IB DP Mathematical Studies 5.3 Paper 2

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

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.
Answer/Explanation

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.

Question

The quadrilateral ABCD shown below represents a sandbox. AB and BC have the same length. AD is \(9{\text{ m}}\) long and CD is \(4.2{\text{ m}}\) long. Angles ADC and ABC are \({95^ \circ }\) and \({130^ \circ }\) respectively.

Find the length of AC.[3]

a.

(i)     Write down the size of angle BCA.

(ii)    Calculate the length of AB.[4]

b.

Show that the area of the sandbox is \(31.1{\text{ }}{{\text{m}}^2}\) correct to 3 s.f.[4]

c.

The sandbox is a prism. Its edges are \(40{\text{ cm}}\) high. The sand occupies one third of the volume of the sandbox. Calculate the volume of sand in the sandbox.[3]

d.
Answer/Explanation

Markscheme

\({\text{A}}{{\text{C}}^2} = {9^2} + {4.2^2} – 2 \times 9 \times 4.2 \times \cos {95^ \circ }\)     (M1)(A1)
\({\text{AC}} = 10.3{\text{ m}}\)     (A1)(G2)

Note: (M1) for correct substituted formula and (A1) for correct substitution. If radians used answer is \(6.59\). Award at most (M1)(A1)(A0).

Note: The final A1 is only awarded if the correct units are present; only penalize once for the lack of units or incorrect units.

a.

(i)     \({\text{B}}\hat{\text{C}}{\text{A}} = {25^ \circ }\)     (A1)

(ii)    \(\frac{{{\text{AB}}}}{{\sin {{25}^ \circ }}} = \frac{{10.258 \ldots }}{{\sin {{130}^ \circ }}}\)     (M1)(A1)

\({\text{AB}} = 5.66{\text{ m}}\)     (A1)(ft)(G2)

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

Follow through with angle \({\text{B}}\)\(\hat{\text{C}}\)\({\text{A}}\) and their AC. Allow \({\text{AB}} = 5.68\) if \({\text{AC}} = 10.3\) used. If radians used answer is \(0.938\) (unreasonable answer). Award at most (M1)(A1)(A0)(ft).

OR

Using that ABC is isosceles

\({\text{cos2}}{{\text{5}}^ \circ } = \frac{{\frac{1}{2} \times 10.258 \ldots }}{{{\text{AB}}}}\) (or equivalent)     (A1)(M1)(ft)

\({\text{AB}} = 5.66{\text{ m}}\)     (A1)(ft)(G2)

Note: (A1) for \(\frac{1}{2}\) of their AB seen, (M1) for correct trigonometric ratio and correct substitution, (A1) for correct answer. If \(\frac{1}{2}{\text{AB}}\) seen and correct answer is given award (A1)(G1). Allow \({\text{AB}} = 5.68\) if \({\text{AC}} = 10.3\) used. If radians used answer is \(3.32\). Award (A1)(M1)(A1)(ft). If \(\sin 65\) and radians used answer is \(3.99\). Award (A1)(M1)(A1)(ft).

Note: The final A1 is only awarded in (ii) if the correct units are present; only penalize once for the lack of units or incorrect units.

b.

Area \( = \frac{1}{2} \times 9 \times 4.2 \times \sin {95^ \circ } + \frac{1}{2} \times {(5.6592 \ldots )^2} \times \sin {130^ \circ }\)    (M1)(M1)(ft)(M1)

\( = 31.095 \ldots  = 31.1{\text{ }}{{\text{m}}^2}\) (correct to 3 s.f.)     (A1)(AG)

Note: (M1)(M1) each for correct substitution in the formula of the area of each triangle, (M1) for adding both areas. (A1) for unrounded answer. Follow through with their length of AB but last mark is lost if they do not reach the correct answer.

c.

Volume of sand \( = \frac{1}{3}(31.09 \ldots  \times 0.4)\)     (M1)(M1)

\( = 4.15{\text{ }}{{\text{m}}^3}\)     (A1)(G2)

Note: (M1) for correct formula of volume of prism and for correct substitution, (M1) for multiplying by \(\frac{1}{3}\) and last (A1) for correct answer only.

Note: The final A1 is only awarded if the correct units are present; only penalize once for the lack of units or incorrect units.

d.

Question

A chocolate bar has the shape of a triangular right prism ABCDEF as shown in the diagram. The ends are equilateral triangles of side 6 cm and the length of the chocolate bar is 23 cm.

Write down the size of angle BAF.[1]

a, i.

Hence or otherwise find the area of the triangular end of the chocolate bar.[3]

a, ii.

Find the total surface area of the chocolate bar.[3]

b.

It is known that 1 cm3 of this chocolate weighs 1.5 g. Calculate the weight of the chocolate bar.[3]

c.

A different chocolate bar made with the same mixture also has the shape of a triangular prism. The ends are triangles with sides of length 4 cm, 6 cm and 7 cm.

Show that the size of the angle between the sides of 6 cm and 4 cm is 86.4° correct to 3 significant figures.[3]

d.

The weight of this chocolate bar is 500 g. Find its length.[4]

e.
Answer/Explanation

Markscheme

60°     (A1)[1 mark]

a, i.

Unit penalty (UP) applies in this part

\({\text{Area}} = \frac{{6 \times 6 \times \sin 60^\circ }}{2}\)     (M1)(A1)

(UP)     = 15.6 cm2   \((9 \sqrt{3})\)     (A1)(ft)(G2) 

Note: Award (M1) for substitution into correct formula, (A1) for correct values. Accept alternative correct methods.[3 marks]

a, ii.

Unit penalty (UP) applies in this part 

\({\text{Surface Area}} =15.58 \times 2 + 23 \times 6 \times 3\)     (M1)(M1)

Note: Award (M1) for two terms with 2 and 3 respectively, (M1) for \(23 \times 6\) (138).

(UP)     Surface Area = 445 cm2     (A1)(ft)(G2)[3 marks]

b.

Unit penalty (UP) applies in this part

\({\text{weight}} = 1.5 \times 15.59 \times 23\)     (M1)(M1)

Note: Award (M1) for finding the volume, (M1) for multiplying their volume by 1.5.

(UP)     weight = 538 g     (A1)(ft)(G3)[3 marks]

c.

\(\cos \alpha  = \frac{{{4^2} + {6^2} – {7^2}}}{{2 \times 4 \times 6}}\)     (M1)(A1)

Note: Award (M1) for using cosine rule with values from the problem, (A1) for correct substitution.

\(\alpha = 86.41…\)     (A1)

\(\alpha = 86.4^{\circ}\)     (AG) 

Note: 86.41… must be seen for final (A1) to be awarded.[3 marks]

d.

