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Geometry and Measurement Study Guide
[198 marks]
The average radius of the orbit of the Earth around the Sun is 150 million kilometres.
1a. Write down this radius, in kilometres, in the form
[2 marks]
a × 10k , where
1 ⩽ a < 10, k ∈ Z.
Markscheme
1.5 × 108 (km)
(A2)
(C2)
Notes: Award (A2) for the correct answer.
Award (A1)(A0) for 1.5 and an incorrect index.
Award (A0)(A0) for answers of the form
15 × 107 .
[2 marks]
The average radius of the orbit of the Earth around the Sun is 150 million kilometres.
1b. The Earth travels around the Sun in one orbit. It takes one year for the Earth to complete one orbit.
Calculate the distance, in kilometres, the Earth travels around the Sun in one orbit, assuming that the orbit is a circle.
[2 marks]
Markscheme
2π1.5 × 108
(M1)
= 942 000 000 (km) (942 477 796.1 … , 3 × 108 π, 9.42 × 108 )
(A1)(ft)
(C2)
Notes: Award (M1) for correct substitution into correct formula. Follow through from part (a).
Do not accept calculator notation
9.42E8.
Do not accept use of
22
7
or
3.14 for
π.
[2 marks]
[2 marks]
1c. Today is Anna’s 17th birthday.
Calculate the total distance that Anna has travelled around the Sun, since she was born.
Markscheme
17 × 942 000 000
(M1)
= 1.60 × 1010 (km) (1.60221 … × 1010 , 1.6014 × 1010 , 16 022 122 530, (5.1 × 109 )π)
(A1)(ft)
(C2)
Note: Follow through from part (b).
[2 marks]
Chocolates in the shape of spheres are sold in boxes of 20.
Each chocolate has a radius of 1 cm.
[2 marks]
2a. Find the volume of 1 chocolate.
Markscheme
The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.
4
π(1)3
3
(M1)
Notes: Award (M1) for correct substitution into correct formula.
= 4.19 (4.18879 … , 43 π) cm3
(A1)
(C2)
[2 marks]
[1 mark]
2b. Write down the volume of 20 chocolates.
Markscheme
The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.
83.8 (83.7758 … ,
80
π)
3
cm3
(A1)(ft)
(C1)
Note: Follow through from their answer to part (a).
[1 mark]
[2 marks]
2c. The diagram shows the chocolate box from above. The 20 chocolates fit perfectly in the box with each chocolate touching the
ones around it or the sides of the box.
Calculate the volume of the box.
Markscheme
The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.
10 × 8 × 2
(M1)
Note: Award (M1) for correct substitution into correct formula.
= 160 cm3
(A1)
(C2)
[2 marks]
2d. The diagram shows the chocolate box from above. The 20 chocolates fit perfectly in the box with each chocolate touching the
ones around it or the sides of the box.
Calculate the volume of empty space in the box.
Markscheme
The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.
76.2 (76.2241 … , (160 − 803 π)) cm3
(A1)(ft)
(C1)
Note: Follow through from their part (b) and their part (c).
[1 mark]
A
m2
[1 mark]
Markscheme
2x(x − 4) or
(A1)
2x2 − 8x
(C1)
Note: Award (A0) for
x − 4 × 2x.
[1 mark]
[3 marks]
3b. The area of the carpet is
10
m2 .
Calculate the value of
x.
Markscheme
2x(x − 4) = 10
(M1)
Note: Award (M1) for equating their answer in part (a) to
10.
x2 − 4x − 5 = 0
(M1)
OR
Sketch of
y = 2x2 − 8x and
y = 10
(M1)
OR
Using GDC solver
x = 5 and
x = −1
(M1)
OR
2(x + 1)(x − 5)
x = 5 (m)
(M1)
(A1)(ft)
(C3)
Notes: Follow through from their answer to part (a).
Award at most (M1)(M1)(A0) if both
5 and
−1 are given as final answer.
Final (A1)(ft) is awarded for choosing only the positive solution(s).
[3 marks]
3c. The area of the carpet is
10
m2 .
Hence, write down the value of the length and of the width of the carpet, in metres.
[2 marks]
Markscheme
2 × 5 = 10 (m)
5 − 4 = 1 (m)
(A1)(ft)
(A1)(ft)
(C2)
Note: Follow through from their answer to part (b).
Do not accept negative answers.
[2 marks]
The diagram shows a wheelchair ramp,
A, designed to descend from a height of
80 cm.
[1 mark]
4a. Use the diagram above to calculate the gradient of the ramp.
Markscheme
80
− 940
(−0.0851,−0.085106 … ,− 474 )
[1 mark]
(A1)
(C1)
[1 mark]
4b. The gradient for a safe descending wheelchair ramp is
− 121 .
Using your answer to part (a), comment on why wheelchair ramp
A is not safe.
Markscheme
−0.0851 (−0.085106 …) < − 121 (−0.083333 …)
Notes: Accept “less than” in place of inequality.
Award (A0) if incorrect inequality seen.
Follow through from part (a).
[1 mark]
(A1)(ft)
(C1)
4c. The equation of a second wheelchair ramp, B, is
[4 marks]
2x + 24y − 1920 = 0.
(i)
Determine whether wheelchair ramp
B is safe or not. Justify your answer.
(ii)
Find the horizontal distance of wheelchair ramp
B.
Markscheme
(i)
ramp
B is safe
(A1)
the gradient of ramp
B is
− 121
(R1)
Notes: Award (R1) for “the gradient of ramp
B is
− 121 ” seen.
Do not award (A1)(R0).
(ii)
2x = 1920
(M1)
Note: Accept alternative methods.
