Last reviewed: 11 November 2026
Updated November 2026. These Grade 9 volume of solids notes show how to find the volume of triangular and rectangular based prisms, triangular, rectangular and square based pyramids, a cone, a frustum and a sphere. Volume measures the space inside a solid, in cubic units such as cm3 or m3. A chalk box, a conical funnel cut from a larger cone, and a football are everyday objects that match this sub-strand.
The examples are original practice for Kenyan junior school learners, not an official paper. They follow the Grade 9 Mathematics measurements sub-strand on volume of solids in the KICD Grade Nine curriculum designs. Surface area is covered separately in the Grade 9 area notes. For the topic list, use the Grade 9 maths revision guide and the Grade 9 notes hub.
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Where Volume of Solids Sits in Grade 9 Maths
By the end of the sub-strand you should be able to work out the volume of a triangular and a rectangular based prism; calculate the volume of a triangular, rectangular and square based pyramid; work out the volume of a cone; determine the volume of a frustum; and calculate the volume of a sphere in real-life situations. Cubes, cuboids and cylinders were met earlier, including in the Grade 7 volume and capacity notes. These notes do not treat mass or density; that is the next measurements topic.
Volume of Prisms and Pyramids
A prism has the same cross-section all along its length, so its volume is the area of that end multiplied by the length. A pyramid tapers to a point, so its volume is one-third of the matching prism: V = (1/3) × base area × perpendicular height. The same splitting-into-known-shapes approach is used in OpenStax Prealgebra’s section on volume and surface area.
| Solid | Volume |
|---|---|
| Rectangular based prism (cuboid) | V = l × w × h |
| Triangular based prism | V = (area of triangular end) × length |
| Pyramid (any of the bases in the design) | V = (1/3) × base area × height |
Worked example: triangular and rectangular prisms
A wooden door wedge is a triangular prism. The right-angled triangular end has sides 3 cm, 4 cm and 5 cm, and the wedge is 10 cm long. Area of the end = (1/2) × 3 × 4 = 6 cm2. Volume = 6 × 10 = 60 cm3.
A closed chalk box measures 8 cm by 5 cm by 4 cm. Volume = 8 × 5 × 4 = 160 cm3.
Worked example: square based pyramid
A decorative model pyramid has a square base of side 6 cm and a perpendicular height of 9 cm. Base area = 36 cm2. Volume = (1/3) × 36 × 9 = 108 cm3. Use the perpendicular height from the apex to the base, not the slant edge along a face.
Volume of a Cone, a Frustum and a Sphere
A cone is a circular based pyramid, so V = (1/3)πr2h. A frustum is the slice that remains when a smaller similar pyramid or cone is cut off parallel to the base. You can subtract the small cone from the large cone, or use V = (1/3)πh(R2 + Rr + r2), where h is the height of the frustum itself, R is the larger radius and r is the smaller radius. A sphere has volume V = (4/3)πr3.
Worked example: cone
A conical paper cup has radius 7 cm and perpendicular height 12 cm. Taking π = 22/7, volume = (1/3) × (22/7) × 72 × 12 = (1/3) × 22 × 7 × 12 = 22 × 28 = 616 cm3.
Worked example: frustum
A lampshade is a frustum of a cone with height 6 cm, lower radius 4 cm and upper radius 2 cm. Volume = (1/3)πh(R2 + Rr + r2) = (1/3) × (22/7) × 6 × (16 + 8 + 4) = (1/3) × (22/7) × 6 × 28. First 6 × 28 = 168, then (22/7) × 168 = 22 × 24 = 528, then 528 ÷ 3 = 176 cm3.
Check by similar cones: the radii 4 cm and 2 cm are in the ratio 2 : 1, so the original cone’s height would be twice the small cone’s height. The frustum height 6 cm is the difference, so the small cone has height 6 cm and the large cone has height 12 cm. Large cone: (1/3)π × 16 × 12 = 64π. Small cone: (1/3)π × 4 × 6 = 8π. Difference = 56π = 56 × 22/7 = 176 cm3, the same answer.
Worked example: sphere
A ball has radius 3 cm. Volume = (4/3) × (22/7) × 27 = (4 × 22 × 9) / 7 = 792/7 = 113 1/7 cm3.
Original Practice Questions With Worked Answers
- A rectangular water tank measures 2 m by 1.5 m by 1 m. Find its volume. 2 × 1.5 × 1 = 3 m3.
- A triangular prism has an equilateral triangular end of area 16 cm2 and length 9 cm. Find the volume. 16 × 9 = 144 cm3.
- A rectangular based pyramid has base 10 cm by 6 cm and height 12 cm. Find the volume. (1/3) × 60 × 12 = 240 cm3.
- A sphere has diameter 14 cm. Taking π = 22/7, find its volume. Radius = 7 cm. V = (4/3) × (22/7) × 343 = (4/3) × 22 × 49 = (88 × 49)/3 = 4312/3 = 1 437 1/3 cm3.
- Why is the height in V = (1/3)Ah the perpendicular height? The one-third formula compares a pyramid with a prism that has the same base and the same upright height. A slant height along a face would over-count the space.
- A learner treats a frustum as “half the large cone”. What is wrong? Cutting parallel to the base does not split the cone into two equal volumes. Subtract the small similar cone, or use the frustum formula.
Cover the answers and try each question before checking. Convert mixed units first with the metric unit converter if needed. For timed practice, open the Grade 9 revision papers or the Grade 9 CBC Challenge. These notes are free; full revision access is available with the KSh 20 Day Pass or the KSh 500 Monthly Pass.
Frequently Asked Questions
What solids are in the Grade 9 volume of solids topic?
The sub-strand covers triangular and rectangular based prisms; triangular, rectangular and square based pyramids; cones; frustums; and spheres. It builds on earlier cubes, cuboids and cylinders rather than repeating them as the main new work.
How do you find the volume of a pyramid?
Use V = (1/3) × base area × perpendicular height. For a square base of side 6 cm and height 9 cm, the volume is 108 cm3.
What is a frustum?
A frustum is the part of a cone or pyramid that remains after a smaller similar solid is cut off by a plane parallel to the base, like a lampshade or a bucket without its lid.
What is the volume of a sphere?
V = (4/3)πr3, where r is the radius. A sphere of radius 3 cm has volume 792/7 cm3, or 113 1/7 cm3, when π = 22/7.