What this guide covers
Selected Sec 2 G5 volume and surface-area formulas for cones, spheres and pyramids. Composite-solid surface-area problems belong in later practice.
Match formula to solid
Cone volume is one third of the cylinder with the same base and perpendicular height: V=⅓πr²h. Sphere surface area is 4πr² and volume is ⁴⁄₃πr³. Pyramid volume is one third of base area times perpendicular height.
Use the perpendicular cone height
A cone has radius 3 cm and perpendicular height 8 cm. Find its volume in terms of π.
- Use V=⅓πr²h, with r=3 and h=8.
- V=⅓π(3²)(8)=24π cm³.
Check: The matching cylinder would have volume 72π cm³; the cone is one third of that.
Try it yourself
Work on paper, then compare the steps. These checks do not establish mastery.
A sphere has radius 4 cm. Find its surface area in terms of π.
Surface area =4πr²=4π(4²)=64π cm². Use square units, not cubic units.
A square-based pyramid has base side 6 cm and perpendicular height 9 cm. Find its volume.
Base area =6×6=36 cm². Volume =⅓×36×9=108 cm³. The given height is perpendicular, as the volume formula requires.
Where this connects
- Earlier: plane area and units →
Area of a base is needed before pyramid volume.
- Later: composite solids →
The Upper Secondary challenge chooses exposed faces on a joined solid.
Sources and scope
Original Mentora solid dimensions and questions aligned to selected Sec 2 G5 scope in the MOE G2/G3 Mathematics syllabus. No school-paper graphic was reproduced.