960 (cm)
(A1)
(C4)
[4 marks]
Günter is at Berlin Tegel Airport watching the planes take off. He observes a plane that is at an angle of elevation of
20∘ from where he is standing at point
G. The plane is at a height of 350 metres. This information is shown in the following diagram.
5a. Calculate the horizontal distance,
GH , of the plane from Günter. Give your answer to the nearest metre.
[3 marks]
Markscheme
350
tan 20∘
(M1)
(A1)
= 961.617 …
= 962 (m)
(A1)(ft)
(C3)
Notes: Award (M1) for correct substitution into correct formula, (A1) for correct answer, (A1)(ft) for correct rounding to the nearest
metre.
Award (M0)(A0)(A0) for
961 without working.
[3 marks]
[3 marks]
5b. The plane took off from a point
T, which is
250 metres from where Günter is standing, as shown in the following diagram.
Using your answer from part (a), calculate the angle
ATH , the takeoff angle of the plane.
Markscheme
961.617 … − 250 = 711.617 …
350
tan−1 ( 711.617…
)
(A1)(ft)
(M1)
= 26.2∘ (26.1896 … )
(A1)(ft)
(C3)
Notes: Accept
26.1774 … from use of 3 sf answer
962 from part (a). Follow through from their answer to part (a).
Accept alternative methods.
[3 marks]
A cross-country running course consists of a beach section and a forest section. Competitors run from
A to
B, then from
B to
C and from
C back to
A.
The running course from
A to
B is along the beach, while the course from
B, through
C and back to
A, is through the forest.
The course is shown on the following diagram.
Angle
ABC is
110∘ .
It takes Sarah
5 minutes and
20 seconds to run from
A to
B at a speed of
3.8 ms−1 .
6a. Using ‘distance = speed
× time’, show that the distance from
A to
B is
1220 metres correct to 3 significant figures.
Markscheme
(A1)
3.8 × 320
Note: Award (A1) for
320 or equivalent seen.
= 1216
(A1)
= 1220 (m)
(AG)
Note: Both unrounded and rounded answer must be seen for the final (A1) to be awarded.
[2 marks]
[2 marks]
[1 mark]
6b. The distance from
B to
C is
850 metres. Running this part of the course takes Sarah
5 minutes and
3 seconds.
Calculate the speed, in
ms−1 , that Sarah runs from
B to
C.
Markscheme
850
303
(m s−1 ) (2.81, 2.80528… )
(A1)(G1)
[1 mark]
[3 marks]
6c. The distance from
B to
C is
850 metres. Running this part of the course takes Sarah
5 minutes and
3 seconds.
Calculate the distance, in metres, from
C to
A.
Markscheme
AC2 = 12202 + 8502 − 2(1220)(850) cos 110∘
(M1)(A1)
Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitutions.
AC = 1710 (m) (1708.87 … )
Notes: Accept
1705 (1705.33 …).
[3 marks]
B
C
850
5
3
(A1)(G2)
Markscheme
1220 + 850 + 1708.87 …
= 3780 (m) (3778.87 … )
(M1)
(A1)(ft)(G1)
Notes: Award (M1) for adding the three sides. Follow through from their answer to part (c). Accept
3771 (3771.33 …).
[2 marks]
[3 marks]
6e. The distance from
B to
C is
850 metres. Running this part of the course takes Sarah
5 minutes and
3 seconds.
Find the size of angle
BCA.
Markscheme
sin C
1220
=
sin 110 ∘
1708.87…
(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∘ (42.1339 … )
(A1)(ft)(G2)
Notes: Accept
41.9∘ , 42.0∘ , 42.2∘ , 42.3∘ .
OR
cos C =
1708.87…2 + 850 2 − 12202
2×1708.87…×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∘ (42.1339 … )
(A1)(ft)(G2)
Notes: Accept
41.2∘ , 41.8∘ , 42.4∘ .
[3 marks]
6f. The distance from
B to
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
AB,
BC and
CA.
[3 marks]
Markscheme
1
2
× 1220 × 850 × sin 110∘
(M1)(A1)(ft)
OR
1
2
× 1708.87 … × 850 × sin 42.1339 …∘
(M1)(A1)(ft)
OR
1
2
× 1220 × 1708.87 … × sin 27.8661 …∘
(M1)(A1)(ft)
Note: Award (M1) for substitution into area formula, (A1)(ft) for correct substitution.
= 487 000 m2 (487 230 … m2 )
(A1)(ft)(G2)
Notes: The answer is
487 000 m2 , units are required.
Accept
486 000 m2 (485 633 … m2 ).
If workings are not shown and units omitted, award (G1) for
487 000 or 486 000.
Follow through from parts (c) and (e).
[3 marks]
A parcel is in the shape of a rectangular prism, as shown in the diagram. It has a length
l cm, width
w cm and height of
20 cm.
The total volume of the parcel is
3000 cm3 .
7a. Express the volume of the parcel in terms of
[1 mark]
l and
w.
Markscheme
20lw OR
V = 20lw
(A1)
[1 mark]
7b. Show that
l=
150
.
w
[2 marks]
w
Markscheme
3000 = 20lw
(M1)
Note: Award (M1) for equating their answer to part (a) to
3000.
l=
3000
20w
(M1)
Note: Award (M1) for rearranging equation to make
l subject of the formula. The above equation must be seen to award (M1).
OR
150 = lw
(M1)
Note: Award (M1) for division by
20 on both sides. The above equation must be seen to award (M1).
l=
150
w
(AG)
[2 marks]
7c. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
Show that the length of string,
S cm, required to tie up the parcel can be written as
S = 40 + 4w +
300
, 0 < w ⩽ 20.
w
Markscheme
S = 2l + 4w + 2(20)
(M1)
Note: Award (M1) for setting up a correct expression for
S.
