Find the size of $\text{AĈB}$.

Find $\text{DE}$.

Find the area of triangle $\text{DCE}$.

Estimate $\text{DF}$. You may assume the highway has a width of zero.

use of cosine rule              (M1)   

$\text{AĈB}={\cos }^{-1}\left(\frac{{1005}^2+{1225}^2-{650}^2}{2\times 1005\times 1225}\right)$             (A1) 

$=32° \left(31.9980\dots \right)$                        A1


[3 marks]

use of sine rule             (M1)   

$\frac{\text{DE}}{\sin 31.9980\dots °}=\frac{210}{\sin 100°}$            (A1) 

$\left(\text{DE}=\right)113 \text{m} \left(112.9937\dots \right)$                    A1

[3 marks]

METHOD 1

$180°-\left(100°+\text{their part} \left(a\right)\right)$            (M1) 

$=48.0019\dots °$  OR  $0.837791\dots$            (A1) 

substituted area of triangle formula            (M1) 

$\frac{1}{2}\times 112.9937\dots \times 210\times \sin 48.002°$            (A1) 

$8820 {\text{m}}^2\left(8817.18\dots \right)$                  A1

METHOD 2

$\frac{\text{CE}}{\sin \left(180-100-\text{their part} \left(a\right)\right)}=\frac{210}{\sin {\displaystyle }{\displaystyle 100}}$            (M1) 

$\left(\text{CE}=\right) 158.472\dots$            (A1) 

substituted area of triangle formula            (M1) 

EITHER

$\frac{1}{2}\times 112.993\dots \times 158.472\dots \times \sin 100$            (A1) 

OR

$\frac{1}{2}\times 210\dots \times 158.472\dots \times \sin \left(\text{their part} \left(a\right)\right)$            (A1) 

THEN

$8820 {\text{m}}^2\left(8817.18\dots \right)$                  A1

METHOD 3

${\text{CE}}^2={210}^2+112.993{\dots }^2-\left(2\times 210\times 112.993\dots \times \cos \left(180-100-\text{their part} \left(a\right)\right)\right)$            (M1) 

$\left(\text{CE}=\right) 158.472\dots$            (A1) 

substituted area of triangle formula            (M1) 

$\frac{1}{2}\times 112.993\dots \times 158.472\dots \times \sin 100$            (A1) 

$8820 {\text{m}}^2\left(8817.18\dots \right)$                  A1

[5 marks]

$1005-210$  OR  $795$            (A1) 

equating answer to part (c) to area of a triangle formula        (M1) 

$8817.18\dots =\frac{1}{2}\times \text{DF}\times \left(1005-210\right)\times \sin 48.002\dots °$            (A1) 

$\left(\text{DF=}\right) 29.8 \text{m }\left(29.8473\dots \right)$                  A1

[4 marks]

Calculate the length of $\text{BD}$.

Show that angle $\text{EDC}=48.0°$, correct to three significant figures.

Calculate the area of triangle $\text{BDC}$.

Pedro draws a circle, with centre at point $\text{E}$, passing through point $\text{C}$. Part of the circle is shown in the diagram.

Show that point $\text{A}$ lies outside this circle. Justify your reasoning.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{\text{BD}}{\sin 51.5°}=\frac{8}{\sin 52.5°}$      (M1)(A1)

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

$\left(\text{BD}=\right) 7.89 \left(\text{cm}\right) \left(7.89164\dots \right)$      (A1)(G2)

Note: If radians are used the answer is $9.58723\dots$ award at most (M1)(A1)(A0).

[3 marks]

$\cos \text{EDC}=\frac{9^2+3.94582{\dots }^2-7^2}{2\times 9\times 3.94582\dots }$      (A1)(ft)(M1)(A1)(ft)

Note: Award (A1) for $3.94582\dots$ or $\frac{7.89164\dots }{2}$ seen, (M1) for substituted cosine rule, (A1)(ft) for correct substitutions.

$\left(\text{EDC}=\right) 47.9515\dots °$      (A1)

$48.0°$ ($3$ sig figures)      (AG)

Note: Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the final (M1) to be awarded.
Award at most (A1)(ft)(M1)(A1)(ft)(A0) if the known angle $48.0°$ is used to validate the result. Follow through from their$\text{ BD}$ in part (a).

[4 marks]

Units are required in this question.

$\left(\text{area}=\right) \frac{1}{2}\times 7.89164\dots \times 9\times \sin 48.0°$      (M1)(A1)(ft)

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

$\left(\text{area}=\right) 26.4 {\text{cm}}^2 \left(26.3908\dots \right)$      (A1)(ft)(G3)

Note: Follow through from part (a).

[3 marks]

${\text{AE}}^2=8^2+{\left(3.94582\dots \right)}^2-2\times 8\times 3.94582\dots \cos \left(76°\right)$        (A1)(M1)(A1)(ft)

Note: Award (A1) for $76°$ seen. Award (M1) for substituted cosine rule to find $\text{AE}$, (A1)(ft) for correct substitutions.

$\left(\text{AE}=\right) 8.02 \left(\text{cm}\right) \left(8.01849\dots \right)$      (A1)(ft)(G3)

Note: Follow through from part (a).

OR

${\text{AE}}^2=9.78424{\dots }^2+{\left(3.94582\dots \right)}^2-2\times 9.78424\dots \times 3.94582\dots \cos \left(52.5°\right)$        (A1)(M1)(A1)(ft)

Note: Award (A1) for $\text{AD} (9.78424\dots )$ or $76°$ seen. Award (M1) for substituted cosine rule to find $\text{AE}$ (do not award (M1) for cosine or sine rule to find $\text{AD}$), (A1)(ft) for correct substitutions.