Unit penalty (UP) applies in this part

\(l \times \frac{{4 \times 6 \times \sin 86.4^\circ }}{2} \times 1.5 = 500\)     (M1)(A1)(M1)

Notes: Award (M1) for finding an expression for the volume, (A1) for correct substitution, (M1) for multiplying the volume by 1.5 and equating to 500, or for equating the volume to \(\frac{500}{1.5}\).

If formula for volume is not correct but consistent with that in (c) award at most (M1)(A0)(ft)(M1)(A0).

(UP)     l = 27.8 cm     (A1)(G3)[4 marks]

e.

Question

A farmer has a triangular field, ABC, as shown in the diagram.

AB = 35 m, BC = 80 m and BÂC = 105°, and D is the midpoint of BC.

Find the size of BĈA.[3]

a.

Calculate the length of AD.[5]

b.

The farmer wants to build a fence around ABD.

Calculate the total length of the fence.[2]

c.

The farmer wants to build a fence around ABD.

The farmer pays 802.50 USD for the fence. Find the cost per metre.[2]

d.

Calculate the area of the triangle ABD.[3]

e.

A layer of earth 3 cm thick is removed from ABD. Find the volume removed in cubic metres.[3]

f.
Answer/Explanation

Markscheme

\(\frac{{\sin {\text{BCA}}}}{{35}} = \frac{{\sin 105^\circ }}{{80}}\)     (M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

\({\text{B}}{\operatorname{\hat C}}{\text{A}} = 25.0^{\circ}\)     (A1)(G2)[3 marks]

a.

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Length BD = 40 m     (A1)

Angle ABC = 180° − 105° − 25° = 50°     (A1)(ft)

Note: (ft) from their answer to (a).

AD2 = 352 + 402 − (2 × 35 × 40 × cos 50°)     (M1)(A1)(ft)

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

(UP)     AD = 32.0 m     (A1)(ft)(G3)

Notes: If 80 is used for BD award at most (A0)(A1)(ft)(M1)(A1)(ft)(A1)(ft) for an answer of 63.4 m.

If the angle ABC is incorrectly calculated in this part award at most (A1)(A0)(M1)(A1)(ft)(A1)(ft).

If angle BCA is used award at most (A1)(A0)(M1)(A0)(A0).[5 marks]

b.

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

length of fence = 35 + 40 + 32     (M1)

(UP)     = 107 m     (A1)(ft)(G2)

Note: (M1) for adding 35 + 40 + their (b).[2 marks]

c.

cost per metre \( = \frac{802.50}{107}\)     (M1)

Note: Award (M1) for dividing 802.50 by their (c).

cost per metre = 7.50 USD (7.5 USD) (USD not required)     (A1)(ft)(G2)[2 marks]

d.

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Area of ABD \( = \frac{1}{2} \times 35 \times 40 \times \sin 50^\circ \)     (M1)

= 536.2311102     (A1)(ft)

(UP)     = 536 m2     (A1)(ft)(G2)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitution, (ft) from their value of BD and their angle ABC in (b).[3 marks]

e.

Volume = 0.03 × 536     (A1)(M1)

= 16.08

= 16.1     (A1)(ft)(G2)

Note: Award (A1) for 0.03, (M1) for correct formula. (ft) from their (e).

If 3 is used award at most (A0)(M1)(A0).[3 marks]

f.

Question

The diagram below shows a square based right pyramid. ABCD is a square of side 10 cm. VX is the perpendicular height of 8 cm. M is the midpoint of BC.

In a mountain region there appears to be a relationship between the number of trees growing in the region and the depth of snow in winter. A set of 10 areas was chosen, and in each area the number of trees was counted and the depth of snow measured. The results are given in the table below.

A path goes around a forest so that it forms the three sides of a triangle. The lengths of two sides are 550 m and 290 m. These two sides meet at an angle of 115°. A diagram is shown below.

Write down the length of XM.[1]

A, a.

Use your graphic display calculator to find the standard deviation of the number of trees.[1]

A, a, ii.

Calculate the length of VM.[2]

A, b.

Calculate the angle between VM and ABCD.[2]

A, c.

Calculate the length of the third side of the triangle. Give your answer correct to the nearest 10 m.[4]

B, a.

Calculate the area enclosed by the path that goes around the forest.[3]

B, b.

Inside the forest a second path forms the three sides of another triangle named ABC. Angle BAC is 53°, AC is 180 m and BC is 230 m.

Calculate the size of angle ACB.[4]

B, c.
Answer/Explanation

Markscheme

UP applies in this question

(UP)     XM = 5 cm     (A1)[1 mark]

A, a.

16.8     (G1)[1 mark]

A, a, ii.

UP applies in this question

VM2 = 52 + 82     (M1)

Note: Award (M1) for correct use of Pythagoras Theorem.

(UP)     VM = \(\sqrt{89}\) = 9.43 cm     (A1)(ft)(G2)[2 marks]

A, b.

\(\tan {\text{VMX}} = \frac{8}{5}\)     (M1)

Note: Other trigonometric ratios may be used.

\({\rm{V\hat MX}} = 58.0^\circ \)     (A1)(ft)(G2)[2 marks]

A, c.

UP applies in this question

l2 = 2902 + 5502 − 2 × 290 × 550 × cos115°     (M1)(A1)

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

l = 722     (A1)(G2)

(UP)     = 720 m     (A1)

Note: If 720 m seen without working award (G3).

The final (A1) is awarded for the correct rounding of their answer.[4 marks]

B, a.

UP applies in this question

\({\text{Area}} = \frac{1}{2} \times 290 \times 550 \times \sin 115\)     (M1)(A1)

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

(UP)     = \(72\,300{\text{ }}{{\text{m}}^2}\)     (A1)(G2)[3 marks]

B, b.

\(\frac{{180}}{{\sin {\text{B}}}} = \frac{{230}}{{\sin 53}}\)     (M1)(A1)

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

B = 38.7°     (A1)(G2)

\({\operatorname{A\hat CB}} = 180 – (53^\circ  + 38.7^\circ )\)

\( = 88.3^\circ \)     (A1)(ft) [4 marks]

B, c.

Question

The diagram shows triangle ABC. Point C has coordinates (4, 7) and the equation of the line AB is x + 2y = 8.

Find the coordinates of A.[1]

a.i.

Find the coordinates of B.[1]

a.ii.

Show that the distance between A and B is 8.94 correct to 3 significant figures.[2]

b.

N lies on the line AB. The line CN is perpendicular to the line AB.

Find the gradient of CN.[3]

c.i.

N lies on the line AB. The line CN is perpendicular to the line AB.

Find the equation of CN.[2]

c.ii.