2 ( 150
) + 4w + 2(20)
w
(M1)
Notes: Award (M1) for correct substitution into the expression for
S. The above expression must be seen to award (M1).
= 40 + 4w + 300
w
[2 marks]
(AG)
[2 marks]
7d. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
Draw the graph of
S for
0 < w ⩽ 20 and
0 < S ⩽ 500, clearly showing the local minimum point. Use a scale of
2 cm to represent
5 units on the horizontal axis
w (cm), and a scale of
2 cm to represent
100 units on the vertical axis
S (cm).
[2 marks]
Markscheme
(A1)(A1)(A1)(A1)
Note: Award (A1) for correct scales, window and labels on axes, (A1) for approximately correct shape, (A1) for minimum point in
approximately correct position, (A1) for asymptotic behaviour at
w = 0.
Axes must be drawn with a ruler and labeled
w and
S.
For a smooth curve (with approximately correct shape) there should be one continuous thin line, no part of which is straight and
no (one-to-many) mappings of
w.
The
S-axis must be an asymptote. The curve must not touch the
S-axis nor must the curve approach the asymptote then deviate away later.
[4 marks]
7e. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
[3 marks]
Find
dS
.
dw
Markscheme
4 − 300
w2
(A1)(A1)(A1)
Notes: Award (A1) for
4, (A1) for
−300, (A1) for
1
or
w2
w−2 . If
extra terms present, award at most (A1)(A1)(A0).
[3 marks]
7f. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
Find the value of
w for which
S is a minimum.
Markscheme
4 − 3002 = 0 OR
w
300
w2
dS
dw
= 4 OR
=0
(M1)
Note: Award (M1) for equating their derivative to zero.
−−
w = 8.66 (√75 , 8.66025 …)
(A1)(ft)(G2)
Note: Follow through from their answer to part (e).
[2 marks]
[2 marks]
7g. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
[1 mark]
Write down the value,
l, of the parcel for which the length of string is a minimum.
Markscheme
17.3 ( 150 , 17.3205 …)
√75
(A1)(ft)
Note: Follow through from their answer to part (f).
[1 mark]
7h. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
Find the minimum length of string required to tie up the parcel.
Markscheme
−−
40 + 4√75 +
300
√75
(M1)
Note: Award (M1) for substitution of their answer to part (f) into the expression for
S.
–
= 110 (cm) (40 + 40√3, 109.282 …)
(A1)(ft)(G2)
Note: Do not accept
109.
Follow through from their answers to parts (f) and (g).
[2 marks]
[2 marks]
The front view of the edge of a water tank is drawn on a set of axes shown below.
The edge is modelled by
y = ax2 + c.
Point
P has coordinates
(−3,1.8), point
O has coordinates
(0,0) and point
Q has coordinates
(3,1.8).
8a. Write down the value of
[1 mark]
c.
Markscheme
(A1)(G1)
0
[1 mark]
[2 marks]
8b. Find the value of
a.
Markscheme
1.8 = a(3)2 + 0
(M1)
OR
1.8 = a(−3)2 + 0
(M1)
Note: Award (M1) for substitution of
y = 1.8 or
x = 3 and their value of
c into equation.
0 may be implied.
a = 0.2
( 15 )
(A1)(ft)(G1)
Note: Follow through from their answer to part (a).
Award (G1) for a correct answer only.
[2 marks]
8c. Hence write down the equation of the quadratic function which models the edge of the water tank.
[1 mark]
Markscheme
y = 0.2x2
(A1)(ft)
Note: Follow through from their answers to parts (a) and (b).
Answer must be an equation.
[1 mark]
8d. The water tank is shown below. It is partially filled with water.
[2 marks]
Calculate the value of y when
x = 2.4 m.
Markscheme
0.2 × (2.4)2 (M1)
= 1.15 (m)
(1.152) (A1)(ft)(G1)
Notes: Award (M1) for correctly substituted formula, (A1) for correct answer. Follow through from their answer to part (c).
Award (G1) for a correct answer only.
[2 marks]
8e. The water tank is shown below. It is partially filled with water.
State what the value of
x and the value of
y represent for this water tank.
[2 marks]
Markscheme
(A1)
y is the height
positive value of
x is half the width (or equivalent)
(A1)
[2 marks]
8f. The water tank is shown below. It is partially filled with water.
Find the value of
x when the height of water in the tank is
0.9 m.
Markscheme
0.9 = 0.2x2
(M1)
Note: Award (M1) for setting their equation equal to
0.9.
x = ±2.12 (m)
–
−−
−
(± 32 √2, ± √4.5 , ± 2.12132 …)
(A1)(ft)(G1)
Note: Accept
2.12. Award (G1) for a correct answer only.
[2 marks]
[2 marks]
8g. The water tank is shown below. It is partially filled with water.
When the water tank is filled to a height of
0.9 m, the front cross-sectional area of the water is
2.55 m2 .
(i)
Calculate the volume of water in the tank.
The total volume of the tank is
36 m3 .
(ii)
Calculate the percentage of water in the tank.
Markscheme
(i)
2.55 × 5
(M1)
Note: Award (M1) for correct substitution in formula.
= 12.8 (m3 )
(12.75 (m3 ))
(A1)(G2)
[2 marks]
(ii)
12.75
36
× 100
(M1)
Note: Award (M1) for correct quotient multiplied by
100.
= 35.4(%)
(35.4166 …)
(A1)(ft)(G2)
Note: Award (G2) for
35.6(%)(35.5555 … (%)).
Follow through from their answer to part (g)(i).
[2 marks]
[2 marks]
In triangle
ABC,
AC = 20 cm,
BC = 12 cm and
^ C = 90∘ .