$\left(\text{AE}=\right) 8.02 \left(\text{cm}\right) \left(8.01849\dots \right)$      (A1)(ft)(G3)

Note: Follow through from part (a).

$8.02>7$.      (A1)(ft)

point $\text{A}$ is outside the circle.      (AG)

Note: Award (A1) for a numerical comparison of $\text{AE}$ and $\text{CE}$. Follow through for the final (A1)(ft) within the part for their $8.02$. The final (A1)(ft) is contingent on a valid method to find the value of $\text{AE}$.
Do not award the final (A1)(ft) if the (AG) line is not stated.
Do not award the final (A1)(ft) if their point $\text{A}$ is inside the circle.

[5 marks]

Find the area of campus managed by office $\text{C}$.

Hence or otherwise find the areas managed by offices $\text{A}$ and $\text{B}$.

State a further assumption that must be made in order to use area covered as a measure of whether or not the manager of office $\text{C}$ is responsible for more students than the managers of offices $\text{A}$ and $\text{B}$.

A new office is to be built within the triangle formed by $\text{A}$, $\text{B}$ and $\text{C}$, at a point as far as possible from the other three offices.

Find the distance of this office from each of the other offices.

Divides area into two appropriate shapes

For example,

Area of triangle $\left(\frac{1}{2}\times 200\times 40=\right) 4000 {\text{m}}^2$      (A1)

Area of rectangle $\left(200\times 97\right)= 19400 {\text{m}}^2$      (A1)

$23400 \left({\text{m}}^2\right)$      A1

Note: The area can be found using different divisions. Award A1 for any two correct areas found and A1 for the final answer.

[3 marks]

EITHER

$\frac{200\times 300-23400}{2}=18300 {\text{m}}^2$      (M1)A1

OR

$\frac{1}{2}\times 100\times \left(203+163\right)=18300 {\text{m}}^2$      (M1)A1

THEN

Area managed by both offices $\text{A}$ and $\text{B}$ is $18300 {\text{m}}^2$      A1

[3 marks]

Density of accommodation/students is uniform      R1

[1 mark]

$250-163=87 \left(\text{m}\right)$        (M1)A1

Note: M1 is for an attempt to find the distance from the intersection point to one of the offices.

[2 marks]

Show that $\text{AC}=\tan \theta$.

Find the value of $\theta$ when the area of the shaded region is equal to the area of sector $\text{OADB}$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct working for $\text{AC}$ (seen anywhere)     A1

eg       $\tan \theta =\frac{\text{AC}}{\text{OA}}, \tan \theta =\frac{\text{AC}}{1}$

$\text{AC}=\tan \theta$      AG   N0

[1 mark]

METHOD 1 (working with half the areas)

area of triangle $\text{OAC}$ or triangle $\text{OBC}$      (A1)

eg      $\frac{1}{2}\times 1\times \tan \theta$

correct sector area      (A1)

eg      $\frac{1}{2}\times \theta \times \left(1^2\right) , \frac{1}{2}\theta$

correct approach using their areas to find the shaded area (seen anywhere)      (A1)

eg      ${\text{A}}_{\text{their triangle}}-{\text{A}}_{\text{their sector}} , \frac{1}{2}\theta -\frac{1}{2}\tan \theta$

correct equation      A1

eg      $\frac{1}{2}\tan \theta -\frac{1}{2}\theta =\frac{1}{2}\theta , \tan \theta =2\theta$

$1.16556$

$1.17$      A2   N4

METHOD 2 (working with entire kite and entire sector)

area of kite $\text{OACB}$      (A1)

eg      $2\times \frac{1}{2}\times 1\times \tan \theta , \frac{1}{2}\times \frac{1}{\cos \theta }\times 2 \sin \theta$

correct sector area      (A1)

eg      $\frac{1}{2}\times 2\theta \times \left(1^2\right) , \theta$

correct approach using their areas to find the shaded area (seen anywhere)      (A1)

eg      ${\text{A}}_{\text{kite OACB}}-{\text{A}}_{\text{sector OADB}} , \theta -\tan \theta$

correct equation      A1

eg      $\tan \theta -\theta =\theta , \tan \theta =2\theta$

$1.16556$

$1.17$      A2   N4

[6 marks]

Find the size of angle $\text{A}\hat{\text{O}}\text{B}$ in degrees.

Find the distance between points $\text{A}$ and $\text{B}$.

Find the size of angle $\theta$ in degrees.

Calculate the length of arc $\text{AC}$.

Calculate the area of the shaded sector, $\text{AOC}$.

Write down the height of the endpoint of the minute hand above the bookshelf at $6:00$ am.

Find the value of $h$ when $\theta =160°$.

Write down the amplitude of $g\left(\theta \right)$.

The endpoints of the minute hand and hour hand meet when $\theta =k$.

Find the smallest possible value of $k$.

$4\times \frac{360°}{12}$  OR  $4\times 30°$                   (M1)

$120°$                  A1

[2 marks]

substitution in cosine rule                   (M1)

${\text{AB}}^2={10}^2+6^2-2\times 10\times 6\times \cos \left(120°\right)$                  (A1)

$\text{AB}=14$ $\text{cm}$                  A1

Note: Follow through marks in part (b) are contingent on working seen.

