 160\,{\text{m}}\)  and angle \({\text{ABW}} = 15^\circ .\)

i)     Calculate the length of \({\text{AW}}\) at mid-afternoon.

ii)    Calculate the area of the shadow, \({\text{ABW}}\), at mid-afternoon.[6]

f.
Answer/Explanation

Markscheme

\(\tan \,(58.3) = \frac{{{\text{VM}}}}{{115}}\)   OR \(115 \times \tan \,(58.3^\circ )\)              (A1)(M1)

Note: Award (A1) for \(115\,\,\left( {ie\,\frac{{230}}{2}} \right)\)   seen, (M1) for correct substitution into trig formula.

\(\left( {{\text{VM}} = } \right)\,\,186.200\,({\text{m}})\)        (A1)

\(\left( {{\text{VM}} = } \right)\,\,186\,({\text{m}})\)           (AG)

Note: Both the rounded and unrounded answer must be seen for the final (A1) to be awarded.

a.

\({\text{V}}{{\text{O}}^2} + {115^2} = {186^2}\)  OR \(\sqrt {{{186}^2} – {{115}^2}} \)       (M1)

Note: Award (M1) for correct substitution into Pythagoras formula. Accept alternative methods.

\({\text{(VO}} = )\,\,146\,({\text{m}})\,\,(146.188…)\)       (A1)(G2)

Note: Use of full calculator display for \({\text{VM}}\) gives \(146.443…\,{\text{(m)}}\).

b.

Units are required in part (c)

\(\frac{1}{3}({230^2} \times 146.188…)\)       (M1)

Note: Award (M1) for correct substitution in volume formula. Follow through from part (b).

\( = 2\,580\,000\,{{\text{m}}^3}\,\,(2\,577\,785…\,{{\text{m}}^3})\)       (A1)(ft)(G2)

Note: The answer is \(2\,580\,000\,{{\text{m}}^3}\) , the units are required. Use of \({\text{OV}} = 146.442\) gives  \(2582271…\,{{\text{m}}^3}\)

Use of \({\text{OV}} = 146\) gives  \(2574466…\,{{\text{m}}^3}.\)

c.

\(2.58 \times {10^6}\,({{\text{m}}^3})\)       (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for \(2.58\) and (A1)(ft) for \( \times {10^6}.\,\)

Award (A0)(A0) for answers of the type: \(2.58 \times {10^5}\,({{\text{m}}^3}).\)

Follow through from part (c).

d.

the volume of a wall would be \(430\,000 \times 5 \times 1\)       (M1)

Note: Award (M1) for correct substitution into volume formula.

\(2150000\,({{\text{m}}^3})\)       (A1)(G2)

which is less than the volume of the pyramid       (R1)(ft)

Ahmad is correct.       (A1)(ft)

OR

the length of the wall would be \(\frac{{{\text{their part (c)}}}}{{5 \times 1 \times 1000}}\)       (M1)

Note: Award (M1) for dividing their part (c) by \(5000.\)

\(516\,({\text{km}})\)          (A1)(ft)(G2)

which is more than the distance from Paris to Amsterdam       (R1)(ft)

Ahmad is correct.       (A1)(ft)

Note: Do not award final (A1) without an explicit comparison. Follow through from part (c) or part (d). Award (R1) for reasoning that is consistent with their working in part (e); comparing two volumes, or comparing two lengths.

e.

Units are required in part (f)(ii).

i)     \({\text{A}}{{\text{W}}^2} = {160^2} + {230^2} – 2 \times 160 \times 230 \times \cos \,(15^\circ )\)       (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution.

\({\text{AW}} = 86.1\,({\text{m}})\,\,\,(86.0689…)\)       (A1)(G2)

Note: Award (M0)(A0)(A0) if \({\text{BAW}}\) or \({\text{AWB}}\)  is considered to be a right angled triangle.

ii)    \({\text{area}} = \frac{1}{2} \times 230 \times 160 \times \sin \,(15^\circ )\)         (M1)(A1)

Note: Award (M1) for substitution into area formula, (A1) for correct substitutions.

\( = 4760\,{{\text{m}}^2}\,\,\,(4762.27…\,{{\text{m}}^2})\)       (A1)(G2)

Note: The answer is \(4760\,{{\text{m}}^2}\) , the units are required.

f.