AB
diagram not to scale
[2 marks]
9a. Find the length of
AB.
Markscheme
(AB2 ) = 202 − 122
(M1)
Note: Award (M1) for correctly substituted Pythagoras formula.
AB = 16 cm
[2 marks]
9b. D is the point on
AB such that
^ B) = 0.6.
tan(DC
Find the length of
DB.
(A1)
(C2)
[2 marks]
Markscheme
DB
12
= 0.6
(M1)
Note: Award (M1) for correct substitution in tangent ratio or equivalent ie seeing
12 × 0.6.
(A1)
DB = 7.2 cm
(C2)
Note: Award (M1)(A0) for using
tan 31 to get an answer of
7.21.
Award (M1)(A0) for
12
sin 59
=
DB
sin 31
to get an answer of
7.2103 … or any other incorrect answer.
[2 marks]
[2 marks]
9c. D is the point on
AB such that
^ B) = 0.6.
tan(DC
Find the area of triangle
ADC.
Markscheme
1
2
× 12 × (16 − 7.2)
(M1)
Note: Award (M1) for their correct substitution in triangle area formula.
OR
1
2
× 12 × 16 − 12 × 12 × 7.2
(M1)
Note: Award (M1) for subtraction of their two correct area formulas.
= 52.8 cm2
(A1)(ft)
(C2)
Notes: Follow through from parts (a) and (b).
Accept alternative methods.
[2 marks]
A child’s wooden toy consists of a hemisphere, of radius 9 cm , attached to a cone with the same base radius. O is the centre of the
base of the cone and V is vertically above O.
Angle OVB is
27.9∘ .
Diagram not to scale.
[2 marks]
10a. Calculate OV, the height of the cone.
Markscheme
tan 27.9∘ =
9
OV
(M1)
Note: Award (M1) for correct substitution in trig formula.
OV = 17.0 (cm) (16.9980 …)
[2 marks]
(A1)
(C2)
10b. Calculate the volume of wood used to make the toy.
[4 marks]
Markscheme
π(9) 2 (16.9980…)
3
+ 12 ×
4π(9) 3
3
(M1)(M1)(M1)
Note: Award (M1) for correctly substituted volume of the cone, (M1) for correctly substituted volume of a sphere divided by two
(hemisphere), (M1) for adding the correctly substituted volume of the cone to either a correctly substituted sphere or hemisphere.
= 2970 cm3 (2968.63 … )
(A1)(ft)
Note: The answer is
2970 cm3 , the units are required.
[4 marks]
(C4)
The diagram shows the points M(a, 18) and B(24, 10) . The straight line BM intersects the y-axis at A(0, 26). M is the midpoint of the
line segment AB.
[1 mark]
11a. Write down the value of
a.
Markscheme
(A1)
12
(C1)
Note: Award (A1) for
(12,18).
[1 mark]
[2 marks]
11b. Find the gradient of the line AB.
Markscheme
26−10
0−24
(M1)
Note: Accept
26−18
0−12
18−10
12−24
or
(or equivalent).
= − 23 (− 16
, − 0.666666 …)
24
(A1)
(C2)
Note: If either of the alternative fractions is used, follow through from their answer to part (a).
The answer is now (A1)(ft).
[2 marks]
11c. Decide whether triangle OAM is a right-angled triangle. Justify your answer.
[3 marks]
Markscheme
gradient of
OM =
3
2
(A1)(ft)
Note: Follow through from their answer to part (b).
− 23 × 32
(M1)
Note: Award (M1) for multiplying their gradients.
Since the product is
−1, OAM is a right-angled triangle
(R1)(ft)
Notes: Award the final (R1) only if their conclusion is consistent with their answer for the product of the gradients.
The statement that OAM is a right-angled triangle without justification is awarded no marks.
OR
(26 − 18)2 + 122 and
122 + 182
(A1)(ft)
( (26 − 18) + 122 ) + (122 + 182 ) = 262
2
(M1)
Note: This method can also be applied to triangle OMB.
Follow through from (a).
Hence a right angled triangle
(R1)(ft)
Note: Award the final (R1) only if their conclusion is consistent with their (M1) mark.
OR
OA = OB = 26 (cm) an isosceles triangle
Note:
(A1)
Award (A1) for
OA = 26 (cm) and
OB = 26 (cm).
Line drawn from vertex to midpoint of base is perpendicular to the base
Conclusion
(R1)
(M1)
(C3)
Note: Award, at most (A1)(M0)(R0) for stating that OAB is an isosceles triangle without any calculations.
[3 marks]
ABC is a triangular field on horizontal ground. The lengths of AB and AC are 70 m and 50 m respectively. The size of angle BCA is
78°.
[3 marks]
12a. Find the size of angle
ABC.
Markscheme
70
sin 78
=
50
^C
sin AB
(M1)(A1)
Note: Award (M1) for substituted sine rule, (A1) for correct substitution.
^ C = 44.3∘ (
AB
44.3209...)
(A1)(G3)
Note: If radians are used the answer is
0.375918..., award at most (M1)(A1)(A0).
[3 marks]
[4 marks]
12b. Find the area of the triangular field.
Markscheme
area ΔABC = 12 × 70 × 50 × sin(57.6790 …)
(A1)(M1)(A1)(ft)
Notes: Award (A1)(ft) for their
57.6790 … seen, (M1) for substituted area formula, (A1)(ft) for correct substitution.
Follow through from part (a).
= 1480 m2
(1478.86 …)
(A1)(ft)(G3)
Notes: The answer is
1480 m2 , units are required.
1479.20 … if 3 sf used.
If radians are used the answer is
1554.11 … m2 , award (A1)(ft)(M1)(A1)(ft)(A1)(ft)(G3).