[3 marks]

$\theta =13\times 6$                   (M1)

$=78°$                A1

[2 marks]

substitution into the formula for arc length                 (M1)

$l=\frac{78}{360}\times 2\times \mathrm{\pi }\times 10$  OR  $l=\frac{13\mathrm{\pi }}{30}\times 10$

$=13.6 \text{cm} \left(13.6135\dots , 4.33\mathrm{\pi }, \frac{13\mathrm{\pi }}{3}\right)$                A1

[2 marks]

substitution into the area of a sector                (M1)

$A=\frac{78}{360}\times \mathrm{\pi }\times {10}^2$  OR  $l=\frac{1}{2}\times \frac{13\mathrm{\pi }}{30}\times {10}^2$

$=68.1 {\text{cm}}^2 \left(68.0678\dots , 21.7\mathrm{\pi }, \frac{65\mathrm{\pi }}{3}\right)$                A1

[2 marks]

$23$              A1

[1 mark]

correct substitution              (M1)

$h=10 \cos \left(160°\right)+13$

$=3.60 \text{cm} \left(3.60307\dots \right)$              A1

[2 marks]

$10$           A1

[1 mark]

EITHER

$10\times \cos \left(\theta \right)+13=-10\times \cos \left(\frac{\theta }{12}\right)+13$              (M1)

OR

                             (M1)

Note: Award M1 for equating the functions. Accept a sketch of $h\left(\theta \right)$ and $g\left(\theta \right)$ with point(s) of intersection marked.

THEN

$k=196° \left(196.363\dots \right)$          A1

Note: The answer $166.153\dots$ is incorrect but the correct method is implicit. Award (M1)A0.

[2 marks]

Given that $\sin 60°=\frac{\sqrt{3}}{2}$, show that the area of the base of the box is equal to $\frac{3\sqrt{3}{x}^2}{2}$.

Given that the total external surface area of the box is $1200 {\text{cm}}^2$, show that the volume of the box may be expressed as $V=300\sqrt{3}x-\frac{9}{4}{x}^3$.

Sketch the graph of $V=300\sqrt{3}x-\frac{9}{4}{x}^3$, for $0\le x\le 16$.

Find an expression for $\frac{dV}{dx}$.

Find the value of $x$ which maximizes the volume of the box.

Hence, or otherwise, find the maximum possible volume of the box.

The box will contain spherical chocolates. The production manager assumes that they can calculate the exact number of chocolates in each box by dividing the volume of the box by the volume of a single chocolate and then rounding down to the nearest integer.

Explain why the production manager is incorrect.

evidence of splitting diagram into equilateral triangles                M1

area $=6\left(\frac{1}{2}{x}^2 \sin 60°\right)$               A1

$=\frac{3\sqrt{3}{x}^2}{2}$               AG

Note: The AG line must be seen for the final A1 to be awarded.

[2 marks]

total surface area of prism $1200=2\left(3x^2\frac{\sqrt{3}}{2}\right)+6xh$               M1A1

Note: Award M1 for expressing total surface areas as a sum of areas of rectangles and hexagons, and A1 for a correctly substituted formula, equated to $1200$.

$h=\frac{400-\sqrt{3}{x}^2}{2x}$               A1

volume of prism $=\frac{3\sqrt{3}}{2}{x}^2\times h$               (M1)

$=\frac{3\sqrt{3}}{2}{x}^2\left(\frac{400-\sqrt{3}{x}^2}{2x}\right)$               A1

$=300\sqrt{3}x-\frac{9}{4}{x}^3$               AG

Note: The AG line must be seen for the final A1 to be awarded.

[5 marks]

               A1A1

Note: Award A1 for correct shape, A1 for roots in correct place with some indication of scale (indicated by a labelled point).

[2 marks]

$\frac{dV}{dx}=300\sqrt{3}-\frac{27}{4}{x}^2$               A1A1

Note: Award A1 for a correct term.

[2 marks]

from the graph of $V$ or $\frac{dV}{dx}$  OR  solving $\frac{dV}{dx}=0$               (M1)

$x=8.88 \left(8.877382\dots \right)$               A1

[2 marks]

from the graph of $V$  OR  substituting their value for $x$ into $V$            (M1)

$V_{\text{max}}=3040 {\text{cm}}^3\left(3039.34\dots \right)$               A1

[2 marks]

EITHER
wasted space / spheres do not pack densely (tesselate)             A1

OR
the model uses exterior values / assumes infinite thinness of materials and hence the modelled volume is not the true volume             A1

[1 mark]

Find the angle of depression from $\text{M}$ to $\text{N}$.

Find $\text{CV}$.

Hence or otherwise, show that the volume of the reservoir is $29 600 {\text{m}}^3$.

By finding an appropriate value, determine whether Joshua is correct.

To avoid water leaking into the ground, the five interior sides of the reservoir have been painted with a watertight material.

Find the area that was painted.

$\tan \left(\theta \right)=\frac{6}{10}$           (M1)

$\left(\theta =\right) 31.0° \left(30.9637\dots °\right)$  OR  $0.540 \left(0.540419\dots \right)$           A1

[2 marks]

$\left(\text{CV}=\right) 40 \tan \left(\theta \right)$  OR  $\left(\text{CV}=\right) 4\times 6$           (M1)

Note: Award (M1) for an attempt at trigonometry or similar triangles (e.g. ratios).

$\left(\text{CV}=\right) 24 \text{m}$           A1

[2 marks]

$\left(V=\right)\frac{1}{3}{80}^2\times 24-\frac{1}{3}{60}^2\times 18$           M1A1A1

Note: Award M1 for finding the difference between the volumes of two pyramids, A1 for each correct volume expression. The final A1 is contingent on correct working leading to the given answer.
If the correct final answer is not seen, award at most M1A1A0. Award M0A0A0 for any height derived from $V=29 600$, including $18.875$ or $13.875$.