N lies on the line AB. The line CN is perpendicular to the line AB.

Calculate the coordinates of N.[3]

d.

It is known that AC = 5 and BC = 8.06.

Calculate the size of angle ACB.[3]

e.

It is known that AC = 5 and BC = 8.06.

Calculate the area of triangle ACB.[3]

f.
Answer/Explanation

Markscheme

A(0, 4)     Accept x = 0, y = 4     (A1)[1 mark]

a.i.

B(8, 0)     Accept x = 8, y = 0     (A1)(ft)

Note: Award (A0) if coordinates are reversed in (i) and (A1)(ft) in (ii).[1 mark]

a.ii.

\({\text{AB}} = \sqrt {{8^2} + {4^2}} {\text{  }} = \sqrt {80} \)     (M1)

AB = 8.944     (A1)

= 8.94     (AG)[2 marks]

b.

y = –0.5x + 4     (M1)
Gradient AB = –0.5     (A1)

Note: Award (A2) if –0.5 seen.

OR

Gradient \({\text{AB}} = \frac{{(0 – 4)}}{{(8 – 0)}}\)     (M1)
\( = -\frac{1}{2}\)     (A1)

Note: Award (M1) for correct substitution in the gradient formula. Follow through from their answers to part (a).

Gradient CN = 2     (A1)(ft)(G2)

Note: Special case: Follow through for gradient CN from their gradient AB.[3 marks]

c.i.

CN: y = 2x + c

7 = 2(4) + c     (M1)

Note: Award (M1)for correct substitution in equation of a line.

y = 2x – 1     (A1)(ft)(G2)

Note: Accept alternative forms for the equation of a line including y – 7 = 2(x – 4) . Follow through from their gradient in (i).

Note: If c = –1 seen but final answer is not given, award (A1)(d).[2 marks]

c.ii.

x + 2(2x – 1) = 8     or equivalent     (M1)

N(2, 3) (x = 2, y = 3)     (A1)(A1)(ft)(G3)

Note: Award (M1) for attempt to solve simultaneous equations or a sketch of the two lines with an indication of the point of intersection.[3 marks]

d.

Cosine rule: \(\cos ({\rm{A\hat CB)}} = \frac{{{5^2} + {{8.06}^2} – {{8.944}^2}}}{{2 \times 5 \times 8.06}}\)     (M1)(A1)

Note: Award (M1) for use of cosine rule with numbers from the problem substituted, (A1) for correct substitution.

\({\rm{A\hat CB  =  82.9^\circ }}\)     (A1)(G2)

Note: If alternative right-angled trigonometry method used award (M1) for use of trig ratio in both triangles, (A1) for correct substitution of their values in each ratio, (A1) for answer.

Note: Accept 82.8° with use of 8.94.[3 marks]

e.

Area \({\text{ACB}} = \frac{{5 \times 8.06\sin (82.9)}}{2}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted area formula, (A1) for correct substitution. Follow through from their angle in part (e).

OR

Area \({\text{ACB}} = \frac{{{\text{AB}} \times {\text{CN}}}}{2} = \frac{{8.94 \times \sqrt {{{(4 – 2)}^2} + {{(7 – 3)}^2}} }}{2}\)     (M1)(M1)(ft)

Note: Award (M1) substituted area formula with their values, (M1) for substituted distance formula. Follow through from
coordinates of N.

Area ACB = 20.0     (A1)(ft)(G2)

Note: Accept 20[3 marks]

f.

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

Pauline owns a piece of land ABCD in the shape of a quadrilateral. The length of BC is \(190{\text{ m}}\) , the length of CD is \(120{\text{ m}}\) , the length of AD is \(70{\text{ m}}\) , the size of angle BCD is \({75^ \circ }\) and the size of angle BAD is \({115^ \circ }\) .

Pauline decides to sell the triangular portion of land ABD . She first builds a straight fence from B to D .

Calculate the length of the fence.[3]

a.

The fence costs \(17\) USD per metre to build.

Calculate the cost of building the fence. Give your answer correct to the nearest USD.[2]

b.

Show that the size of angle ABD is \({18.8^ \circ }\) , correct to three significant figures.[3]

c.

Calculate the area of triangle ABD .[4]

d.

She sells the land for \(120\) USD per square metre.

Calculate the value of the land that Pauline sells. Give your answer correct to the nearest USD.[2]

e.

Pauline invests \(300 000\) USD from the sale of the land in a bank that pays compound interest compounded annually.

Find the interest rate that the bank pays so that the investment will double in value in 15 years.[4]

f.
Answer/Explanation

Markscheme

\({\text{B}}{{\text{D}}^2} = {190^2} + {120^2} – 2(190)(120)\cos {75^ \circ }\)     (M1)(A1)

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

\(= 197\) m     (A1)(G2)

Note: If radians are used award a maximum of (M1)(A1)(A0).[3 marks]

a.

\({\text{cost}} = 196.717 \ldots  \times 17\)     (M1)

\( = 3344{\text{ USD}}\)     (A1)(ft)(G2)

Note: Accept \(3349\) from \(197\).[2 marks]

b.

\(\frac{{\sin ({\text{ABD}})}}{{70}} = \frac{{\sin ({{115}^ \circ })}}{{196.7}}\)     (M1)(A1)

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

\( = {18.81^ \circ } \ldots \)     (A1)(ft)
\( = {18.8^ \circ } \)     (AG)

Notes: Both the unrounded and rounded answers must be seen for the final (A1) to be awarded. Follow through from their (a). If 197 is used the unrounded answer is \( = {18.78^ \circ } \ldots \)[3 marks]

c.

\({\text{angle BDA}} = {46.2^ \circ }\)     (A1)
\({\text{Area}} = \frac{{70 \times (196.717 \ldots ) \times \sin ({{46.2}^ \circ })}}{2}\)     (M1)(A1)

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

\({\text{Area ABD}} = 4970{\text{ }}{{\text{m}}^2}\)     (A1)(ft)(G2)

Notes: If \(197\) used answer is \(4980\).

Notes: Follow through from (a) only. Award (G2) if there is no working shown and \({46.2^ \circ }\) not seen. If \({46.2^ \circ }\) seen without subsequent working, award (A1)(G2).[4 marks]

d.

\(4969.38 \ldots  \times 120\)     (M1)
\( = 596 327{\text{ USD}}\)     (A1)(ft)(G2)

Notes: Follow through from their (d).[2 marks]

e.

\(300000{\left( {1 + \frac{r}{{100}}} \right)^{15}} = 600000\) or equivalent     (A1)(M1)(A1)

Notes: Award (A1) for \(600 000\) seen or implied by alternative formula, (M1) for substituted CI formula, (A1) for correct substitutions.

