Write down the name of a solid that has 4 triangular faces, 4 vertices and 6 edges.
(1)
2
State how many faces, edges and vertices a triangular prism has.
(1)
3
A cone has height 9 cm and base radius 4 cm. State how many curved faces, flat faces and edges it has.
(2)
4
A net is shown for a triangular prism made from two equilateral triangular faces and three rectangular faces. The triangular face has side 8 cm and each rectangle has height 10 cm. Calculate the total surface area of the prism.
(2)
5
Draw the front elevation and a top plan for a cuboid with a small square hole cut through the centre from the top to the bottom. The cuboid measures 8 cm by 5 cm by 3 cm (length by width by height). The hole is a square of side 1 cm centred on the top face. Describe the shapes seen in the front elevation and top plan.
(2)
6
A cone and a cylinder have the same base radius 5 cm and the same height 12 cm. State whether the cone has the same volume as the cylinder. Give a brief reason.
(2)
7
Identify the net of a tetrahedron (triangular-based pyramid) from these descriptions and give a reason: (i) four squares attached in a strip, (ii) one triangle with three rectangles attached, (iii) one triangle with three triangles attached.
(2)
8
A solid is made by attaching a hemisphere of radius 6 cm to the flat face of a cylinder of radius 6 cm and height 15 cm. Write down the shapes you would see in the top plan and the front elevation.
(2)
9
A cube has side length 5 cm. Calculate the length of the diagonal across one face and give the units.
(2)
10
A rectangular block measures 10 cm by 7 cm by 4 cm. Give the three different plans (top, front, side) in terms of rectangle dimensions. List them as width x depth.
(2)
11
Write down the name of the solid that has one circular flat face, an apex and one curved surface.
(1)
12
State the number of vertices of a cube.
(1)
13
A solid is a right triangular prism with triangular cross-section sides 5 cm, 12 cm, 13 cm and length 9 cm. Calculate the volume of the prism. (Area of triangle = 1/2 x base x height, using the two shorter sides as base and height.)
(3)
14
A cube of side 8 cm has a smaller cube removed from one corner. The small cube has side 3 cm. Calculate the surface area of the remaining solid. (Remember new faces from the cut may be exposed.)
(3)
15
A model is made from a cuboid 10 cm by 6 cm by 4 cm with a triangular prism cut out along its length. The triangular prism removed has right triangle cross-section with legs 3 cm and 4 cm and length 6 cm. Calculate the volume of the remaining solid.
(3)
16
A solid consists of a right circular cone of height 7 cm and base radius 3 cm sitting on top of a cylinder of height 10 cm and the same base radius. Calculate the total surface area excluding the base of the cylinder that sits on the table (include the curved surfaces and the top of the cone). Give your answer in cm2. Use π as π.
(3)
17
A shape is built from 8 identical small cubes of side 3 cm arranged to form a larger cube, with one small cube missing from a corner. Draw the top plan and front elevation and calculate the total surface area of the resulting shape.
(4)
18
A castle model has a rectangular tower 16 cm by 9 cm base and height 25 cm. On top of the tower sits a crenelated parapet formed by removing four equal square prisms of side 2 cm from the corners of a thin slab of thickness 2 cm. Calculate the surface area added to the tower by the parapet (include exposed top and sides of parapet and the new exposed faces on the tower where the parapet sits).