Question

A playground, when viewed from above, is shaped like a quadrilateral, \({\text{ABCD}}\), where \({\text{AB}} = 21.8\,{\text{m}}\) and \({\text{CD}} = 11\,{\text{m}}\) . Three of the internal angles have been measured and angle \({\text{ABC}} = 47^\circ \) , angle \({\text{ACB}} = 63^\circ \) and angle \({\text{CAD}} = 30^\circ \) . This information is represented in the following diagram.

Calculate the distance \({\text{AC}}\).[3]

a.

Calculate angle \({\text{ADC}}\).[3]

b.

There is a tree at \({\text{C}}\), perpendicular to the ground. The angle of elevation to the top of the tree from \({\text{D}}\) is \(35^\circ \).

Calculate the height of the tree.[2]

c.

Chavi estimates that the height of the tree is \(6\,{\text{m}}\).

Calculate the percentage error in Chavi’s estimate.[2]

d.

Chavi is celebrating her birthday with her friends on the playground. Her mother brings a \(2\,\,{\text{litre}}\) bottle of orange juice to share among them. She also brings cone-shaped paper cups.

Each cup has a vertical height of \(10\,{\text{cm}}\) and the top of the cup has a diameter of \(6\,{\text{cm}}\).

Calculate the volume of one paper cup.[3]

e.

Calculate the maximum number of cups that can be completely filled with the \(2\,\,{\text{litre}}\) bottle of orange juice.[3]

f.
Answer/Explanation

Markscheme

\(\frac{{21.8}}{{\sin 63^\circ }} = \frac{{{\text{AC}}}}{{\sin 47^\circ }}\)          (M1)(A1)

Note: Award (M1) for substitution into the sine rule formula, (A1) for correct substitution.

\(({\text{AC}} = )\,\,17.9\,({\text{m}})\,\,\,(17.8938…\,({\text{m}}))\)           (A1)(G2)

a.

\(\frac{{11}}{{\sin 30}} = \frac{{17.8938…}}{{\sin {\text{ADC}}}}\)          (M1)(A1)(ft)

Note: Award (M1) for substitution into the sine rule formula, (A1) for correct substitution.

\(\left( {{\text{Angle ADC}} = } \right)\,\,\,54.4^\circ \,\,\,(54.4250…^\circ )\)          (A1)(ft)(G2)

Note: Accept \(54.5\,\,(54.4527…)\) or \(126\,\,(125.547…)\) from using their \(3\) sf answer.
Follow through from part (a). Accept \(125.575…\)

b.

\(11 \times \tan 35^\circ \)  (or equivalent)          (M1)

Note: Award (M1) for correct substitution into trigonometric ratio.

\(7.70\,({\text{m}})\,\,\,(7.70228…\,({\text{m}}))\)          (A1)(G2)

c.

\(\left| {\frac{{6 – 7.70228…}}{{7.70228…}}} \right| \times 100\,\% \)          (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

OR

\(100 – \left| {\frac{{6 \times 100}}{{7.70228…}}} \right|\)          (M1)

Note: Award (M1) for the alternative method.

\(22.1\,(\% )\,\,\,(22.1009…\,(\% ))\)          (A1)(ft)(G2)

Note: Award at most (M1)(A0) for a final answer that is negative. Follow through from part (c).

d.

\(\frac{1}{3}\pi  \times {3^2} \times 10\)          (A1)(M1)

Note: Award (A1) for \(3\) seen, (M1) for their correct substitution into volume of a cone formula.

\(94.2\,{\text{c}}{{\text{m}}^3}\,\,\,(30\pi \,{\text{c}}{{\text{m}}^3},\,\,94.2477…\,{\text{c}}{{\text{m}}^3})\)          (A1)(G3)

Note: The answer is \(94.2\,{\text{c}}{{\text{m}}^3}\), units are required. Award at most (A0)(M1)(A0) if an incorrect value for \(r\) is used.

e.

\(\frac{{2000}}{{94.2477…}}\) OR \(\frac{2}{{0.0942477…}}\)          (M1)(M1)

Note: Award (M1) for correct conversion (litres to \({\text{c}}{{\text{m}}^3}\) or \({\text{c}}{{\text{m}}^3}\) to litres), (M1) for dividing by their part (e) (or their converted part (e)).