\(r = 4.73\)     (A1)(ft)(G3)

Notes: Award (G3) for \(4.73\) with no working. Award (G2) for \(4.7\) with no working.[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

The diagram shows triangle ABC in which \({\text{AB}} = 28{\text{ cm}}\), \({\text{BC}} = 13{\text{ cm}}\), \({\text{BD}} = 12{\text{ cm}}\) and \({\text{AD}} = 20{\text{ cm}}\).

Calculate the size of angle ADB.[3]

a.

Find the area of triangle ADB.[3]

b.

Calculate the size of angle BCD.[4]

c.

Show that the triangle ABC is not right angled.[4]

d.
Answer/Explanation

Markscheme

\(\cos {\text{ADB}} = \frac{{{{12}^2} + {{20}^2} – {{28}^2}}}{{2(12)(20)}}\)     (M1)(A1)

Notes: Award (M1) for substituted cosine rule formula, (A1) for correct substitutions.

\(\angle {\text{ADB}} = 120\)     (A1)(G2)[3 marks]

a.

\({\text{Area}} = \frac{{(12)(20)\sin {{120}^ \circ }}}{2}\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted area formula, (A1)(ft) for their correct substitutions.

\( = 104{\text{ c}}{{\text{m}}^2}\) (\(103.923 \ldots {\text{ c}}{{\text{m}}^2}\))     (A1)(ft)(G2)

Note: The final answer is \(104{\text{ c}}{{\text{m}}^2}\) , the units are required. Accept \(100{\text{ c}}{{\text{m}}^2}\) .[3 marks]

b.

\(\frac{{\sin {\text{BCD}}}}{{12}} = \frac{{\sin {{60}^ \circ }}}{{13}}\)     (A1)(ft)(M1)(A1)

Note: Award (A1)(ft) for their 60 seen, (M1) for substituted sine rule formula, (A1) for correct substitutions.

\({\text{BCD}} = {53.1^ \circ }\) (\(53.0736 \ldots \))     (A1)(G3)

Note: Accept \(53\), do not accept \(50\) or \(53.0\).[4 marks]

c.

Using triangle ABC

\(\frac{{\sin {\text{BAC}}}}{{13}} = \frac{{\sin {{53.1}^ \circ }}}{{28}}\)     (M1)(A1)(ft)

OR

Using triangle ABD

\(\frac{{\sin {\text{BAD}}}}{{12}} = \frac{{\sin {{120}^ \circ }}}{{28}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula (one of the above), (A1)(ft) for their correct substitutions. Follow through from (a) or (c) as appropriate.

\({\text{BAC}} = {\text{BAD}} = {21.8^ \circ }\) (\(21.7867 \ldots \))     (A1)(ft)(G2)

Notes: Accept \(22\), do not accept \(20\) or \(21.7\). Accept equivalent methods, for example cosine rule.

\({180^ \circ } – ({53.1^ \circ } + {21.8^ \circ }) \ne {90^ \circ }\), hence triangle ABC is not right angled     (R1)(AG)

OR

\(\frac{{{\text{CD}}}}{{\sin {{66.9}^ \circ }}} = \frac{{13}}{{\sin {{60}^ \circ }}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula, (A1)(ft) for their correct substitutions. Follow through from (a) and (c).

\({\text{CD}} = 13.8{\text{ }}(13.8075 \ldots )\)     (A1)(ft)

\({13^3} + {28^2} \ne {33.8^2}\), hence triangle ABC is not right angled.     (R1)(ft)(AG)

Note: The complete statement is required for the final (R1) to be awarded.[4 marks]

d.

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 contractor is building a house. He first marks out three points A , B and C on the ground such that AB = 5 m , AC = 7 m and angle BAC = 112°.

Find the length of BC.[3]

a.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

Find the size of angle DBC .[4]

b.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

 

Find the area of the quadrilateral ABDC.[4]

c.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

The contractor digs up and removes the soil under the quadrilateral ABDC to a depth of 50 cm for the foundation of the house.

Find the volume of the soil removed. Give your answer in m3 .[3]

d.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

The contractor digs up and removes the soil under the quadrilateral ABDC to a depth of 50 cm for the foundation of the house.

To transport the soil removed, the contractor uses cylindrical drums with a diameter of 30 cm and a height of 40 cm. 

(i) Find the volume of a drum. Give your answer in m3 .

(ii) Find the minimum number of drums required to transport the soil removed.[5]

e.
Answer/Explanation

Markscheme

Units are required in part (c) only. 

BC2 = 52 + 72 − 2(5)(7)cos112°     (M1)(A1)

Note: Award (M1) for substitution in cosine formula, (A1) for correct substitutions.

BC = 10.0 (m) (10.0111…)     (A1)(G2)

Note: If radians are used, award at most (M1)(A1)(A0).[3 marks]

a.

Units are required in part (c) only.


\(\frac{{\sin 40^\circ }}{{10.0111…}} = \frac{{\sin {\text{D}}{\operatorname{\hat C}}{\text{B}}}}{6}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution in sine formula, (A1)(ft) for their correct substitutions. Follow through from their part (a).

\({\text{D}}{\operatorname {\hat C}}{\text{B}}\) = 22.7° (22.6589…)     (A1)(ft) 

Notes: Award (A2) for 22.7° seen without working. Use of radians results in unrealistic answer. Award a maximum of (M1)(A1)(ft)(A0)(ft). Follow through from their part (a).

\({\text{D}}{\operatorname {\hat C}}{\text{B}}\) = 117° (117.341…)     (A1)(ft)(G3) 

Notes: Do not penalize if use of radians was already penalized in part (a). Follow through from their answer to part (a).

OR

From use of cosine formula

DC = 13.8(m)   (13.8346…)     (A1)(ft)

Note: Follow through from their answer to part (a).

\(\frac{{\sin \alpha }}{{13.8346…}} = \frac{{\sin 40^\circ }}{{10.0111…}}\)     (M1)

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

α = 62.7°  (62.6589)     (A1)(ft)

Note: Accept 62.5° from use of 3sf.

\({\text{D}}{\operatorname {\hat B}}{\text{C}}\) = 117(117.341…)     (A1)(ft) 

Note: Follow through from their part (a). Use of radians results in unrealistic answer, award a maximum of (A1)(M1)(A0)(A0).[4 marks]

b.

Units are required in part (c) only.

\({\text{ABDC}} = \frac{1}{2}(5)(7)\sin 112^\circ  + \frac{1}{2}(6)(10.0111…)\sin 117.341…^\circ \)     (M1)(A1)(ft)(M1)N

Note: Award (M1) for substitution in both triangle area formulae, (A1)(ft) for their correct substitutions, (M1) for seen or implied addition of their two triangle areas. Follow through from their answer to part (a) and (b).