$\left(V=\right) 29 600 {\text{m}}^3$           AG

[3 marks]

METHOD 1

$\left(\frac{29 600}{80}=\right) 370$ (days)             A1

$\left(370>366\right)$  Joshua is correct             A1

Note: Award A0A0 for unsupported answer of “Joshua is correct”. Accept $1.01\dots >1$ for the first A1 mark.

METHOD 2

$80\times 366=29280 {\text{m}}^3$  OR  $80\times 365=29200 {\text{m}}^3$             A1

$\left(29280<29600\right)$  Joshua is correct             A1

Note: The second A1 can be awarded for an answer consistent with their result.

[2 marks]

height of trapezium is $\sqrt{{10}^2+6^2} \left(=11.6619\dots \right)$            (M1)

area of trapezium is $\frac{80+60}{2}\times \sqrt{{10}^2+6^2} \left(=816.333\dots \right)$            (M1)(A1)

$\left(SA=\right) 4\times \left(\frac{80+60}{2}\times \sqrt{{10}^2+6^2}\right)+{60}^2$            (M1)

Note: Award M1 for adding $4$ times their ($\text{MNOP}$) trapezium area to the area of the ($60\times 60$) base.

$\left(SA=\right) 6870 {\text{m}}^2 \left(6865.33 {\text{m}}^2\right)$            A1

Note: No marks are awarded if the correct shape is not identified.

[5 marks]

Finding the angle of depression from $\mathrm{M}$ to $\mathrm{N}$ proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle $\mathrm{NMP}$. Most candidates attempted to find $\mathrm{CV}$ through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of $29600 {\mathrm{m}}^3$ accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify $360$ as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid $\mathrm{MPON}$ as $6$.

Finding the angle of depression from $\mathrm{M}$ to $\mathrm{N}$ proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle $\mathrm{NMP}$. Most candidates attempted to find $\mathrm{CV}$ through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of $29600 {\mathrm{m}}^3$ accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify $360$ as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid $\mathrm{MPON}$ as $6$.

Finding the angle of depression from $\mathrm{M}$ to $\mathrm{N}$ proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle $\mathrm{NMP}$. Most candidates attempted to find $\mathrm{CV}$ through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of $29600 {\mathrm{m}}^3$ accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify $360$ as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid $\mathrm{MPON}$ as $6$.

Finding the angle of depression from $\mathrm{M}$ to $\mathrm{N}$ proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle $\mathrm{NMP}$. Most candidates attempted to find $\mathrm{CV}$ through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of $29600 {\mathrm{m}}^3$ accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify $360$ as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid $\mathrm{MPON}$ as $6$.

Finding the angle of depression from $\mathrm{M}$ to $\mathrm{N}$ proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle $\mathrm{NMP}$. Most candidates attempted to find $\mathrm{CV}$ through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of $29600 {\mathrm{m}}^3$ accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify $360$ as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid $\mathrm{MPON}$ as $6$.

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.

Use Giovanni's diagram to calculate the length of AX.

Use Giovanni's diagram to find the length of BX, the horizontal displacement of the Tower.

Find the percentage error on Giovanni’s diagram.

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.

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

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

angle $\text{B}\stackrel{\wedge }{\text{A}}\text{C}$ = 34.9034…°    (A1)

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

angle $\text{A}\stackrel{\wedge }{\text{B}}\text{C}$ = 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 $\text{A}\stackrel{\wedge }{\text{B}}\text{C}$, 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.

sin 85… × 56     (M1)

= 55.8 (55.7869…) (m)     (A1)(G2)

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

$\sqrt{{56}^2-55.7869{\dots }^2}$     (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]

$\left|\frac{4.88-3.9}{3.9}\right|\times 100$     (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]

$\text{ta}{\text{n}}^{-1}\left(\frac{55.7869\dots }{40.11927\dots }\right)$     (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…

$\frac{\text{sin 120}}{68.7226\dots }=\frac{\text{sin A}\stackrel{\wedge }{\text{D}}\text{C}}{64.3725\dots }$    (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]

Find the angle $\text{AÔB}$.

Find the area of the shaded segment.

Find the area of a circle with radius $4.5 \text{m}$.

Find the area of the field that can be reached by the goat.

Find the value of $t$ at which the goat is eating grass at the greatest rate.

$\left(\frac{1}{2}\text{AÔB}=\right) \text{arccos}\left(\frac{4}{4.5}\right)=27.266\dots$        (M1)(A1)

$\text{AÔB}=54.532\dots \approx 54.5°$  ($0.951764\dots \approx 0.952$ radians)        A1 

Note: Other methods may be seen; award (M1)(A1) for use of a correct trigonometric method to find an appropriate angle and then A1 for the correct answer.

[3 marks]

finding area of triangle

EITHER

area of triangle $=\frac{1}{2}\times 4.5^2\times \sin \left(54.532\dots \right)$        (M1)

Note: Award M1 for correct substitution into formula.

