1.
A firm makes cone shaped containers out of card.
The card is in the shape of a sector of a circle of radius 12 cm.
The angle of the sector is 270°.
The straight edges are brought together to make the cone.
12 cm
270º
12 cm
12 cm
(a)
Find the arc length of the card used to make the cone.
Give your answer in terms of .
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Answer .......................................................... cm
(2)
(b)
Calculate the radius of the base of the cone.
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Answer .......................................................... cm
(2)
(Total 4 marks)
2.
A container consists of a cylinder on top of a cone.
The container is full of oil.
4m
1.5 m
0.9 m
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1
The diameter of both the cylinder and the cone is 4 m.
The height of the cone is 0.9 m and the height of the cylinder is 1.5 m.
Calculate the volume of oil in the container.
Give your answer in terms of .
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Answer .................................................... m3
(Total 3 marks)
3.
A marble paperweight consists of a cuboid and a hemisphere as shown in the diagram.
The hemisphere has a radius of 4 cm.
4 cm
5 cm
10 cm
10 cm
Not to scale
Calculate the volume of the paperweight.
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Answer ………………….……………..
(Total 4 marks)
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2
4.
A child’s toy is in the shape of a cone on top of a hemisphere.
The diameter of the hemisphere is 15 cm and the overall height of the toy is 26 cm.
Not to scale
26 cm
15 cm
Calculate the volume of this toy.
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Answer ................................................................. cm3
(Total 5 marks)
5.
The diagram shows a float made from two cones with dimensions as shown.
Not to scale
12 cm
6 cm
10 cm
Calculate the total surface area of the float.
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Answer .......................................................... cm2
(Total 5 marks)
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6.
A hemispherical bowl of radius 6 cm has the same volume as a cone of perpendicular
height 27 cm.
Not drawn accurately
27 cm
6 cm
r
Calculate the base radius, r, of the cone.
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Answer ................................................................... cm
(Total 4 marks)
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4
1.
3
(a)
4
× (2 ×  × 12)M1
18
(b)
A1
Not 18, unless notation previously penalised
 × 18 is acceptable
2 ×  × r = their 18
Or their 18 ÷ 2
M1
r=9
A1ft
r=
3
4
of 12 = 9 scores 2 marks
[4]
2.
 × 22 × 1.5 or 6 
1
 × 22 × 0.9 or 1.2
3
7.2 
M1
M1
A1
[3]
3.
Use of
4
π  43  2
3
M1
Must use 4 or 8 as radius.
(Volume hemisphere =) 133.9
to 134.1 (inclusive)
133.97 if  = 3.14 used.
Al
(Volume paperweight =)
500+(their 134) (=634)
If Ml awarded.
Alf
cm3
B1
This mark is independent
[4]
4.
4  7.53 or 2  7.53
3
3
M1
883.1 to 883.6
A1*
18.5 used as height
B1
1  7.52 18.5
3
M1
Allow 26 here
1089.1 to 1089.75
A1*
(*) score one of these only
1972 to 1973.35
A1
Use of r as 15 throughout gives 9660
SC2
[5]
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5.
Slant height top cone = 13 cm
B1
Slant height bottom cone (52+62)
Must have 
M1
Slant height = 7.8(1…)
A1
Area =  × 5 × 13
or  × 5 × their ‘7.8’
Either slant height can be used but M1 must be awarded for
calculating slant height of bottom cone
= 327 or 326.9 or 326.89 or 326.88(…)
NB. Anything involving the area of the two circular ”bases‘ can
be ignored except if it affects the final answer
DM1
A1
[5]
6.
Vol Hemisphere =
1
2
1
2

4
3
4
3
   63
M1
   6 3  13    r 2  27 Ml, M1
144
A1
(3 × 4 × 6 × )  (2 × 3 × 27 × ) = r or...
3
1
3
2
   r 2  27  their 144
M1
...(4 × 6 )  (2 × 27) = r A1
3
(r =) 4
2
A1
(r =) 4
[4]
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