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How do you find perimeter, area, surface area and volume of plane shapes and solids?

Find the perimeter and area of triangles, quadrilaterals and compound shapes, the surface area and volume of prisms, cylinders, pyramids, cones and spheres, and convert between units of area and volume.

A CCEA GCSE Mathematics answer on mensuration, covering perimeter and area of plane and compound shapes, surface area and volume of prisms cylinders pyramids cones and spheres, and converting between units of area and volume.

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  1. What this dot point is asking
  2. Perimeter and area of plane shapes
  3. Volume and surface area of solids
  4. Converting units of area and volume
  5. Why this matters

What this dot point is asking

Mensuration is the measurement of length, area and volume in the CCEA Geometry and Measures strand. You must find the perimeter and area of triangles, quadrilaterals and compound shapes, the surface area and volume of prisms, cylinders, pyramids, cones and spheres, and convert between units of area and volume. The compound-shape and composite-solid questions, where you add or subtract parts, are where the multi-step reasoning marks lie.

Perimeter and area of plane shapes

Perimeter is the total distance around a shape, found by adding the side lengths. Area is the space inside, with a standard formula for each common shape.

For a compound shape, split it into shapes you know, find each area, and add them (or subtract a removed piece). Keeping the working organised piece by piece protects the method marks. A common figure is an L-shape, which you can either split into two rectangles and add, or treat as a large rectangle with a smaller rectangle removed; both routes give the same answer, so choose whichever uses the dimensions you are given. When a question gives a shape with a circular part, such as a rectangle with a semicircular end, work out the radius carefully (often half a given side) before applying the circle formula.

Volume and surface area of solids

A prism has the same cross-section throughout its length, so its volume is the cross-section area times the length. A cylinder is a prism with a circular cross-section, giving V=πr2hV = \pi r^2 h.

The surface area of a solid is found by adding the areas of every face. For a cylinder this is two circles plus the curved surface 2πrh2\pi r h; for a cone it is the base πr2\pi r^2 plus the curved surface πrl\pi r l, using the slant height ll. A sphere has surface area 4πr24\pi r^2. The reliable approach is to imagine the solid opened out into a flat net, then find the area of each piece and add them, which makes it much harder to forget a hidden face. For a composite solid, such as a cone sitting on a cylinder, include only the surfaces that are actually exposed: the join between the two parts is internal and is not counted, a subtlety that often costs a mark.

Converting units of area and volume

Unit conversions for area and volume are not the same as for length, because area is two-dimensional and volume three-dimensional.

Why this matters

Mensuration is the most directly practical part of the course, used in construction, packaging, capacity and design, and it carries plenty of AO3 problem-solving marks through compound shapes and composite solids. It links to circles, to Pythagoras (for slant heights and missing lengths) and to the similar-shapes scale-factor rules. CCEA rewards organised working with correct units, so always state the unit and keep area in squared and volume in cubed units.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20193 marksA cylinder has radius 44 cm and height 1010 cm. Find its volume, to 1 decimal place. (Calculator.)
Show worked answer →

The volume of a cylinder is the area of the circular cross-section times the height.

Cross-section area: πr2=π×42=16π\pi r^2 = \pi \times 4^2 = 16\pi.

Volume: 16π×10=160π=502.7 cm316\pi \times 10 = 160\pi = 502.7 \text{ cm}^3 (to 1 decimal place).

Marks are for the circle area, for multiplying by the height, and for the final value. Using the diameter instead of the radius, or forgetting to square the radius, are the usual errors.

CCEA 20224 marksA trapezium has parallel sides 66 cm and 1010 cm and a height of 55 cm. A semicircle of diameter 44 cm is removed from it. Find the remaining area to 1 decimal place. (Calculator.)
Show worked answer →

Find the trapezium area: 12(a+b)h=12(6+10)(5)=12×16×5=40 cm2\tfrac{1}{2}(a + b)h = \tfrac{1}{2}(6 + 10)(5) = \tfrac{1}{2} \times 16 \times 5 = 40 \text{ cm}^2.

Find the semicircle area: radius is 22 cm, so 12πr2=12π×22=2π=6.28 cm2\tfrac{1}{2}\pi r^2 = \tfrac{1}{2}\pi \times 2^2 = 2\pi = 6.28 \text{ cm}^2.

Subtract: 406.28=33.7 cm240 - 6.28 = 33.7 \text{ cm}^2 (to 1 decimal place). Marks are for each area and for the subtraction. Forgetting to halve the semicircle, or using the diameter as the radius, lose marks.

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