(4)
Mark scheme · KS3.M-G20D 3D Shapes, Plans and Elevations: Fluency and Exam Drill
Question 1
B1 Tetrahedron (or triangular-based pyramid) cao
Answer: Tetrahedron (triangular-based pyramid)
Question 2
B1 all three correct: faces = 5, edges = 9, vertices = 6 cao
Answer: Faces = 5, Edges = 9, Vertices = 6
Question 3
M1 identifies two categories correctly (eg curved face = 1 and flat face = 1)
A1 all three correct: 1 curved face, 1 flat face, 1 edge cao
M1 area of two equilateral triangles found (2 x √3/4 x 82 = 32sqrt(3)) or area of three rectangles found (3 x 8 x 10 = 240)
A1 total surface area = 240 + 32sqrt(3) cm2, approximately 295.4 cm2 cao
Answer: 240 + 32sqrt(3) cm2 (approx 295.4 cm2)
Question 5
M1 identifies front elevation shows an 8 cm by 3 cm rectangle
A1 identifies top plan shows an 8 cm by 5 cm rectangle with a central 1 cm square hole cao
Answer: Front elevation: an 8 cm by 3 cm rectangle (no hole visible from the front); Top plan: an 8 cm by 5 cm rectangle with a central 1 cm square hole
Question 6
M1 some correct idea about the relationship of cone and cylinder volumes, e.g. cone volume = (1/3) cylinder volume if same height
A1 answer stating cone could not have same volume unless its height was 36 cm (cao) or explaining the ratio 1:3
Answer: No. A cone with the same base radius and height as a cylinder has volume 1/3 of the cylinder's volume; to equal the cylinder the cone would need height 36 cm (1/3 rule), so with height 12 cm it is smaller
Question 7
B1 chooses (iii) as the correct net
B1 gives reason that a tetrahedron has four triangular faces meeting at a point (oe)
Answer: (iii) because a tetrahedron is made entirely of four triangular faces
Question 8
M1 identifies top plan shows a circle of radius 6 cm
A1 identifies front elevation shows a rectangle (cylinder height 15) topped by a semicircle of radius 6 cm cao
Answer: Top plan: circle radius 6 cm. Front elevation: rectangle 15 cm high with a semicircle of radius 6 cm on top
Question 9
M1 uses Pythagoras correctly: diagonal2 = 52 + 52
A1 diagonal = 5sqrt(2) cm or 7.071... cm cao
Answer: 5sqrt(2) cm
Question 10
M1 gives two of the three plans correct
A1 all three correct: 10 x 7, 10 x 4, 7 x 4 cao
Answer: Top: 10 x 7, Front: 10 x 4, Side: 7 x 4
Question 11
B1 Cone cao
Answer: Cone
Question 12
B1 8 cao
Answer: 8
Question 13
M1 finds area of triangular cross-section using base 5 and height 12, right triangle area = 1/2 x 5 x 12 = 30
M1 multiplies area by length 9 correctly to get volume method
A1 volume = 270 cm3 cao
Answer: 270 cm3
Question 14
M1 calculates original cube area 6 x 82 = 384 cm2 or calculates area removed 3 x 32 = 27 cm2
M1 accounts for newly exposed faces from the cut (three faces of size 3x3 appear) or equivalent method
A1 final surface area = 384 - 3*(32) + 3*(32) = 384 cm2 cao (the removed and newly exposed corner faces are equal in area, so the total is unchanged)
Answer: 384 cm2
Question 15
M1 calculates volume of cuboid = 10 x 6 x 4 = 240 cm3 or area of triangular cross-section = 1/2 x 3 x 4 = 6
M1 calculates volume removed = 6 x 6 = 36 cm3
A1 volume remaining = 240 - 36 = 204 cm3 cao
Answer: 204 cm3
Question 16
M1 finds curved surface area of cylinder = 2 π r h = 2 π x 3 x 10 = 60 π
M1 finds slant height of cone l = √72 + 32 = √58 and curved surface area of cone = π r l = π x 3 x √58
A1 adds areas: total = 60 π + 3 π √58 (cao) written as (60 + 3sqrt(58)) π cm2
Answer: (60 + 3sqrt(58)) π cm2
Question 17
M1 correct description or sketch of top plan and front elevation showing a 2x2 arrangement with one corner square missing (a step visible in one corner)
M1 finds surface area of the complete 2x2x2 cube (side 6 cm) = 6 x 62 = 216 cm2
M1 recognises that removing a corner cube takes away 3 exterior faces (3 x 32 = 27 cm2) but exposes 3 new faces of equal total area (3 x 32 = 27 cm2), leaving the surface area unchanged
A1 final surface area = 216 cm2 cao
Answer: 216 cm2
Question 18
M1 calculates area of top slab before cuts = 16 x 9 = 144 and area removed by four corner squares = 4 x 22 = 16, so exposed top of parapet = 128
M1 calculates exposed vertical wall area of the parapet: each corner notch removes as much edge length as it adds, so the perimeter is unchanged at 2 x (16+9) = 50 cm, giving wall area = 50 x 2 = 100 cm2
M1 recognises the 128 cm2 of exposed slab top exactly replaces the 128 cm2 of tower top it now covers (net zero change there), and the 16 cm2 of tower top seen through the notches was already exposed before the parapet was added (also no change), so neither contributes to the total added
A1 gives final added surface area = 100 cm2 cao (the parapet's own vertical wall area is the only genuinely new surface)