[4 marks]
12c. M is the midpoint of
AC.
Find the length of
BM.
[3 marks]
Markscheme
BM2 = 702 + 252 − 2 × 70 × 25 × cos(57.6790 …)
(M1)(A1)(ft)
Notes: Award (M1) for substituted cosine rule, (A1)(ft) for correct substitution. Follow through from their angle in part (b).
BM = 60.4 (m)
(60.4457 …)
(A1)(ft)(G2)
Notes: If the 3 sf answer is used the answer is
60.5 (m).
If radians are used the answer is
62.5757 … (m), award (M1)(A1)(ft)(A1)(ft)(G2).
[3 marks]
12d. A vertical mobile phone mast,
TB, is built next to the field with its base at
B. The angle of elevation of
T from
M is
63.4∘ .
N is the midpoint of the mast.
Calculate the angle of elevation of
N from
M.
[5 marks]
Markscheme
tan 63.4∘ =
TB
60.4457…
(M1)
Note: Award (M1) for their correctly substituted trig equation.
(A1)(ft)
TB = 120.707 …
Notes: Follow through from part (c). If 3 sf answers are used throughout,
TB = 120.815 …
If
TB = 120.707 … is seen without working, award (A2).
( 120.707… )
2
^B=
tan NM
60.4457…
(A1)(ft)(M1)
Notes: Award (A1)(ft) for their
TB divided by
2 seen, (M1) for their correctly substituted trig equation.
Follow through from part (c) and within part (d).
^ B = 45.0∘
NM
(44.9563 … )
Notes:
(A1)(ft)(G3)
If 3 sf are used throughout, answer is
∘
45 .
If radians are used the answer is
0.308958 …, and if full working is shown, award at most (M1)(A1)(ft)(A1)(ft)(M1)(A0).
If no working is shown for radians answer, award (G2).
OR
^B=
tan NM
∘
tan 63.4 =
NB
BM
2×NB
BM
(M1)
(A1)(M1)
Note: Award (A1) for
2 × NB seen.
^ B = 1 tan 63.4∘
tan NM
2
^ B = 45.0∘
NM
(44.9563 …)
(M1)
(A1)(G3)
Notes: If radians are used the answer is
0.308958 …, and if full working is shown, award at most (M1)(A1)(M1)(M1)(A0). If no working is shown for radians answer,
award (G2).
[5 marks]
A lobster trap is made in the shape of half a cylinder. It is constructed from a steel frame with netting pulled tightly around it. The
steel frame consists of a rectangular base, two semicircular ends and two further support rods, as shown in the following diagram.
The semicircular ends each have radius
r and the support rods each have length
l.
Let
T be the total length of steel used in the frame of the lobster trap.
13a. Write down an expression for
[3 marks]
T in terms of
r,
l and
π.
Markscheme
2πr + 4r + 4l
(A1)(A1)(A1)
Notes: Award (A1) for
2πr (“
π” must be seen), (A1) for
4r, (A1) for
4l. Accept equivalent forms. Accept
T = 2πr + 4r + 4l. Award a maximum of (A1)(A1)(A0) if extra terms are seen.
[3 marks]
13b. The volume of the lobster trap is
0.75 m3 .
Write down an equation for the volume of the lobster trap in terms of
r,
l and
π.
[3 marks]
Markscheme
2
0.75 = πr2 l
Notes:
(A1)(A1)(A1)
Award (A1) for their formula equated to
0.75, (A1) for
l substituted into volume of cylinder formula, (A1) for volume of cylinder formula divided by
2.
If “
π” not seen in part (a) accept use of
3.14 or greater accuracy. Award a maximum of (A1)(A1)(A0) if extra terms are seen.
[3 marks]
13c. The volume of the lobster trap is
0.75
[2 marks]
m3 .
Show that
T = (2π + 4)r +
6
.
πr2
Markscheme
T = 2πr + 4r + 4 ( 1.52 )
πr
= (2π + 4)r + πr62
(A1)(ft)(A1)
(AG)
Notes: Award (A1)(ft) for correct rearrangement of their volume formula in part (b) seen, award (A1) for the correct substituted
formula for
T. The final line must be seen, with no incorrect working, for this second (A1) to be awarded.
[2 marks]
13d. The volume of the lobster trap is
0.75 m3 .
Find
dT
.
dr
Markscheme
dT
dr
= 2π + 4 −
12
πr3
(A1)(A1)(A1)
Note: Award (A1) for
2π + 4, (A1) for
−12
,
π
r−3 .
(A1) for
Accept 10.3 (10.2832…) for
2π + 4, accept
– 3.82
– 3.81971 … for
−12
.
π
Award a maximum of (A1)(A1)(A0) if extra terms are seen.
[3 marks]
[3 marks]
13e. The lobster trap is designed so that the length of steel used in its frame is a minimum.
[2 marks]
Show that the value of
r for which
T is a minimum is
0.719 m, correct to three significant figures.
Markscheme
2π + 4 − πr123 = 0 OR
dT
dr
(M1)
=0
Note: Award (M1) for setting their derivative equal to zero.
r = 0.718843 … OR
−−−−−−−−−
3−
√
0.371452 … OR
−−−12−−−
3
OR
√
π(2π+4)
−−
−
−
−
−
−
3.81971
3
√
(A1)
10.2832…
r = 0.719 (m)
(AG)
Note: The rounded and unrounded or formulaic answers must be seen for the final (A1) to be awarded. The use of
3.14 gives an unrounded answer of
r = 0.719039 ….
[2 marks]
13f. The lobster trap is designed so that the length of steel used in its frame is a minimum.
Calculate the value of
l for which
T is a minimum.