\(21\)          (A1)(ft)(G2)

Note: The final (A1) is not awarded if the final answer is not an integer. Follow through from part (e), but only if the answer is rounded down.

f.

Question

The base of an electric iron can be modelled as a pentagon ABCDE, where:

\[\begin{array}{*{20}{l}} {{\text{BCDE is a rectangle with sides of length }}(x + 3){\text{ cm and }}(x + 5){\text{ cm;}}} \\ {{\text{ABE is an isosceles triangle, with AB}} = {\text{AE and a height of }}x{\text{ cm;}}} \\ {{\text{the area of ABCDE is 222 c}}{{\text{m}}^{\text{2}}}{\text{.}}} \end{array}\]

M17/5/MATSD/SP2/ENG/TZ1/02

Insulation tape is wrapped around the perimeter of the base of the iron, ABCDE.

F is the point on AB such that \({\text{BF}} = {\text{8 cm}}\). A heating element in the iron runs in a straight line, from C to F.

Write down an equation for the area of ABCDE using the above information.[2]

a.i.

Show that the equation in part (a)(i) simplifies to \(3{x^2} + 19x – 414 = 0\).[2]

a.ii.

Find the length of CD.[2]

b.

Show that angle \({\rm{B\hat AE}} = 67.4^\circ \), correct to one decimal place.[3]

c.

Find the length of the perimeter of ABCDE.[3]

d.

Calculate the length of CF.[4]

e.
Answer/Explanation

Markscheme

\(222 = \frac{1}{2}x(x + 3) + (x + 3)(x + 5)\)     (M1)(M1)(A1)

Note:     Award (M1) for correct area of triangle, (M1) for correct area of rectangle, (A1) for equating the sum to 222.

OR

\(222 = (x + 3)(2x + 5) – 2\left( {\frac{1}{4}} \right)x(x + 3)\)     (M1)(M1)(A1)

Note:     Award (M1) for area of bounding rectangle, (M1) for area of triangle, (A1) for equating the difference to 222.[2 marks]

a.i.

\(222 = \frac{1}{2}{x^2} + \frac{3}{2}x + {x^2} + 3x + 5x + 15\)     (M1)

Note:     Award (M1) for complete expansion of the brackets, leading to the final answer, with no incorrect working seen. The final answer must be seen to award (M1).

\(3{x^2} + 19x – 414 = 0\)     (AG)[2 marks]

a.ii.

\(x = 9{\text{ }}\left( {{\text{and }}x =  – \frac{{46}}{3}} \right)\)     (A1)

\({\text{CD}} = 12{\text{ (cm)}}\)     (A1)(G2)[2 marks]

b.

\(\frac{1}{2}({\text{their }}x + 3) = 6\)     (A1)(ft)

Note:     Follow through from part (b).

\(\tan \left( {\frac{{{\rm{B\hat AE}}}}{2}} \right) = \frac{6}{9}\)     (M1)

Note:     Award (M1) for their correct substitutions in tangent ratio.

\({\rm{B\hat AE}} = 67.3801 \ldots ^\circ \)     (A1)

\( = 67.4^\circ \)     (AG)

Note:     Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.

OR

\(\frac{1}{2}({\text{their }}x + 3) = 6\)     (A1)(ft)

\(\tan ({\rm{A\hat BE}}) = \frac{9}{6}\)     (M1)

Note:     Award (M1) for their correct substitutions in tangent ratio.

\({\rm{B\hat AE}} = 180^\circ  – 2({\rm{A\hat BE}})\)

\({\rm{B\hat AE}} = 67.3801 \ldots ^\circ \)     (A1)

\( = 67.4^\circ \)     (AG)

Note:     Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.[3 marks]

c.

\(2\sqrt {{9^2} + {6^2}}  + 12 + 2(14)\)     (M1)(M1)

Note:     Award (M1) for correct substitution into Pythagoras. Award (M1) for the addition of 5 sides of the pentagon, consistent with their \(x\).

\(61.6{\text{ (cm) }}\left( {61.6333 \ldots {\text{ (cm)}}} \right)\)     (A1)(ft)(G3)

Note:     Follow through from part (b).[3 marks]

d.