$=8.24621\dots \approx 8.25 {\text{m}}^2$        (A1)

OR

$\text{AB}=2\times \sqrt{4.5^2-4^2}=4.1231\dots$

area triangle $=\frac{4.1231\dots \times 4}{2}$        (M1)

$=8.24621\dots \approx 8.25 {\text{m}}^2$        (A1)

finding area of sector

EITHER

area of sector $=\frac{54.532\dots }{360}\times \mathrm{\pi }\times 4.5^2$        (M1)

$=9.63661\dots \approx 9.64 {\text{m}}^2$        (A1)

OR

area of sector $=\frac{1}{2}\times 0.9517641\dots \times 4.5^2$        (M1)

$=9.63661\dots \approx 9.64 {\text{m}}^2$        (A1)

THEN

area of segment $=9.63661\dots -8.24621\dots$

$=1.39 {\text{m}}^2 \left(1.39040\dots \right)$        A1 

[5 marks]

$\mathrm{\pi }\times 4.5^2$         (M1)

$63.6 {\text{m}}^2 \left(63.6172\dots {\text{m}}^2\right)$        A1 

[2 marks]

METHOD 1

$4\times 1.39040... (5.56160)$         (A1)

subtraction of four segments from area of circle         (M1)

$=58.1 {\text{m}}^2 \left(58.055\dots \right)$       A1 

METHOD 2

$4\left(0.5\times 4.5^2\times \sin 54.532\dots \right)+4\left(\frac{35.4679}{360}\times \mathrm{\pi }\times 4.5^2\right)$         (M1)

$= 32.9845\dots +25.0707$         (A1)

$=58.1 {\text{m}}^2 \left(58.055\dots \right)$       A1 

[3 marks]

sketch of $\frac{dV}{dt}$   OR   $\frac{dV}{dt}=0.110363\dots$   OR   attempt to find where $\frac{d^{2}V}{d{t}^2}=0$         (M1)

$t=1$ hour        A1 

[2 marks]

Part (a)(i) proved to be difficult for many candidates. About half of the candidates managed to correctly find the angle $\text{A}\hat{O}B$. A variety of methods were used: cosine to find half of $\text{A}\hat{O}B$ then double it; sine to find angle $\text{A}\hat{O}B$ , then find half of A$\text{A}\hat{O}B$ and double it; Pythagoras to find half of AB and then sine rule to find half of angle $\text{A}\hat{O}B$ then double it; Pythagoras to find half of AB, then double it and use cosine rule to find angle $\text{A}\hat{O}B$ . Many candidates lost a mark here due to premature rounding of an intermediate value and hence the final answer was not correct (to three significant figures).

In part (a)(ii) very few candidates managed to find the correct area of the shaded segment and include the correct units. Some only found the area of the triangle or the area of the sector and then stopped.

In part (b)(i), nearly all candidates managed to find the area of a circle.

In part (b)(ii), finding the area of the field reached by the goat proved troublesome for most of the candidates. It appeared as if the candidates did not fully understand the problem. Very few candidates realized the connection to part (a)(ii).

Part (c) was accessed by only a handful of candidates. The candidates could simply have graphed the function on their GDC to find the greatest value, but most did not realize this.

Sketch a Voronoi diagram to represent this information.

Sketch a new Voronoi diagram to represent this new situation.

State what the shape of the land, owned by the youngest child, is.

Find the area of the youngest child’s land.

Find how much land an older child has lost.

State, with a reason, if all five children now own an equal amount of land.

  A2

[2 marks]

 A2

[2 marks]

By symmetry a square      A1

[1 mark]

Distance from (2, 2) to (1, 3) is $\sqrt{1^2+1^2}=\sqrt{2}$        M1A1

So length of youngest child’s square is $2\frac{\sqrt{2}}{2}=\sqrt{2}$ and thus area is 2.        M1A1

[4 marks]

By symmetry each older child must lose $\frac{2}{4}=\frac{1}{2}$      A1

[1 mark]

No, youngest child has less as each older child has $2\times 2-\frac{1}{2}=3\frac{1}{2}$      A1R1

[2 marks]

Find the midpoint of $\text{[BD]}$.

Find the equation of $\text{(XZ)}$.

Find the coordinates of $\text{X}$.

Determine the exact length of $\text{[YZ]}$.

Given that the exact length of $\text{[XY]}$ is $\sqrt{32}$, find the size of $\text{XŶZ}$ in degrees.

Hence find the area of triangle $\text{XYZ}$.

State one criticism of this interpretation.

$\left(\frac{2+6}{2}, \frac{2+0}{2}\right)$          (M1)

$\left(4, 1\right)$         A1

Note: Award A0 if parentheses are omitted in the final answer.

[2 marks]

attempt to substitute values into gradient formula          (M1)

$\left(\frac{0-2}{6-2}=\right)-\frac{1}{2}$          (A1)

therefore the gradient of perpendicular bisector is $2$          (M1)

so $y-1=2\left(x-4\right) \left(y=2x-7\right)$         A1

[4 marks]

identifying the correct equations to use:          (M1)

$y=2-x$  and  $y=2x-7$

evidence of solving their correct equations or of finding intersection point graphically          (M1)

$\left(3, -1\right)$           A1

Note: Accept an answer expressed as “$x=3, y=-1$”.

[3 marks]

attempt to use distance formula          (M1)

$\text{YZ}=\sqrt{{\left(7-\left(-1\right)\right)}^2+{\left(7-3\right)}^2}$

$=\sqrt{80} \left(4\sqrt{5}\right)$           A1

[2 marks]

METHOD 1 (cosine rule)

length of $\text{XZ}$ is $\sqrt{80} \left(4\sqrt{5}, 8.94427\dots \right)$          (A1)

Note: Accept $8.94$ and $8.9$.

attempt to substitute into cosine rule          (M1)

$\cos \text{XŶZ}=\frac{80+32-80}{2\times \sqrt{80}\sqrt{32}} \left(=0.316227\dots \right)$          (A1)

Note: Award A1 for correct substitution of $\text{XZ}$, $\text{YZ}$, $\sqrt{32}$ values in the cos rule. Exact values do not need to be used in the substitution.