Markscheme
0.75 =
π× (0.719) 2 l
2
(M1)
Note: Award (M1) for substituting
0.719 into their volume formula. Follow through from part (b).
l = 0.924 (m)
(0.923599 …)
[2 marks]
T
(A1)(ft)(G2)
[2 marks]
Markscheme
T = (2π + 4) × 0.719 +
6
π(0.719) 2
(M1)
Notes: Award (M1) for substituting
0.719 in their expression for
T. Accept alternative methods, for example substitution of their
l and
0.719 into their part (a) (for which the answer is
11.08961024). Follow through from their answer to part (a).
= 11.1 (m)
(11.0880 …)
(A1)(ft)(G2)
The straight line, L1, has equation
2y − 3x = 11. The point A has coordinates (6, 0).
14a. Give a reason why L1 does not pass through A.
[1 mark]
Markscheme
2 × 0 − 3 × 6 ≠ 11
(R1)
Note: Stating
2 × 0 − 3 × 6 = −18 without a conclusion is not sufficient.
OR
Clear sketch of L1 and A.
(R1)
OR
Point A is (6, 0) and
2y − 3x = 11 has x-intercept at
− 113 or the line has only one x-intercept which occurs when x is negative.
14b. Find the gradient of L1.
(R1)
(C1)
[2 marks]
Markscheme
2y = 3x + 11 or
y − 32 x = 112
(M1)
Note: Award (M1) for a correct first step in making y the subject of the equation.
(gradient equals) = 32 (1.5)
(A1)
(C2)
Note: Do not accept 1.5x.
L is a line perpendicular to L1. The equation of L2 is
14c. 2
[1 mark]
y = mx + c.
Write down the value of m.
Markscheme
(m =) − 23
(A1)(ft)
(C1)
Notes: Follow through from their part (b).
14d. L2 does pass through A.
Find the value of c.
Markscheme
0 = − 23 (6) + c
(M1)
Note: Award (M1) for correct substitution of their gradient and (6, 0) into any form of the equation.
(c =) 4
(A1)(ft)
(C2)
Note: Follow through from part (c).
[2 marks]
A greenhouse ABCDPQ is constructed on a rectangular concrete base ABCD and is made of glass. Its shape is a right prism, with
cross section, ABQ, an isosceles triangle. The length of BC is 50 m, the length of AB is 10 m and the size of angle QBA is 35°.
15a. Write down the size of angle AQB.
[1 mark]
Markscheme
(A1)
110°
15b. Calculate the length of AQ.
[3 marks]
Markscheme
AQ
sin 35∘
=
10
sin 110 ∘
(M1)(A1)
Note: Award (M1) for substituted sine rule formula, (A1) for their correct substitutions.
OR
AQ =
5
cos 35∘
(A1)(M1)
Note: Award (A1) for 5 seen, (M1) for correctly substituted trigonometric ratio.
AQ = 6.10 (6.10387...)
(A1)(ft)(G2)
Notes: Follow through from their answer to part (a).
15c. Calculate the length of AC.
[2 marks]
Markscheme
AC 2 = 102 + 502
(M1)
Note: Award (M1) for correctly substituted Pythagoras formula.
−−−−
AC = 51.0(√2600 ,50.9901...)
(A1)(G2)
15d. Show that the length of CQ is 50.37 m, correct to 4 significant figures.
[2 marks]
Markscheme
QC 2 = (6.10387...)2 + (50)2
(M1)
Note: Award (M1) for correctly substituted Pythagoras formula.
QC = 50.3711...
= 50.37
(A1)
(AG)
Note: Both the unrounded and rounded answers must be seen to award (A1).
If 6.10 is used then 50.3707... is the unrounded answer.
For an incorrect follow through from part (b) award a maximum of (M1)(A0) – the given answer must be reached to award the
final (A1)(AG).
15e. Find the size of the angle AQC.
[3 marks]
Markscheme
cos AQC =
(6.10387...) 2 + (50.3711...) 2 − (50.9901...) 2
2(6.10387...)(50.3711...)
(M1)(A1)(ft)
Note: Award (M1) for substituted cosine rule formula, (A1)(ft) for their correct substitutions.
= 92.4° (
92.3753...∘ )
(A1)(ft)(G2)
Notes: Follow through from their answers to parts (b), (c) and (d). Accept 92.2 if the 3 sf answers to parts (b), (c) and (d) are used.
Accept 92.5° (
92.4858...∘ ) if the 3 sf answers to parts (b), (c) and 4 sf answers to part (d) used.
15f.
Calculate the total area of the glass needed to construct
(i) the two rectangular faces of the greenhouse;
(ii) the two triangular faces of the greenhouse.
[5 marks]
Markscheme
(i)
2(50 × 6.10387...)
(M1)
Note: Award (M1) for their correctly substituted rectangular area formula, the area of one rectangle is not sufficient.
= 610 m2 (610.387...)
(A1)(ft)(G2)
Notes: Follow through from their answer to part (b).
The answer is 610 m2. The units are required.
(ii) Area of triangular face
= 12 × 10 × 6.10387... × sin 35∘
(M1)(A1)(ft)
OR
Area of triangular face
= 12 × 6.10387... × 6.10387... × sin 110∘
(M1)(A1)(ft)
= 17.5051...
Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correct substitutions.
OR
(Height of triangle)
= (6.10387...)2 − 52
= 3.50103...
Area of triangular face
= 12 × 10 × their height
= 17.5051...
Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correctly substituted area formula. If 6.1 is used, the height is
3.49428... and the area of both triangular faces 34.9 m2
Area of both triangular faces = 35.0 m2 (35.0103...)
(A1)(ft)(G2)
Notes: The answer is 35.0 m2. The units are required. Do not penalize if already penalized in part (f)(i). Follow through from their
part (b).