\({\rm{F\hat BC}} = 90 + \left( {\frac{{180 – 67.4}}{2}} \right){\text{ }}( = 146.3^\circ )\)     (M1)

OR

\(180 – \frac{{67.4}}{2}\)     (M1)

\({\rm{C}}{{\text{F}}^2} = {8^2} + {14^2} – 2(8)(14)\cos (146.3^\circ )\)     (M1)(A1)(ft)

Note:     Award (M1) for substituted cosine rule formula and (A1) for correct substitutions. Follow through from part (b).

\({\text{CF}} = 21.1{\text{ (cm) }}(21.1271 \ldots )\)     (A1)(ft)(G3)

OR

\({\rm{G\hat BF}} = \frac{{67.4}}{2} = 33.7^\circ \)     (A1)

Note:     Award (A1) for angle \({\rm{G\hat BF}} = 33.7^\circ \), where G is the point such that CG is a projection/extension of CB and triangles BGF and CGF are right-angled triangles. The candidate may use another variable.

M17/5/MATSD/SP2/ENG/TZ1/02.e/M

\({\text{GF}} = 8\sin 33.7^\circ  = 4.4387 \ldots \)\(\,\,\,\)AND\(\,\,\,\)\({\text{BG}} = 8\cos 33.7^\circ  = 6.6556 \ldots \)     (M1)

Note:     Award (M1) for correct substitution into trig formulas to find both GF and BG.

\({\text{C}}{{\text{F}}^2} = {(14 + 6.6556 \ldots )^2} + {(4.4387 \ldots )^2}\)     (M1)

Note:     Award (M1) for correct substitution into Pythagoras formula to find CF.

\({\text{CF}} = 21.1{\text{ (cm) }}(21.1271 \ldots )\)     (A1)(ft)(G3)[4 marks]

e.

Question

Farmer Brown has built a new barn, on horizontal ground, on his farm. The barn has a cuboid base and a triangular prism roof, as shown in the diagram.

The cuboid has a width of 10 m, a length of 16 m and a height of 5 m.
The roof has two sloping faces and two vertical and identical sides, ADE and GLF.
The face DEFL slopes at an angle of 15° to the horizontal and ED = 7 m .

The roof was built using metal supports. Each support is made from five lengths of metal AE, ED, AD, EM and MN, and the design is shown in the following diagram.

ED = 7 m , AD = 10 m and angle ADE = 15° .
M is the midpoint of AD.
N is the point on ED such that MN is at right angles to ED.

Farmer Brown believes that N is the midpoint of ED.

Calculate the area of triangle EAD.[3]

a.

Calculate the total volume of the barn.[3]

b.

Calculate the length of MN.[2]

c.

Calculate the length of AE.[3]

d.

Show that Farmer Brown is incorrect.[3]

e.

Calculate the total length of metal required for one support.[4]

f.
Answer/Explanation

Markscheme

(Area of EAD =) \(\frac{1}{2} \times 10 \times 7 \times {\text{sin}}15\)    (M1)(A1)

Note: Award (M1) for substitution into area of a triangle formula, (A1) for correct substitution. Award (M0)(A0)(A0) if EAD or AED is considered to be a right-angled triangle.

= 9.06 m2  (9.05866… m2)     (A1)   (G3)[3 marks]

a.

(10 × 5 × 16) + (9.05866… × 16)     (M1)(M1)

Note: Award (M1) for correct substitution into volume of a cuboid, (M1) for adding the correctly substituted volume of their triangular prism.

= 945 m3  (944.938… m3)     (A1)(ft)  (G3)

Note: Follow through from part (a).[3 marks]

b.

\(\frac{{{\text{MN}}}}{5} = \,\,\,{\text{sin}}15\)     (M1)

Note: Award (M1) for correct substitution into trigonometric equation.

(MN =) 1.29(m) (1.29409… (m))     (A1) (G2)[2 marks]

c.

(AE2 =) 102 + 72 − 2 × 10 × 7 × cos 15     (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, and (A1) for correct substitution.

(AE =) 3.71(m)  (3.71084… (m))     (A1) (G2)[3 marks]

d.