$\left(\text{XŶZ}=\right) 71.6° \left(71.5650\dots °\right)$           A1

Note: Last A1 mark may be lost if prematurely rounded values of $\text{XZ}$, $\text{YZ}$ and/or $\text{XY}$ are used.

METHOD 2 (splitting isosceles triangle in half)

length of $\text{XZ}$ is $\sqrt{80} \left(4\sqrt{5}, 8.94427\dots \right)$          (A1)

Note: Accept $8.94$ and $8.9$.

required angle is ${\cos }^{-1}\left(\frac{\sqrt{32}}{2\sqrt{80}}\right)$          (M1)(A1)

Note: Award A1 for correct substitution of $\text{XZ}$ (or $\text{YZ}$), $\frac{\sqrt{32}}{2}$ values in the cos rule. Exact values do not need to be used in the substitution.

$\left(\text{XŶZ}=\right) 71.6° \left(71.5650\dots °\right)$           A1

Note: Last A1 mark may be lost if prematurely rounded values of $\text{XZ}$, $\text{YZ}$ and/or $\text{XY}$ are used.

[4 marks]

(area =) $\frac{1}{2}\sqrt{80}\sqrt{32} \sin 71.5650\dots$   OR  (area =) $\frac{1}{2}\sqrt{32}\sqrt{72}$          (M1)

$=24 {\text{km}}^2$           A1

[2 marks]

Any sensible answer such as:

There might be factors other than proximity which influence shopping choices.

A larger area does not necessarily result in an increase in population.

The supermarkets might be specialized / have a particular clientele who visit even if other shops are closer.

Transport links might not be represented by Euclidean distances.

etc.          R1

[1 mark]

Part (a) was answered very well and demonstrated that the candidates had good knowledge of the midpoint formula. Some candidates did not write the midpoint they found as a coordinate pair and lost a mark there. Part (b) was answered well. Most candidates were able to find the gradient of [BD] and then the gradient of the perpendicular [XZ] and its equation. Candidates who lost marks in (b) were able to collect follow through marks in parts (c), (d), and (e). In part (c), not all candidates were able to identify the correct equations that they needed to find the coordinates of point X. In part (d), many candidates overlooked the fact that the question called for the exact length of [YZ] – the majority of candidates gave the answer correct to three significant figures and hence lost a mark. In part (d) most candidates were able to correctly use the cosine rule. Marks were lost here if the candidates calculated the length of [XZ] incorrectly, or substituted the [XZ], [YZ] or [XY] values incorrectly. Often candidates used rounded [XZ], [YZ] or [XY] values prematurely, for which they lost the last mark. Part (f) was mostly well answered. If candidates found an angle in part (e), they were usually able to find the area in (f) correctly. Again, the contextual question in part (g) was challenging to many candidates. Some answers showed misconceptions about Voronoi diagrams, for example some candidates stated that “if the cell were larger, then some people living there will be much closer to supermarkets A, B, C than to supermarket D.” Many candidates offered explanations, which were long and unclear, which often did not address the issue they were asked to comment on, nor displayed a sound understanding of the topic.

Calculate the volume of this pan.

Find the radius of the sphere in cm, correct to one decimal place.

Find the value of $a$.

Find the temperature that the pizza will be 5 minutes after it is taken out of the oven.

Calculate, to the nearest second, the time since the pizza was taken out of the oven until it can be eaten.

In the context of this model, state what the value of 19 represents.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(V=)\pi \times {\text{(17.5)}}^2\times 0.5$     (A1)(M1)

Notes:     Award (A1) for 17.5 (or equivalent) seen.

Award (M1) for correct substitutions into volume of a cylinder formula.

$=481\text{ c}{\text{m}}^3(481.056\dots \text{ c}{\text{m}}^3,153.125\pi \text{ c}{\text{m}}^3)$     (A1)(G2)

[3 marks]

$\frac{4}{3}\times \pi \times r^3=481.056\dots$     (M1)

Note:     Award (M1) for equating their answer to part (a) to the volume of sphere.

$r^3=\frac{3\times 481.056\dots }{4\pi }(=114.843\dots )$     (M1)

Note:     Award (M1) for correctly rearranging so $r^3$ is the subject.

$r=4.86074\dots \text{ (cm)}$     (A1)(ft)(G2)

Note:     Award (A1) for correct unrounded answer seen. Follow through from part (a).

$=4.9\text{ (cm)}$     (A1)(ft)(G3)

Note:     The final (A1)(ft) is awarded for rounding their unrounded answer to one decimal place.

[4 marks]

$230=a(2.06{)}^0+19$     (M1)

Note:     Award (M1) for correct substitution.

$a=211$     (A1)(G2)

[2 marks]

$(P=)211\times (2.06{)}^{-5}+19$      (M1)

Note:     Award (M1) for correct substitution into the function, $P(t)$. Follow through from part (c). The negative sign in the exponent is required for correct substitution.

$=24.7$ (°C) $(24.6878\dots$ (°C))     (A1)(ft)(G2)

[2 marks]

$45=211\times (2.06{)}^{-t}+19$     (M1)

Note:     Award (M1) for equating 45 to the exponential equation and for correct substitution (follow through for their $a$ in part (c)).

$(t=)2.89711\dots$     (A1)(ft)(G1)

$174\text{ (seconds) }\left(173.826\dots \text{ (seconds)}\right)$     (A1)(ft)(G2)

Note:     Award final (A1)(ft) for converting their $\text{2.89711}\dots$ minutes into seconds.