= 42.9 m2 (42.9039…)     (A1)(ft)(G3)

Notes: Answer is 42.9 m2 i.e. the units are required for the final (A1)(ft) to be awarded. Accept 43.0 m2 from using 3sf answers to parts (a) and (b). Do not penalize if use of radians was previously penalized.[4 marks]

c.

Units are required in part (c) only.

42.9039… × 0.5     (M1)(M1)

Note: Award (M1) for 0.5 seen (or equivalent), (M1) for multiplication of their answer in part (c) with their value for depth.

= 21.5 (m3) (21.4519…)     (A1)(ft)(G3)

Note: Follow through from their part (c) only if working is seen. Do not penalize if use of radians was previously penalized. Award at most (A0)(M1)(A0)(ft) for multiplying by 50.[3 marks]

d.

Units are required in part (c) only.

(i) π(0.15)2(0.4)     (M1)(A1)

OR

π × 152 × 40  (28274.3…)     (M1)(A1)

Notes: Award (M1) for substitution in the correct volume formula. (A1) for correct substitutions.

= 0.0283 (m3) (0.0282743…, 0.09π)

(ii) \(\frac{{21.4519…}}{{0.0282743…}}\)     (M1)

Note: Award (M1) for correct division of their volumes.

= 759     (A1)(ft)(G2)

Notes: Follow through from their parts (d) and (e)(i). Accept 760 from use of 3sf answers. Answer must be a positive integer for the final (A1)(ft) mark to be awarded.[5 marks]

e.

Question

A manufacturer has a contract to make \(2600\) solid blocks of wood. Each block is in the shape of a right triangular prism, \({\text{ABCDEF}}\), as shown in the diagram.

\({\text{AB}} = 30{\text{ cm}},{\text{ BC}} = 24{\text{ cm}},{\text{ CD}} = 25{\text{ cm}}\) and angle \({\rm{A\hat BC}} = 35^\circ {\text{ }}\).

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

a.

Calculate the area of triangle \({\text{ABC}}\).[3]

b.

Assuming that no wood is wasted, show that the volume of wood required to make all \(2600\) blocks is \({\text{13}}\,{\text{400}}\,{\text{000 c}}{{\text{m}}^3}\), correct to three significant figures.[2]

c.

Write \({\text{13}}\,{\text{400}}\,{\text{000}}\) in the form \(a \times {10^k}\) where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\).[2]

d.

Show that the total surface area of one block is \({\text{2190 c}}{{\text{m}}^2}\), correct to three significant figures.[3]

e.

The blocks are to be painted. One litre of paint will cover \(22{\text{ }}{{\text{m}}^2}\).

Calculate the number of litres required to paint all \(2600\) blocks.[3]

f.
Answer/Explanation

Markscheme

\({\text{A}}{{\text{C}}^2} = {30^2} + {24^2} – 2 \times 30 \times 24 \times \cos 35^\circ \)     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula,

     (A1) for correct substitutions.

\({\text{AC}} = 17.2{\text{ cm}}\)   \((17.2168…)\)     (A1)(G2)

Notes: Use of radians gives \(52.7002…\) Award (M1)(A1)(A0).

     No marks awarded in this part of the question where candidates assume that angle \({\text{ACB}} = 90^\circ \).[3 marks]

a.

Units are required in part (b).

Area of triangle \({\text{ABC = }}\frac{1}{2} \times 24 \times 30 \times \sin 35^\circ \)     (M1)(A1)

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

     Special Case: Where a candidate has assumed that angle \({\text{ACB}} = 90^\circ \) in part (a), award (M1)(A1) for a correct alternative substituted formula for the area of the triangle \(\left( {ie{\text{ }}\frac{1}{2} \times {\text{base}} \times {\text{height}}} \right)\).

\( = 206{\text{ c}}{{\text{m}}^2}\)   \((206.487 \ldots {\text{c}}{{\text{m}}^2})\)     (A1)(G2)

Notes: Use of radians gives negative answer, \(–154.145…\) Award (M1)(A1)(A0).

     Special Case: Award (A1)(ft) where the candidate has arrived at an area which is correct to the standard rounding rules from their lengths (units required).[3 marks]

b.

\(206.487… \times 25 \times 2600\)     (M1)

Note: Award (M1) for multiplication of their answer to part (b) by \(25\) and \(2600\).

\({\text{13}}\,{\text{421}}\,{\text{688.61}}\)     (A1)

Note: Accept unrounded answer of \({\text{13}}\,{\text{390}}\,{\text{000}}\) for use of \(206\).

\({\text{13}}\,{\text{400}}\,{\text{000}}\)     (AG)

Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.[2 marks]

c.

\(1.34 \times {10^7}\)     (A2)

Notes: Award (A2) for the correct answer.

     Award (A1)(A0) for \(1.34\) and an incorrect index value.

     Award (A0)(A0) for any other combination (including answers such as \(13.4 \times {10^6}\)).[2 marks]

d.

\(2 \times 206.487 \ldots  + 24 \times 25 + 30 \times 25 + 17.2168 \ldots  \times 25\)     (M1)(M1)

Note: Award (M1) for multiplication of their answer to part (b) by \(2\) for area of two triangular ends, (M1) for three correct rectangle areas using \(24\), \(30\) and their \(17.2\).

\(2193.26…\)     (A1)

Note: Accept \(2192\) for use of 3 sf answers.

\(2190\)     (AG)

Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.[3 marks]

e.

\(\frac{{2190 \times 2600}}{{22 \times 10\,000}}\)     (M1)(M1)

Notes: Award (M1) for multiplication by \(2600\) and division by \(22\), (M1) for division by \({10\,000}\).

     The use of \(22\) may be implied ie division by \(2200\) would be acceptable.

\(25.9\) litres   \((25.8818…)\)     (A1)(G2)

Note: Accept \(26\).[3 marks]

f.

Question

A cross-country running course consists of a beach section and a forest section. Competitors run from \({\text{A}}\) to \({\text{B}}\), then from \({\text{B}}\) to \({\text{C}}\) and from \({\text{C}}\) back to \({\text{A}}\).

The running course from \({\text{A}}\) to \({\text{B}}\) is along the beach, while the course from \({\text{B}}\), through \({\text{C}}\) and back to \({\text{A}}\), is through the forest.

The course is shown on the following diagram.



Angle \({\text{ABC}}\) is \(110^\circ\).

It takes Sarah \(5\) minutes and \(20\) seconds to run from \({\text{A}}\) to \({\text{B}}\) at a speed of \(3.8{\text{ m}}{{\text{s}}^{ – 1}}\).