ND2 = 52 − (1.29409…)2     (M1)

Note: Award (M1) for correct substitution into Pythagoras theorem.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

\(\frac{{1.29409 \ldots }}{{{\text{ND}}}} = {\text{tan}}\,15^\circ \)     (M1)

Note: Award (M1) for correct substitution into tangent.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

\(\frac{{{\text{ND}}}}{5} = {\text{cos }}15^\circ \)     (M1)

Note: Award (M1) for correct substitution into cosine.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

ND2 = 1.29409…2 + 52 − 2 × 1.29409… × 5 × cos 75°     (M1)

Note: Award (M1) for correct substitution into cosine rule.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

4.82962… ≠ 3.5   (ND ≠ 3.5)     (R1)(ft)

OR

4.82962… ≠ 2.17038…   (ND ≠ NE)     (R1)(ft)

(hence Farmer Brown is incorrect)

Note: Do not award (M0)(A0)(R1)(ft). Award (M0)(A0)(R0) for a correct conclusion without any working seen.[3 marks]

e.

(EM2 =) 1.29409…2 + (7 − 4.82962…)2     (M1)

Note: Award (M1) for their correct substitution into Pythagoras theorem.

OR

(EM2 =) 52 + 72 − 2 × 5 × 7 × cos 15     (M1)

Note: Award (M1) for correct substitution into cosine rule formula.

(EM =) 2.53(m) (2.52689…(m))     (A1)(ft) (G2)(ft)

Note: Follow through from parts (c), (d) and (e).

(Total length =) 2.52689… + 3.71084… + 1.29409… +10 + 7     (M1)

Note: Award (M1) for adding their EM, their parts (c) and (d), and 10 and 7.

= 24.5 (m)    (24.5318… (m))     (A1)(ft) (G4)

Note: Follow through from parts (c) and (d).[4 marks]

f.

Question

The Tower of Pisa is well known worldwide for how it leans.

Giovanni visits the Tower and wants to investigate how much it is leaning. He draws a diagram showing a non-right triangle, ABC.

On Giovanni’s diagram the length of AB is 56 m, the length of BC is 37 m, and angle ACB is 60°. AX is the perpendicular height from A to BC.

Giovanni’s tourist guidebook says that the actual horizontal displacement of the Tower, BX, is 3.9 metres.

Use Giovanni’s diagram to show that angle ABC, the angle at which the Tower is leaning relative to the
horizontal, is 85° to the nearest degree.[5]

a.i.

Use Giovanni’s diagram to calculate the length of AX.[2]

a.ii.

Use Giovanni’s diagram to find the length of BX, the horizontal displacement of the Tower.[2]

a.iii.

Find the percentage error on Giovanni’s diagram.[2]

b.

Giovanni adds a point D to his diagram, such that BD = 45 m, and another triangle is formed.

Find the angle of elevation of A from D.[3]

c.
Answer/Explanation

Markscheme

\(\frac{{{\text{sin BAC}}}}{{37}} = \frac{{{\text{sin 60}}}}{{56}}\)    (M1)(A1)

Note: Award (M1) for substituting the sine rule formula, (A1) for correct substitution.

angle \({\text{B}}\mathop {\text{A}}\limits^ \wedge  {\text{C}}\) = 34.9034…°    (A1)

Note: Award (A0) if unrounded answer does not round to 35. Award (G2) if 34.9034… seen without working.

angle \({\text{A}}\mathop {\text{B}}\limits^ \wedge  {\text{C}}\) = 180 − (34.9034… + 60)     (M1)

Note: Award (M1) for subtracting their angle BAC + 60 from 180.

85.0965…°    (A1)

85°     (AG)

Note: Both the unrounded and rounded value must be seen for the final (A1) to be awarded. If the candidate rounds 34.9034…° to 35° while substituting to find angle \({\text{A}}\mathop {\text{B}}\limits^ \wedge  {\text{C}}\), the final (A1) can be awarded but only if both 34.9034…° and 35° are seen.
If 85 is used as part of the workings, award at most (M1)(A0)(A0)(M0)(A0)(AG). This is the reverse process and not accepted.

a.i.

sin 85… × 56     (M1)

= 55.8 (55.7869…) (m)     (A1)(G2)

Note: Award (M1) for correct substitution in trigonometric ratio.

a.ii.

\(\sqrt {{{56}^2} – 55.7869{ \ldots ^2}} \)     (M1)

Note: Award (M1) for correct substitution in the Pythagoras theorem formula. Follow through from part (a)(ii).