[3 marks]

the temperature of the (dining) room     (A1)

OR

the lowest final temperature to which the pizza will cool     (A1)

[1 mark]

Show that $\text{BD}=93\text{ m}$ correct to the nearest metre.

Calculate angle $\mathrm{B}\hat{\mathrm{C}}\mathrm{D}$.

Find the area of ABCD.

Calculate Abdallah’s estimate for the area.

Find the percentage error in Abdallah’s estimate.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

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

Accept correct substitution into cosine rule.

$\text{BD}=93.0376\dots$     (A1)

$=93$     (AG)

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

[2 marks]

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

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

$={53.7}^{\circ }(53.6679{\dots }^{\circ })$     (A1)(G2)

[3 marks]

$\frac{1}{2}(40)(84)+\frac{1}{2}(115)(60)\sin (53.6679\dots )$     (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.

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

Notes:     Follow through from part (b).

[4 marks]

$\frac{(40+60)(84+115)}{4}$     (M1)

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

$=4980{\text{m}}^2(4975{\text{m}}^2)$     (A1)(G2)

[2 marks]

$\left|\frac{4975-4459.30\dots }{4459.30\dots }\right|\times 100$     (M1)

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

$=11.6(\mathrm{\%})(11.5645\dots )$     (A1)(ft)(G2)

Notes:    Follow through from parts (c) and (d)(i).

[2 marks]

Write down a formula for $A$, the surface area to be coated.

Express this volume in $\text{c}{\text{m}}^3$.

Write down, in terms of $r$ and $h$, an equation for the volume of this water container.

Show that $A=\pi r^2+\frac{1000000}{r}$.

Find $\frac{\text{d}A}{\text{d}r}$.

Using your answer to part (e), find the value of $r$ which minimizes $A$.

Find the value of this minimum area.

Find the least number of cans of water-resistant material that will coat the area in part (g).

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(A=)\pi r^2+2\pi rh$    (A1)(A1)

Note:     Award (A1) for either $\pi r^2$ OR $2\pi rh$ seen. Award (A1) for two correct terms added together.

[2 marks]

$500000$    (A1)

Notes:     Units not required.

[1 mark]

$500000=\pi r^{2}h$    (A1)(ft)

Notes:     Award (A1)(ft) for $\pi r^{2}h$ equating to their part (b).

Do not accept unless $V=\pi r^{2}h$ is explicitly defined as their part (b).

[1 mark]

$A=\pi r^2+2\pi r\left(\frac{500000}{\pi r^2}\right)$    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their $\frac{500000}{\pi r^2}$ seen.

Award (M1) for correctly substituting only $\frac{500000}{\pi r^2}$ into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to $\pi rh=\frac{500000}{r}$ and substituting for $\pi rh$ in expression for $A$.

$A=\pi r^2+\frac{1000000}{r}$    (AG)

Notes:     The conclusion, $A=\pi r^2+\frac{1000000}{r}$, must be consistent with their working seen for the (A1) to be awarded.

Accept ${10}^6$ as equivalent to $1000000$.

[2 marks]

$2\pi r-\frac{\text{1}\text{000}\text{000}}{r^2}$    (A1)(A1)(A1)

Note:     Award (A1) for $2\pi r$, (A1) for $\frac{1}{r^2}$ or $r^{-2}$, (A1) for $-\text{1}\text{000}\text{000}$.

[3 marks]

$2\pi r-\frac{1000000}{r^2}=0$    (M1)

Note:     Award (M1) for equating their part (e) to zero.

$r^3=\frac{1000000}{2\pi }$ OR $r=\sqrt[3]{\frac{1000000}{2\pi }}$     (M1)

Note:     Award (M1) for isolating $r$.

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

$(r=)54.2(\text{cm})(54.1926\dots )$    (A1)(ft)(G2)

[3 marks]

$\pi (54.1926\dots {)}^2+\frac{1000000}{(54.1926\dots )}$    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

$=27700(\text{c}{\text{m}}^2)(27679.0\dots )$    (A1)(ft)(G2)

[2 marks]

$\frac{27679.0\dots }{2000}$    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

$=13.8395\dots$    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their $13.8395\dots$ to the next integer.

[3 marks]

Find the slant height of the cone-shaped container.

Find the slant height of the cone-shaped container.

Show that the total surface area of the cone-shaped container is 314 cm², correct to three significant figures.

Find the height, $h$, of this cylinder-shaped container.

The factory director wants to increase the volume of coconut water sold per container.

State whether or not they should replace the cone-shaped containers with cylinder‑shaped containers. Justify your conclusion.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{\pi {\left(5.2\right)}^2\times 13}{3}$    (M1)

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

368  (368.110…) cm³     (A1)(G2)

Note: Accept 117.173…$\pi$ cm³ or $\frac{8788}{75}\pi$ cm³.

[2 marks]

(slant height²) = (5.2)² + 13²   (M1)

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

14.0  (14.0014…) (cm)     (A1)(G2)

[2 marks]

14.0014… × (5.2) × $\pi$ + (5.2)² × $\pi$     (M1)(M1)

Note: Award (M1) for their correct substitution in the curved surface area formula for cone; (M1) for adding the correct area of the base. The addition must be explicitly seen for the second (M1) to be awarded. Do not accept rounded values here as may come from working backwards.

313.679… (cm²)     (A1)

Note: Use of 3 sf value 14.0 gives an unrounded answer of 313.656….