15g. The cost of one square metre of glass used to construct the greenhouse is 4.80 USD.
Calculate the cost of glass to make the greenhouse. Give your answer correct to the nearest 100 USD.
[3 marks]
Markscheme
(610.387... + 35.0103...) × 4.80
= 3097.90...
(M1)
(A1)(ft)
Notes: Follow through from their answers to parts (f)(i) and (f)(ii).
Accept 3096 if the 3 sf answers to part (f) are used.
(A1)(ft)(G2)
= 3100
Notes: Follow through from their unrounded answer, irrespective of whether it is correct. Award (M1)(A2) if working is shown and
3100 seen without the unrounded answer being given.
The quadrilateral ABCD has AB = 10 cm, AD = 12 cm and CD = 7 cm.
The size of angle ABC is 100°​ and the size of angle ACB is 50​°.
16a.
Find the length of AC in centimetres.
[3 marks]
Markscheme
AC
sin 100 ∘
=
10
sin 50∘
(M1)(A1)
Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.
= 12.9(12.8557...)
(A1)
(C3)
Note: Radian answer is 19.3, award (M1)(A1)(A0).
16b.
Find the size of angle ADC.
[3 marks]
Markscheme
122 + 72 − 12.8557... 2
2×12×7
(M1)(A1)(ft)
Note: Award (M1) for substitution in the cosine rule formula, (A1)(ft) for correct substitutions.
= 80.5° (80.4994...°)
(A1)(ft)
(C3)
Notes: Follow through from their answer to part (a). Accept 80.9° for using 12.9. Using the radian answer from part (a) leads to an
impossible triangle, award (M1)(A1)(ft)(A0).
The equation of a line L1 is
2x + 5y = −4.
17a.
Write down the gradient of the line L1.
[1 mark]
Markscheme
−2
5
(A1)
(C1)
17b. A second line L2 is perpendicular to L1.
[1 mark]
Write down the gradient of L2.
Markscheme
5
2
(A1)(ft)
(C1)
Note: Follow through from their answer to part (a).
17c. The point (5, 3) is on L2.
[2 marks]
Determine the equation of L2.
Markscheme
3 = 52 × 5 + c
(M1)
Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).
y = 52 x − 192
(A1)(ft)
OR
y − 3 = 52 (x − 5)
(M1)(A1)(ft)
(C2)
Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).
17d. Lines L1 and L2 intersect at point P.
Using your graphic display calculator or otherwise, find the coordinates of P.
[2 marks]
Markscheme
(3, −2)
(A1)(ft)(A1)(ft)
(C2)
Notes: If parentheses not seen award at most (A0)(A1)(ft). Accept x = 3, y = −2. Follow through from their answer to part (c), even if
no working is seen. Award (M1)(A1)(ft) for a sensible attempt to solve
2x + 5y = −4 and their
y = 52 x − 192 or equivalent, simultaneously.
The diagram below represents a rectangular flag with dimensions 150 cm by 92 cm. The flag is divided into three regions A, B and
C.
18a.
Write down the total area of the flag.
[1 mark]
Markscheme
Units are required in this question for full marks to be awarded.
13800 cm2
(A1)
(C1)
18b. Write down the value of y.
[1 mark]
Markscheme
75
(A1)
(C1)
18c. The areas of regions A, B, and C are equal.
[1 mark]
Write down the area of region A.
Markscheme
Units are required in this question for full marks to be awarded.
4600 cm2
(A1)(ft)
(C1)
Notes: Units are required unless already penalized in part (a). Follow through from their part (a).
18d. Using your answers to parts (b) and (c), find the value of x.
[3 marks]
Markscheme
0.5(x + 92) × 75 = 4600
(M1)(A1)(ft)
OR
0.5 × 150 × (92 − x) = 4600
(M1)(A1)(ft)
Note: Award (M1) for substitution into area formula, (A1)(ft) for their correct substitution.
(= 30.7 (cm)(30.6666...(cm))
(A1)(ft)
(C3)
Note: Follow through from their parts (b) and (c).
A tent is in the shape of a triangular right prism as shown in the diagram below.
The tent has a rectangular base PQRS .
PTS and QVR are isosceles triangles such that PT = TS and QV = VR .
PS is 3.2 m , SR is 4.7 m and the angle TSP is 35​°.
19a.
Show that the length of side ST is 1.95 m, correct to 3 significant figures.
[3 marks]
Markscheme
ST =
1.6
cos 35∘
(M1)(A1)
Note: Award (M1) for correctly substituted trig equation, (A1) for 1.6 seen.
OR
ST
sin 35∘
=
3.2
sin 110 ∘
(M1)(A1)
Note: Award (M1) for substituted sine rule equation, (A1) for correct substitutions.
ST = 1.95323...
= 1.95 (m)
(A1)
(AG)
Notes: Both unrounded and rounded answer must be seen for final (A1) to be awarded.
19b.
Calculate the area of the triangle PTS.
[3 marks]
Markscheme
1
2
1
2
× 3.2 × 1.95323... × sin 35∘ or
× 1.95323... × 1.95323... × sin 110∘
(M1)(A1)
Note: Award (M1) for substituted area formula, (A1) for correct substitutions. Do not award follow through marks.
= 1.79 m2 (1.79253...m2)
(A1)(G2)
Notes: The answer is 1.79 m2, units are required. Accept 1.78955... from using 1.95.
OR
1
2
× 3.2 × 1.12033...
(A1) (M1)
Note: Award (A1) for the correct value for TM (1.12033...) OR correct expression for TM (i.e. 1.6tan35°,
−−−−−−−−−−−−−−−
√(1.95323...) 2 − 1.62 ), (M1) for correctly substituted formula for triangle area.