OR

cos(85) × 56     (M1)

Note: Award (M1) for correct substitution in trigonometric ratio.

= 4.88 (4.88072…) (m)     (A1)(ft)(G2)

Note: Accept 4.73 (4.72863…) (m) from using their 3 s.f answer. Accept equivalent methods.[2 marks]

a.iii.

\(\left| {\frac{{4.88 – 3.9}}{{3.9}}} \right| \times 100\)     (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

= 25.1  (25.1282) (%)     (A1)(ft)(G2)

Note: Follow through from part (a)(iii).[2 marks]

b.

\({\text{ta}}{{\text{n}}^{ – 1}}\left( {\frac{{55.7869 \ldots }}{{40.11927 \ldots }}} \right)\)     (A1)(ft)(M1)

Note: Award (A1)(ft) for their 40.11927… seen. Award (M1) for correct substitution into trigonometric ratio.

OR

(37 − 4.88072…)2 + 55.7869…2

(AC =) 64.3725…

64.3726…2 + 82 − 2 × 8 × 64.3726… × cos120

(AD =) 68.7226…

\(\frac{{{\text{sin 120}}}}{{68.7226 \ldots }} = \frac{{{\text{sin A}}\mathop {\text{D}}\limits^ \wedge  {\text{C}}}}{{64.3725 \ldots }}\)    (A1)(ft)(M1)

Note: Award (A1)(ft) for their correct values seen, (M1) for correct substitution into the sine formula.

= 54.3°  (54.2781…°)     (A1)(ft)(G2)

Note: Follow through from part (a). Accept equivalent methods.[3 marks]

c.

Question

An office block, ABCPQR, is built in the shape of a triangular prism with its “footprint”, ABC, on horizontal ground. \({\text{AB}} = 70{\text{ m}}\), \({\text{BC}} = 50{\text{ m}}\) and \({\text{AC}} = 30{\text{ m}}\). The vertical height of the office block is \(120{\text{ m}}\) .

Calculate the size of angle ACB.[3]

a.

Calculate the area of the building’s footprint, ABC.[3]

b.

Calculate the volume of the office block.[2]

c.

To stabilize the structure, a steel beam must be made that runs from point C to point Q.

Calculate the length of CQ.[2]

d.

Calculate the angle CQ makes with BC.[2]

e.
Answer/Explanation

Markscheme

\(\cos {\text{ACB}} = \frac{{{{30}^2} + {{50}^2} – {{70}^2}}}{{2 \times 30 \times 50}}\)     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula, (A1) for correct substitution.

\({\text{ACB}} = {120^ \circ }\)     (A1)(G2)

a.

\({\text{Area of triangle ABC}} = \frac{{30(50)\sin {{120}^ \circ }}}{2}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted area formula, (A1)(ft) for correct substitution.

\( = 650{\text{ }}{{\text{m}}^2}\) \((649.519 \ldots {\text{ }}{{\text{m}}^2})\)     (A1)(ft)(G2)

Notes: The answer is \(650{\text{ }}{{\text{m}}^2}\) ; the units are required. Follow through from their answer in part (a).

b.

\({\text{Volume}} = 649.519 \ldots \times 120\)     (M1)
\( = 77900{\text{ }}{{\text{m}}^3}\) (\(77942.2 \ldots {\text{ }}{{\text{m}}^3}\))     (A1)(G2)

Note: The answer is \(77900{\text{ }}{{\text{m}}^3}\) ; the units are required. Do not penalise lack of units if already penalized in part (b). Accept \(78000{\text{ }}{{\text{m}}^3}\) from use of 3sf answer \(650{\text{ }}{{\text{m}}^2}\) from part (b).

c.

\({\text{C}}{{\text{Q}}^2} = {50^2} + {120^2}\)     (M1)
\({\text{CQ}} = 130{\text{ (m)}}\)     (A1)(G2)

Note: The units are not required.

d.

\(\tan {\text{QCB}} = \frac{{120}}{{50}}\)     (M1)

Note: Award (M1) for correct substituted trig formula.

\({\text{QCB}} = {67.4^ \circ }\) (\(67.3801 \ldots \))     (A1)(G2)

Note: Accept equivalent methods.

e.
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