Using ‘distance = speed \( \times \) time’, show that the distance from \({\text{A}}\) to \({\text{B}}\) is \(1220\) metres correct to 3 significant figures.[2]

a.

The distance from \({\text{B}}\) to \({\text{C}}\) is \(850\) metres. Running this part of the course takes Sarah \(5\) minutes and \(3\) seconds.

Calculate the speed, in \({\text{m}}{{\text{s}}^{ – 1}}\), that Sarah runs from \({\text{B}}\) to \({\text{C}}\).[1]

b.

The distance from \({\text{B}}\) to \({\text{C}}\) is \(850\) metres. Running this part of the course takes Sarah \(5\) minutes and \(3\) seconds.

Calculate the distance, in metres, from \({\mathbf{C}}\) to \({\mathbf{A}}\).[3]

c.

The distance from \({\text{B}}\) to \({\text{C}}\) is \(850\) metres. Running this part of the course takes Sarah \(5\) minutes and \(3\) seconds.

Calculate the total distance, in metres, of the cross-country running course.[2]

d.

The distance from \({\text{B}}\) to \({\text{C}}\) is \(850\) metres. Running this part of the course takes Sarah \(5\) minutes and \(3\) seconds.

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

e.

The distance from \({\text{B}}\) to \({\text{C}}\) is \(850\) metres. Running this part of the course takes Sarah \(5\) minutes and \(3\) seconds.

Calculate the area of the cross-country course bounded by the lines \({\text{AB}}\), \({\text{BC}}\) and \({\text{CA}}\).[3]

f.
Answer/Explanation

Markscheme

\(3.8 \times 320\)     (A1)

Note: Award (A1) for \(320\) or equivalent seen.

\( = 1216\)     (A1)

\( = 1220{\text{ (m)}}\)     (AG)

Note: Both unrounded and rounded answer must be seen for the final (A1) to be awarded.[2 marks]

a.

\(\frac{{850}}{{303}}{\text{ (m}}{{\text{s}}^{ – 1}}){\text{ (2.81, 2.80528}} \ldots {\text{)}}\)     (A1)(G1)[1 mark]

b.

\({\text{A}}{{\text{C}}^2} = {1220^2} + {850^2} – 2(1220)(850)\cos 110^\circ \)     (M1)(A1)

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

\({\text{AC}} = 1710{\text{ (m) (1708.87}} \ldots {\text{)}}\)     (A1)(G2)

Notes: Accept \(1705{\text{ }} (1705.33…)\).[3 marks]

c.

\(1220 + 850 + {\text{1708.87}} \ldots \)     (M1)

\( = {\text{3780 (m) (3778.87}} \ldots {\text{)}}\)     (A1)(ft)(G1)

Notes: Award (M1) for adding the three sides. Follow through from their answer to part (c). Accept \(3771{\text{ }} (3771.33…)\).[2 marks]

d.

\(\frac{{\sin C}}{{1220}} = \frac{{\sin 110^\circ }}{{{\text{1708.87}} \ldots }}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into sine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).

\(C = 42.1^\circ {\text{ (42.1339}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Notes: Accept \(41.9^{\circ}, 42.0^{\circ}, 42.2^{\circ}, 42.3^{\circ}\).

OR

\(\cos C = \frac{{{\text{1708.87}}{ \ldots ^2} + {{850}^2} – {{1220}^2}}}{{2 \times {\text{1708.87}} \ldots  \times 850}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into cosine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).

\(C = 42.1^\circ {\text{ (42.1339}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Notes: Accept \(41.2^{\circ}, 41.8^{\circ}, 42.4^{\circ}\).[3 marks]

e.

\(\frac{1}{2} \times 1220 \times 850 \times \sin 110^\circ \)     (M1)(A1)(ft)

OR

\(\frac{1}{2} \times {\text{1708.87}} \ldots  \times 850 \times \sin {\text{42.1339}} \ldots ^\circ \)     (M1)(A1)(ft)

OR

\(\frac{1}{2} \times 1220 \times {\text{1708.87}} \ldots  \times \sin {\text{27.8661}} \ldots ^\circ \)     (M1)(A1)(ft)

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

\( = 487\,000{\text{ }}{{\text{m}}^2}{\text{ (487}}\,{\text{230}} \ldots {\text{ }}{{\text{m}}^2})\)     (A1)(ft)(G2)

Notes: The answer is \(487\,000{\text{ }}{{\text{m}}^2}\), units are required.

     Accept \(486\,000{\text{ }}{{\text{m}}^2}{\text{ (485}}\,{\text{633}} \ldots {\text{ }}{{\text{m}}^2})\).

     If workings are not shown and units omitted, award (G1) for \(487\,000{\text{ or }}486\,000\).

     Follow through from parts (c) and (e).[3 marks]

f.

Question

A surveyor has to calculate the area of a triangular piece of land, DCE.

The lengths of CE and DE cannot be directly measured because they go through a swamp.

AB, DE, BD and AE are straight paths. Paths AE and DB intersect at point C.

The length of AB is 15 km, BC is 10 km, AC is 12 km, and DC is 9 km.

The following diagram shows the surveyor’s information.

(i)     Find the size of angle \({\rm{ACB}}\).

(ii)     Show that the size of angle \({\rm{DCE}}\) is \(85.5^\circ\), correct to one decimal place.[4]

a.

The surveyor measures the size of angle \({\text{CDE}}\) to be twice that of angle \({\text{DEC}}\).

(i)     Using angle \({\text{DCE}} = 85.5^\circ \), find the size of angle \({\text{DEC}}\).

(ii)     Find the length of \({\text{DE}}\).[5]

b.

Calculate the area of triangle \({\text{DEC}}\).[4]

c.
Answer/Explanation

Markscheme

(i)     \(\cos {\rm{A\hat CB}} = \frac{{{{10}^2} + {{12}^2} – {{15}^2}}}{{2 \times 10 \times 12}}\)     (M1)(A1)

Note: Award (M1) for substituted cosine rule,

(A1) for correct substitution.

\({\rm{A\hat CB}} = 85.5^\circ \;\;\;({\text{85.4593}} \ldots {\text{)}}\)     (A1)(G2)

(ii)     \({\rm{D\hat CE}} = {\rm{A\hat CB}}\;\;\;{\text{and}}\;\;\;{\rm{A\hat CB}} = 85.5^\circ \;\;\;({\text{85.4593}} \ldots ^\circ {\text{)}}\)     (A1)

OR

\({\rm{B\hat CE}} = 180^\circ  – 85.5^\circ  = 94.5^\circ \;\;\;{\text{and}}\;\;\;{\rm{D\hat CE}} = 180^\circ  – 94.5^\circ  = 85.5^\circ \)     (A1)

Notes: Both reasons must be seen for the (A1) to be awarded.

