A cylindrical tank has radius 0.9 m and height 2.5 m. Calculate the volume of the tank. Give your answer to 3 significant figures.
(Total for Question 1 is 3 marks)
2
A triangular prism has a triangular cross-section with base 8 cm and height 6 cm. The length of the prism is 40 cm. Calculate the volume of the prism in cubic centimetres.
(Total for Question 2 is 3 marks)
3
A right triangular prism has a triangular cross-section with base 5 m and height 3 m. The volume of the prism is 150 m3. Work out the length of the prism.
(Total for Question 3 is 3 marks)
4
The cross-section of a prism is a trapezium with parallel sides 6x cm and (x + 4) cm and distance between the parallel sides 5 cm. The length of the prism is 12 cm. Express the volume of the prism in terms of x and simplify.
(Total for Question 4 is 3 marks)
5
Show that the volume, V cm3, of a prism whose cross-section is an isosceles triangle with sides 2x, 2x and base 4 is V = 4*√x2 - 1 * L, where L is the length of the prism.
(Total for Question 5 is 4 marks)
6
A metal prism has a triangular cross-section with base 9 cm and height 6 cm. The prism is 80 cm long. The metal has density 7.8 g/cm3. Calculate the mass of the prism in kilograms. Give your answer to 3 significant figures.
(Total for Question 6 is 5 marks)
7
A cylindrical column has height 3.6 m and volume 40.715 m3. Calculate the radius of the column. Give your answer correct to 2 decimal places.
(Total for Question 7 is 4 marks)
8
Two similar triangular prisms A and B have corresponding linear dimensions in the ratio 1 : k. Prism A has cross-sectional area 24 cm2 and length 15 cm. The volume of prism B is 2880 cm3. Find the scale factor k and the cross-sectional area of prism B.
(Total for Question 8 is 5 marks)
Mark scheme · 4.13H Volume of a Prism: Higher Tier Practice
Question 1
M1 Use of formula V = π r2 h with r = 0.9 and h = 2.5 or equivalent method
M1 substitution and evaluation to a correct intermediate value, e.g. π * 0.92 * 2.5 = 6.362... m3
A1 6.36 m3 awrt 3 s.f.
Answer: 6.36 m3 (3 s.f.)
Question 2
M1 Use of area triangle = 1/2 * base * height = 1/2 * 8 * 6 = 24 cm2
M1 Multiply cross-sectional area by length: 24 * 40
A1 960 cm3 cao
Answer: 960 cm3
Question 3
M1 Use V = area cross-section * length and area = 1/2 * 5 * 3 = 7.5 m2
M1 Set up length = V / area = 150 / 7.5
A1 20 m cao
Answer: 20 m
Question 4
M1 Use area trapezium = 1/2*(sum parallel sides)*distance = 1/2*(6x + x + 4)*5
dM1 split the isosceles triangle into two right-angled triangles using the altitude from the apex to the base, each with hypotenuse 2x and base 2 (half of the base 4)
M1 use Pythagoras to find the altitude: √(2x)2 - 22 = √4x2 - 4 = 2*√x2 - 1
M1 area = 1/2 * base * height = 1/2 * 4 * 2*√x2 - 1 = 4*√x2 - 1
A1 multiply area by length L to get V = 4*√x2 - 1 * L, as required cso
Answer: V = 4*√x2 - 1 * L cso
Question 6
M1 Find area triangle = 1/2 * 9 * 6 = 27 cm2
M1 Find volume = 27 * 80 = 2160 cm3
M1 Find mass in grams = volume * density = 2160 * 7.8 = 16848 g
A1 Convert to kilograms: 16.848 kg
A1 Give answer to 3 s.f.: 16.8 kg awrt
Answer: 16.8 kg (3 s.f.)
Question 7
M1 Use V = π r2 h and rearrange r = √V/(&π; h)
M1 Substitute values r = √40.715 / (&π; * 3.6)
M1 Evaluate to a correct numerical radius
A1 1.90 m to 2 d.p. cao
Answer: 1.90 m
Question 8
M1 Use similarity: areas scale as k2 and lengths scale as k, so volumes scale as k3
M1 Volume of A = areaA * lengthA = 24 * 15 = 360 cm3; set up 360 * k3 = 2880