Pyramids, Cones, Spheres and Composite Solids
Not every solid is a prism. Pyramids and cones taper to an apex, while spheres are curved in every direction. This lesson develops the volume and surface-area methods for these solids and then combines them in multi-step problems.
Foundation & Higher. Use perpendicular heights for volume and distinguish curved, total and exposed surface area.
Pyramids
A pyramid has a polygon base and triangular faces meeting at an apex. Its perpendicular height is measured from the base to the apex at 90°.
V = ⅓ × base area × perpendicular height.
Worked example: a square-based pyramid has base side 9 cm and perpendicular height 10 cm. Base area = 81 cm², so V = ⅓×81×10 = 270 cm³.
Worked example — pyramid surface area
A square-based pyramid has base side 6 cm and each triangular face has perpendicular face height 5 cm. One face has area ½×6×5=15 cm². Add the four faces and the square base: total surface area=4×15+6²=96 cm². The face height lies in a sloping triangular face; it differs from the vertical height used for volume.
Cones
A cone is the circular counterpart of a pyramid. Its volume is V = ⅓πr²h, where h is the perpendicular height.
The curved surface area is πrl, where l is the slant height. Total surface area of a closed cone is πr² + πrl.
Worked example: radius 5 cm and perpendicular height 12 cm gives V = ⅓π×25×12 = 100π cm³. If the slant height is 13 cm, total surface area = 25π+65π = 90π cm².
Spheres and hemispheres
For a sphere of radius r: volume = ⁴⁄₃πr³ and surface area = 4πr².
A hemisphere has half the volume of a sphere: ⅔πr³. Its curved surface area is 2πr², but its total surface area includes the circular base, giving 3πr².
Worked example: a sphere of radius 6 cm has volume ⁴⁄₃π×6³ = 288π cm³ and surface area 4π×6² = 144π cm².
Composite solids
Break a composite solid into familiar parts. For volume, add the volumes of joined parts or subtract a removed part. For surface area, count only surfaces that are exposed: a joining circle inside the solid is not part of the external surface.
Worked example: a cylinder of radius 3 cm and height 8 cm has a hemisphere of radius 3 cm on top. Volume = 72π + 18π = 90π cm³. The shared circular face is internal, so it is not counted in the exposed surface area.
For its external surface area, add the cylinder wall 2π×3×8=48π cm², the bottom circle 9π cm², and the hemisphere curved surface 18π cm². Total=75π cm². Neither copy of the joined circle is exposed.
Reverse problems
Sometimes the volume is known and a radius or height is missing. Substitute the known values, rearrange, and only then take a square root or cube root if required.
Worked example: a sphere has volume 36π cm³. ⁴⁄₃πr³=36π, so r³=27 and r=3 cm.
Frustums when dimensions are supplied
A frustum is what remains when the top of a cone or pyramid is cut off parallel to its base. When the dimensions of the full solid and removed smaller solid are supplied, find the frustum volume by subtraction: large solid − small solid.
Do not assume missing dimensions. Problems requiring similar-shape scale factors are developed separately in Similar Shapes: Length, Area and Volume.
Think Like a Mathematician
Sketch the solid as separate components before calculating. Label each dimension and decide whether it belongs to a radius, perpendicular height or slant height. This prevents most formula-selection errors.
Common Mistakes
Height vs slant height: cone volume uses perpendicular height h; curved surface area uses slant height l.
Hemisphere surface area: decide whether the question wants curved area only (2πr²) or total area including the base (3πr²).
Composite surfaces: do not count faces hidden where two solids join.
Practice — curved and composite solids
1. Find the volume of a square-based pyramid with base side 6 cm and height 9 cm.
⅓×36×9 = 108 cm³.
2. A cone has radius 4 cm and height 9 cm. Give its volume exactly.
⅓π×16×9 = 48π cm³.
3. A cone has radius 5 cm and slant height 12 cm. Find its curved surface area.
πrl = π×5×12 = 60π cm².
4. Find the volume of a sphere of radius 3 cm.
⁴⁄₃π×27 = 36π cm³.
5. Find the total surface area of a hemisphere of radius 4 cm.
3πr² = 3π×16 = 48π cm².
6. A sphere has surface area 100π cm². Find its radius.
4πr²=100π, so r²=25 and r=5 cm.
7. A solid consists of a cylinder radius 2 cm, height 7 cm, topped by a hemisphere radius 2 cm. Find its volume exactly.
Cylinder = 28π. Hemisphere = ⅔π×8 = 16π/3. Total = 84π/3+16π/3 = 100π/3 cm³.
8. A large cone has volume 200π cm³. A smaller cone of volume 25π cm³ is removed from the top. Find the frustum volume.
200π−25π = 175π cm³.
Frequently Asked Questions
Why do pyramids and cones have a factor of one third?
For the same base area and perpendicular height, their volume is one third of the corresponding prism or cylinder.
Does a hemisphere always have surface area 3πr²?
Only when the flat circular base is included. Its curved surface alone has area 2πr².
Continue Learning
Previous: Surface Area and Volume of Prisms and Cylinders
Next: Similar Shapes: Length, Area and Volume
Foundation route: continue to Pythagoras and Coordinate Distances. The similar-shape area and volume lesson above is Higher.