Edexcel GCSE Maths (1MA1), Higher tier · Geometry and measures
Practise Volume and surface area. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
The diagram shows a prism. The cross-section of the prism is a trapezium. Diagram NOT accurately drawn
[object Object]
Work out the volume of the prism.[3]
Show the answer and mark scheme
Answer: 782 cm³
M1 for ½ × (9 + 14) × 4 (= 46)
M1 for 46 × 17 (dep on a method for the area of the cross-section)
A1 for 782
Worked solution: Area of trapezium \(= \frac{1}{2}(9 + 14) \times 4 = 46\) cm² Volume \(= 46 \times 17 = 782\) cm³
Question 2Medium3 marks
A tank in the shape of a cuboid measures 40 cm by 30 cm by 25 cm. It is full of water. All of the water is poured into an empty cylinder with radius 15 cm and height 40 cm.
Will the water overflow? You must show your working.[3]
Show the answer and mark scheme
Answer: Yes: the water has volume 30 000 cm³ but the cylinder holds only \(\pi \times 15^2 \times 40 = 28\,274\) cm³.
M1 for \(40 \times 30 \times 25\) (= 30 000)
M1 for \(\pi \times 15^2 \times 40\) (= 28 274...)
C1 for 'yes', with 30 000 compared with 28 274 (or 28 300)
Worked solution: Water: \(40 \times 30 \times 25 = 30\,000\) cm3. Cylinder: \(\pi \times 225 \times 40 = 28\,274.3\ldots\) cm3. 30 000 > 28 274, so the water overflows.
Question 3Hard6 marks
A solid is made from a hemisphere and a cone. They have the same radius, \(r\) cm. The flat face of the hemisphere is joined to the base of the cone. The volume of the solid is equal to the volume of a sphere of radius \(r\) cm. Volume of sphere \(= \frac{4}{3}\pi r^3\) Volume of cone \(= \frac{1}{3}\pi r^2 h\) Surface area of sphere \(= 4\pi r^2\) Curved surface area of cone \(= \pi r l\)
(a) Find the height of the cone in terms of \(r\).[3]
(b) The radius is 3 cm. Show that the total surface area of the solid is \(9\pi(2 + \sqrt{5})\) cm2.[3]
Show the answer and mark scheme
(a)Answer: \(h = 2r\)
M1 for the hemisphere volume \(\frac{2}{3}\pi r^3\)
M1 for \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi r^3\)
A1 for \(h = 2r\)
Worked solution: \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi r^3\), so \(\frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3\) and \(h = 2r\).