= 1.79 m2 (1.79253...m2)
(A1)(G2)
Notes: The answer is 1.79 m2, units are required. Accept 1.78 m2 from using 1.95.
19c.
Write down the area of the rectangle STVR.
[1 mark]
Markscheme
9.18 m2 (9.18022 m2)
(A1)(G1)
Notes: The answer is 9.18 m2, units are required. Do not penalize if lack of units was already penalized in (b). Do not award follow
through marks here. Accept 9.17 m2 (9.165 m2) from using 1.95.
19d. Calculate the total surface area of the tent, including the base.
[3 marks]
Markscheme
2 × 1.79253... + 2 × 9.18022... + 4.7 × 3.2
(M1)(A1)(ft)
Note: Award (M1) for addition of three products, (A1)(ft) for three correct products.
= 37.0 m2 (36.9855...m2)
(A1)(ft)(G2)
Notes: The answer is 37.0 m2, units are required. Accept 36.98 m2 from using 3sf answers. Follow through from their answers to
(b) and (c). Do not penalize if lack of units was penalized earlier in the question.
19e. Calculate the volume of the tent.
[2 marks]
Markscheme
1.79253... × 4.7
(M1)
Note: Award (M1) for their correctly substituted volume formula.
= 8.42 m3 (8.42489...m3)
(A1)(ft)(G2)
Notes: The answer is 8.42 m3, units are required. Accept 8.41 m3 from use of 1.79. An answer of 8.35, from use of TM = 1.11,
will receive follow-through marks if working is shown. Follow through from their answer to part (b). Do not penalize if lack of units
was penalized earlier in the question.
19f.
A pole is placed from V to M, the midpoint of PS.
[4 marks]
Find in metres,
(i) the height of the tent, TM;
(ii) the length of the pole, VM.
Markscheme
(i)
TM = 1.6 tan 35∘
(M1)
Notes: Award (M1) for their correct substitution in trig ratio.
OR
−−−−−−−−−−−−−−−
TM = √(1.95323...) 2 − 1.62
(M1)
Note: Award (M1) for correct substitution in Pythagoras’ theorem.
OR
3.2×TM
2
= 1.79253...
(M1)
Note: Award (M1) for their correct substitution in area of triangle formula.
= 1.12 (m) (1.12033...)
(A1)(ft)(G2)
Notes: Follow through from their answer to (b) if area of triangle is used. Accept 1.11 (1.11467) from use of ST = 1.95.
(ii)
−−−−−−−−−−−−−−
VM = √1.12033...2 + 4.72
(M1)
Note: Award (M1) for their correct substitution in Pythagoras’ theorem.
= 4.83 (m) (4.83168 )
(A1)(ft)(G2)
Notes: Follow through from (f)(i).
19g. Calculate the angle between VM and the base of the tent.
[2 marks]
Markscheme
sin−1 ( 1.12033...
)
4.83168...
(M1)
OR
4.7
cos−1 ( 4.83168...
)
(M1)
OR
tan−1 ( 1.12033...
)
4.7
(M1)
Note: Award (M1) for correctly substituted trig equation.
OR
cos−1 (
4.72 + (4.83168...) 2 − (1.12033...) 2
2×4.7×4.83168...
)
(M1)
Note: Award (M1) for correctly substituted cosine formula.
= 13.4° (13.4073...)
(A1)(ft)(G2)
Notes: Accept 13.3°. Follow through from part (f).
A manufacturer has a contract to make
2600 solid blocks of wood. Each block is in the shape of a right triangular prism,
ABCDEF, as shown in the diagram.
AB = 30 cm, BC = 24 cm, CD = 25 cm and angle
^ C = 35∘ .
AB
20a. Calculate the length of
AC.
[3 marks]
Markscheme
AC2 = 302 + 242 − 2 × 30 × 24 × cos 35∘
(M1)(A1)
Note: Award (M1) for substituted cosine rule formula,
(A1) for correct substitutions.
AC = 17.2 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
ACB = 90∘ .
[3 marks]
[3 marks]
20b. Calculate the area of triangle
ABC.
Markscheme
Units are required in part (b).
Area of triangle
ABC = 12 × 24 × 30 × sin 35∘
(M1)(A1)
Notes: Award (M1) for substitution into area formula, (A1) for correct substitutions.
Special Case: Where a candidate has assumed that angle
ACB = 90∘ in part (a), award (M1)(A1) for a correct alternative substituted formula for the area of the triangle
(ie 12 × base × height).
= 206 cm2
(206.487 … cm2 )
(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]
20c. Assuming that no wood is wasted, show that the volume of wood required to make all
2600 blocks is
13 400 000 cm3 , correct to three significant figures.
[2 marks]
Markscheme
206.487 … × 25 × 2600
(M1)
Note: Award (M1) for multiplication of their answer to part (b) by
25 and
2600.
13 421 688.61
(A1)
Note: Accept unrounded answer of
13 390 000 for use of
206.
13 400 000
(AG)
Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.
[2 marks]
[2 marks]
20d. Write
13 400 000 in the form
a × 10k where
1 ⩽ a < 10 and
k ∈ Z.
Markscheme
1.34 × 107
(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 × 106 ).
[2 marks]
20e. Show that the total surface area of one block is
2190
cm2 ,
correct to three significant figures.
[3 marks]
Markscheme
2 × 206.487 … + 24 × 25 + 30 × 25 + 17.2168 … × 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.
(A1)
2193.26 …
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]
20f. The blocks are to be painted. One litre of paint will cover
[3 marks]
22 m2 .
Calculate the number of litres required to paint all
2600 blocks.
Markscheme
2190×2600
22×10 000
(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]
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