314 (cm²)     (AG)

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

[3 marks]

2 × $\pi$ × (5.2) × $h$ + 2 × $\pi$ × (5.2)² = 314     (M1)(M1)(M1)

Note: Award (M1) for correct substitution in the curved surface area formula for cylinder; (M1) for adding two correct base areas of the cylinder; (M1) for equating their total cylinder surface area to 314 (313.679…). For this mark to be awarded the areas of the two bases must be added to the cylinder curved surface area and equated to 314. Award at most (M1)(M0)(M0) for cylinder curved surface area equated to 314.

($h$ =) 4.41 (4.41051…) (cm)     (A1)(G3)

[4 marks]

$\pi$ × (5.2)² × 4.41051…     (M1)

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

375  (374.666…) (cm³)     (A1)(ft)(G2)

Note: Follow through from part (d).

375 (cm³) > 368 (cm³)      (R1)(ft)

OR

“volume of cylinder is larger than volume of cone” or similar    (R1)(ft)

Note: Follow through from their answer to part (a). The verbal statement should be consistent with their answers from parts (e) and (a) for the (R1) to be awarded.

replace with the cylinder containers     (A1)(ft)

Note: Do not award (A1)(ft)(R0). Follow through from their incorrect volume for the cylinder in this question part but only if substitution in the volume formula shown.

[4 marks]

Find the size of angle $\text{C}\stackrel{\wedge }{\text{A}}\text{D}$.

Find the size of angle $\text{A}\stackrel{\wedge }{\text{C}}\text{D}$.

(CAD =) 53.1°  (53.0521…°)       (A1)(ft)

Note: Follow through from their part (b)(i) only if working seen.

[1 mark]

(ACD = ) 70° − (180° − 125° − 31.9478°…)      (M1)

Note: Award (M1) for subtracting their angle $\text{A}\stackrel{\wedge }{\text{C}}\text{B}$ from 70°.

OR

(ADC =) 360 − (85 + 70 + 125) = 80

(ACD =) 180 − 80 − 53.0521...      (M1)

46.9°  (46.9478…°)      (A1)(ft)(G2)

Note: Follow through from part (b)(i).

[2 marks]

Find the gradient of $L_1$.

Find the coordinates of point $\text{C}$.

Find the equation of $L_2$. Give your answer in the form $ax+by+d=0$, where $a, b, d\in \mathrm{\mathbb{Z}}$.

Find the value of $k$.

Find the distance between points $\text{M}$ and $\text{N}$.

Given that the length of $\text{AM}$ is $\sqrt{45}$, find the area of triangle $\text{ANC}$.

$\frac{2-\left(-1\right)}{-3-\left(-9\right)}$       (M1)

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

$=\frac{1}{2} \left(\frac{3}{6}, 0.5\right)$       (A1)(G2)

[2 marks]

$-3=\frac{-9+x}{2} \left(-6+9=x\right)$  and  $2=\frac{-1+y}{2} \left(4+1=y\right)$       (M1)

Note: Award (M1) for correct substitution into the midpoint formula for both coordinates.

OR

       (M1)

Note: Award (M1) for a sketch showing the horizontal displacement from $\text{M}$ to $\text{C}$ is $6$ and the vertical displacement is $3$ and the coordinates at $\text{M}$.

OR

$-3+6=3$  and  $2+3=5$       (M1)

Note: Award (M1) for correct equations seen.

$\left(3, 5\right)$      (A1)(G1)(G1)

Note: Accept $x=3, y=5$. Award at most (M1)(A0) or (G1)(G0) if parentheses are missing.

[2 marks]

gradient of the normal $=-2$      (A1)(ft)

Note: Follow through from their gradient from part (a).

$y-2=-2\left(x+3\right)$  OR  $2=-2\left(-3\right)+c$        (M1)

Note: Award (M1) for correct substitution of $\text{M}$ and their gradient of normal into straight line formula.

$2x+y+4=0$ (accept integer multiples)        (A1)(ft)(G3)

[3 marks]

$2\left(k\right)+4+4=0$      (M1)

Note: Award (M1) for substitution of $y=4$ into their equation of normal line or substitution of $\text{M}$ and $\left(k, 4\right)$ into equation of gradient of normal.

$k=-4$        (A1)(ft)(G2)

Note: Follow through from part (c).

[2 marks]

$\sqrt{{\left(-4+3\right)}^2+{\left(4-2\right)}^2}$      (M1)

Note: Award (M1) for correctly substituting point $\text{M}$ and their $\text{N}$ into distance formula.

$\sqrt{5} \left(2.24, 2.23606\dots \right)$        (A1)(ft)

Note: Follow through from part (d).

[2 marks]

$\frac{1}{2}\times \left(2\times \sqrt{45}\right)\times \sqrt{5}$      (M1)

Note: Award (M1) for their correct substitution into area of a triangle formula. Award (M0) for their $\frac{1}{2}\times \left(\sqrt{45}\right)\times \sqrt{5}$ without any evidence of multiplication by $2$ to find length $\text{AC}$. Accept any other correct method to find the area.

$15$        (A1)(ft)(G2)

Note: Accept $15.02637\dots$ from use of a $3$ sf value for $\sqrt{5}$. Follow through from part (e).

[2 marks]

Calculate the area of triangle EAD.

Calculate the total volume of the barn.

Calculate the length of MN.

Calculate the length of AE.

Show that Farmer Brown is incorrect.

Calculate the total length of metal required for one support.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(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]

(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]

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

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

(MN =) 1.29(m) (1.29409… (m))     (A1) (G2)

[2 marks]

(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]

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

$\frac{1.29409\dots }{\text{ND}}=\text{tan}{15}^{\circ }$     (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)