\({\rm{D\hat CE}} = 85.5^\circ \)     (AG)

a.

(i)     \({\rm{D\hat EC}} = \frac{{180^\circ  – 85.5^\circ }}{3}\)     (M1)

\({\rm{D\hat EC}} = 31.5^\circ \)     (A1)(G2)

(ii)     \(\frac{{\sin (31.5^\circ )}}{9} = \frac{{\sin (85.5^\circ )}}{{{\text{DE}}}}\)     (M1)(A1)(ft)

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

\({\text{DE}} = 17.2{\text{ (km)}}(17.1718 \ldots )\).     (A1)(ft)(G2)

b.

\(0.5 \times 17.1718 \ldots  \times 9 \times \sin (63^\circ )\)     (A1)(ft)(M1)(A1)(ft)

Note: Award (A1)(ft) for \(63\) seen, (M1) for substituted triangle area formula, (A1)(ft) for \(0.5 \times 17.1718 \ldots  \times 9 \times \sin ({\text{their angle CDE}})\).

OR

\({\text{(triangle height}} = ){\text{ }}9 \times \sin (63^\circ )\)     (A1)(ft)(A1)(ft)

\({\text{0.5}} \times {\text{17.1718}} \ldots  \times {\text{9}} \times {\text{sin(their angle CDE)}}\)     (M1)

Note: Award (A1)(ft) for \(63\) seen, (A1)(ft) for correct triangle height with their angle \({\text{CDE}}\), (M1) for \({\text{0.5}} \times {\text{17.1718}} \ldots  \times {\text{9}} \times {\text{sin(their angle CDE)}}\).

\( = 68.9{\text{ k}}{{\text{m}}^2}\;\;\;(68.8509 \ldots )\)     (A1)(ft)(G3)

Notes: Units are required for the last (A1)(ft) mark to be awarded.

Follow through from parts (b)(i) and (b)(ii).

Follow through from their angle \({\text{CDE}}\) within this part of the question.

c.

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 farmer owns a plot of land in the shape of a quadrilateral ABCD.

\({\text{AB}} = 105{\text{ m, BC}} = 95{\text{ m, CD}} = 40{\text{ m, DA}} = 70{\text{ m}}\) and angle \({\text{DCB}} = 90^\circ \).

N16/5/MATSD/SP2/ENG/TZ0/05

The farmer wants to divide the land into two equal areas. He builds a fence in a straight line from point B to point P on AD, so that the area of PAB is equal to the area of PBCD.

Calculate

the length of BD;[2]

a.

the size of angle DAB;[3]

b.

the area of triangle ABD;[3]

c.

the area of quadrilateral ABCD;[2]

d.

the length of AP;[3]

e.

the length of the fence, BP.[3]

f.
Answer/Explanation

Markscheme

\(({\text{BD}} = ){\text{ }}\sqrt {{{95}^2} + {{40}^2}} \)    (M1)

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

\( = 103{\text{ }}({\text{m}}){\text{ }}\left( {103.077 \ldots ,{\text{ }}25\sqrt {17} } \right)\)    (A1)(G2)[2 marks]

a.

\(\cos {\rm{B\hat AD}} = \frac{{{{105}^2} + {{70}^2} – {{(103.077 \ldots )}^2}}}{{2 \times 105 \times 70}}\)     (M1)(A1)(ft)

Note:     Award (M1) for substitution into cosine rule, (A1)(ft) for their correct substitutions. Follow through from part (a).

\(({\rm{B\hat AD}}) = 68.9^\circ {\text{ }}(68.8663 \ldots )\)    (A1)(ft)(G2)

Note:     If their 103 used, the answer is \(68.7995 \ldots \)[3 marks]

b.

\(({\text{Area of ABD}} = )\frac{1}{2} \times 105 \times 70 \times \sin (68.8663 \ldots )\)    (M1)(A1)(ft)

Notes:     Award (M1) for substitution into the trig form of the area of a triangle formula.

Award (A1)(ft) for their correct substitutions.

Follow through from part (b).

If 68.8° is used the area \( = 3426.28 \ldots {\text{ }}{{\text{m}}^2}\).

\( = 3430{\text{ }}{{\text{m}}^2}{\text{ }}(3427.82 \ldots )\)    (A1)(ft)(G2)[3 marks]

c.

\({\text{area of ABCD}} = \frac{1}{2} \times 40 \times 95 + 3427.82 \ldots \)    (M1)

Note:     Award (M1) for correctly substituted area of triangle formula added to their answer to part (c).

\( = 5330{\text{ }}{{\text{m}}^2}{\text{ }}(5327.83 \ldots )\)    (A1)(ft)(G2)[2 marks]

d.

\(\frac{1}{2} \times 105 \times {\text{AP}} \times \sin (68.8663 \ldots ) = 0.5 \times 5327.82 \ldots \)    (M1)(M1)

Notes:     Award (M1) for the correct substitution into triangle formula.

Award (M1) for equating their triangle area to half their part (d).

\(({\text{AP}} = ){\text{ }}54.4{\text{ }}({\text{m}}){\text{ }}(54.4000 \ldots )\)    (A1)(ft)(G2)

Notes:     Follow through from parts (b) and (d).[3 marks]

e.

\({\text{B}}{{\text{P}}^2} = {105^2} + {(54.4000 \ldots )^2} – 2 \times 105 \times (54.4000 \ldots ) \times \cos (68.8663 \ldots )\)    (M1)(A1)(ft)

Notes:     Award (M1) for substituted cosine rule formula.

Award (A1)(ft) for their correct substitutions. Accept the exact fraction \(\frac{{53}}{{147}}\) in place of \(\cos (68.8663 \ldots )\).

Follow through from parts (b) and (e).

\(({\text{BP}} = ){\text{ }}99.3{\text{ }}({\text{m}}){\text{ }}(99.3252 \ldots )\)    (A1)(ft)(G2)

Notes:     If 54.4 and \(\cos (68.9)\) are used the answer is \(99.3567 \ldots \)[3 marks]

f.

Question

The quadrilateral ABCD represents a park, where \({\text{AB}} = 120{\text{ m}}\), \({\text{AD}} = 95{\text{ m}}\) and \({\text{DC}} = 100{\text{ m}}\). Angle DAB is 70° and angle DCB is 110°. This information is shown in the following diagram.

M17/5/MATSD/SP2/ENG/TZ2/04

A straight path through the park joins the points B and D.

A new path, CE, is to be built such that E is the point on BD closest to C.

The section of the park represented by triangle DCE will be used for a charity race. A track will be marked along the sides of this section.

Find the length of the path BD.[3]

a.

