0580

Volume and Surface Area

Lengths, Areas and Volumes

Joining and cutting

  • A compound solid is two shapes joined, and a part of a solid is a piece cut from one, such as a hemisphere or a frustum
A cone sitting on top of a hemisphere so the two share a circular join: the slant edge of the cone is 12 cm and the radius of the hemisphere is 5 cm, and the flat circle where they meet is inside the solid rather than on its surface
Source: Surface Area by Save My Exams
  • Volumes simply add or subtract, so a cylinder with a hemisphere on top has the two volumes added
  • Surface areas do not simply add, because the faces where the two solids meet are inside the finished solid and are no longer surfaces
  • Work along the outside of the solid, listing each surface you would actually touch, then add only those
  • A frustum is a cone with its top cut off, so its volume is the whole cone minus the small cone removed, and the syllabus names it directly
A frustum drawn as a cone with its tip cut off: the base radius is 20 cm, the radius of the flat top is 10 cm, the frustum itself is 30 cm tall and the dashed removed tip adds a further 15 cm
Source: Problem-solving with volumes by Save My Exams
A step-shaped solid 10 cm deep whose cross-section is an L: 7 cm up the left side, 4 cm across the top step, then down to a 2 cm lower step, over a 9 cm base
Source: Problem-solving with volumes by Save My Exams
  • Answers here are frequently wanted in terms of π, so keep π symbolic throughout and collect at the end
Worked example

Surface area from a net

The net of a solid square-based pyramid is shown. Find its total surface area.

The net of a square-based pyramid: a square of side 15 cm in the middle with an identical triangle folded out from each of its four edges, and the perpendicular height of one triangle marked 23 cm
Source: Surface area by Save My Exams

Solution:

  • A net shows every face laid flat, so the surface area is just the total area of the shapes on it
  • There is one square base and four identical triangles
  • Square: 15 × 15 = 225 cm²
  • Each triangle has base 15 cm and perpendicular height 23 cm
  • One triangle: ½ × 15 × 23 = 172.5 cm²
  • Four triangles: 4 × 172.5 = 690 cm²
  • Total surface area = 225 + 690 = 915 cm²
  • The 23 cm is the height of the triangle, not the height of the pyramid — the two are different lengths, joined by Pythagoras
  • No formula for this is printed on the paper; it has to be built from the faces
Exam tip

Every volume formula you need is printed, but only the curved surface areas are — the totals for a cylinder and a cone, and every surface area of a cuboid, prism or pyramid, must be built from the faces. Use the perpendicular height in a volume formula and the sloping edge in a cone's curved surface area, converting between them with Pythagoras. On a joined solid, leave out the faces hidden where the parts meet.

Worked example

Surface area of a joined solid

A solid is made by joining a hemisphere of radius 5 cm to the top of a cylinder of radius 5 cm and height 14 cm. Find its total surface area in terms of π.

A solid made of an upright cylinder with a dome sitting on its top rim. A dashed radius on the join circle is labelled 5 cm and the straight side of the cylinder is labelled 14 cm. The circle where the two solids meet is drawn, showing that it is inside the finished solid. The figure is marked NOT TO SCALE.

Solution:

  • Walk round the outside and list the surfaces that are actually on show
  • The curved surface of the cylinder: 2π × 5 × 14 = 140π cm²
  • The flat circular base of the cylinder: π × 5² = 25π cm²
  • The curved surface of the hemisphere, which is half a sphere: ½ × 4π × 5² = 50π cm²
  • The top circle of the cylinder is not counted, because the hemisphere sits on it and it is no longer on the outside
  • Total = 140π + 25π + 50π
  • Total surface area = 215π cm²