Show that angle DBC is 48.7°, correct to three significant figures.[3]

b.

Find the area of the park.[4]

c.

Find the length of the path CE.[2]

d.

Calculate the total length of the track.[3]

e.
Answer/Explanation

Markscheme

\(({\text{B}}{{\text{D}}^2} = ){\text{ }}{95^2} + {120^2} – 2 \times 95 \times 120 \times \cos 70^\circ \)     (M1)(A1)

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

\(({\text{BD}} = ){\text{ }}125{\text{ (m) }}\left( {125.007 \ldots {\text{ (m)}}} \right)\)     (A1)(G2)[3 marks]

a.

\(\frac{{\sin {\text{DBC}}}}{{100}} = \frac{{\sin 110^\circ }}{{125.007 \ldots }}\)     (M1)(A1)(ft)

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

Follow through from their answer to part (a).

\(({\text{DBC}} = ){\text{ }}48.7384 \ldots ^\circ \)     (A1)(ft)

\(({\text{DBC}} = ){\text{ }}48.7^\circ \)     (AG)

Notes:     Award the final (A1)(ft) only if both their unrounded answer and 48.7° is seen. Follow through from their answer to part (a), only if their unrounded answer rounds to 48.7°.[3 marks]

b.

\(\frac{1}{2} \times 125.007 \ldots  \times 100 \times \sin 21.3^\circ  + \frac{1}{2} \times 95 \times 120 \times \sin 70^\circ \)     (A1)(M1)(M1)

Note:     Award (A1) for 21.3° (21.2615…) seen, (M1) for substitution into (at least) one area of triangle formula in the form \(\frac{1}{2}ab\sin c\), (M1) for their correct substitutions and adding the two areas.

\(7630{\text{ }}{{\text{m}}^2}{\text{ }}(7626.70 \ldots {{\text{m}}^2})\)     (A1)(ft)(G3)

Notes:     Follow through from their answers to part (a). Accept \(7620{\text{ }}{{\text{m}}^2}{\text{ }}(7622.79 \ldots {{\text{m}}^2})\) from use of 48.7384…[4 marks]

c.

\(({\text{CE}} = ){\text{ }}100 \times \sin 21.3^\circ \)     (M1)

\(({\text{CE}} = ){\text{ }}36.3{\text{ (m) }}\left( {36.3251 \ldots {\text{ (m)}}} \right)\)     (A1)(ft)(G2)

Note:     Follow through from their angle 21.3° in part (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.

OR

\({\text{area of BCD}} = \frac{1}{2}{\text{BD}} \times {\text{CE}}\)     (M1)

\(({\text{CE}} = ){\text{ }}36.3{\text{ (m) }}\left( {36.3251 \ldots {\text{ (m)}}} \right)\)     (A1)(ft)(G2)

Note:     Follow through from parts (a) and (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.[2 marks]

d.

\(\sqrt {{{100}^2} – 36.3251{ \ldots ^2}}  + 100 + 36.3251 \ldots \)     (M1)(M1)

Note:     Award (M1) for correct use of Pythagoras to find DE (or correct trigonometric equation, \(100 \times \cos 21.3\), to find DE), (M1) for the sum of 100, their DE and their CE.

\(229{\text{ (m) }}\left( {229.494 \ldots {\text{ (m)}}} \right)\)     (A1)(ft)(G2)

Note:     Follow through from part (d). Use of 3 sf values gives an answer of \(230{\text{ (m) }}\left( {229.5{\text{ (m)}}} \right)\).[3 marks]

e.

Question

Abdallah owns a plot of land, near the river Nile, in the form of a quadrilateral ABCD.

The lengths of the sides are \({\text{AB}} = {\text{40 m, BC}} = {\text{115 m, CD}} = {\text{60 m, AD}} = {\text{84 m}}\) and angle \({\rm{B\hat AD}} = 90^\circ \).

This information is shown on the diagram.

N17/5/MATSD/SP2/ENG/TZ0/03

The formula that the ancient Egyptians used to estimate the area of a quadrilateral ABCD is

\({\text{area}} = \frac{{({\text{AB}} + {\text{CD}})({\text{AD}} + {\text{BC}})}}{4}\).

Abdallah uses this formula to estimate the area of his plot of land.

Show that \({\text{BD}} = 93{\text{ m}}\) correct to the nearest metre.[2]

a.

Calculate angle \({\rm{B\hat CD}}\).[3]

b.

Find the area of ABCD.[4]

c.

Calculate Abdallah’s estimate for the area.[2]

d.i.

Find the percentage error in Abdallah’s estimate.[2]

d.ii.
Answer/Explanation

Markscheme

\({\text{B}}{{\text{D}}^2} = {40^2} + {84^2}\)     (M1)

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

Accept correct substitution into cosine rule.

\({\text{BD}} = 93.0376 \ldots \)     (A1)

\( = 93\)     (AG)

Note:     Both the rounded and unrounded value must be seen for the (A1) to be awarded.[2 marks]

a.

\(\cos C = \frac{{{{115}^2} + {{60}^2} – {{93}^2}}}{{2 \times 115 \times 60}}{\text{ }}({93^2} = {115^2} + {60^2} – 2 \times 115 \times 60 \times \cos C)\)     (M1)(A1)

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

\( = 53.7^\circ {\text{ }}(53.6679 \ldots ^\circ )\)     (A1)(G2)[3 marks]

b.

\(\frac{1}{2}(40)(84) + \frac{1}{2}(115)(60)\sin (53.6679 \ldots )\)     (M1)(M1)(A1)(ft)

Note:     Award (M1) for correct substitution into right-angle triangle area. Award (M1) for substitution into area of triangle formula and (A1)(ft) for correct substitution.

\( = 4460{\text{ }}{{\text{m}}^2}{\text{ }}(4459.30 \ldots {\text{ }}{{\text{m}}^2})\)     (A1)(ft)(G3)

Notes:     Follow through from part (b).[4 marks]

c.

\(\frac{{(40 + 60)(84 + 115)}}{4}\)     (M1)

Note:     Award (M1) for correct substitution in the area formula used by ‘Ancient Egyptians’.

\( = 4980{\text{ }}{{\text{m}}^2}{\text{ }}(4975{\text{ }}{{\text{m}}^2})\)     (A1)(G2)[2 marks]

d.i.

\(\left| {\frac{{4975 – 4459.30 \ldots }}{{4459.30 \ldots }}} \right| \times 100\)     (M1)

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

\( = 11.6{\text{ }}(\% ){\text{ }}(11.5645 \ldots )\)     (A1)(ft)(G2)

Notes:    Follow through from parts (c) and (d)(i).[2 marks]

d.ii.

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

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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