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How do you find the area and perimeter of 2D shapes and the surface area and volume of 3D solids?

Calculate the area and perimeter of rectangles, triangles, parallelograms, trapezia, circles and sectors; and the surface area and volume of prisms, cylinders, pyramids, cones and spheres.

A focused answer to the OCR GCSE Mathematics geometry content on area and volume, covering the area and perimeter of 2D shapes including circles and sectors, and the surface area and volume of prisms, cylinders, pyramids, cones and spheres.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Area and perimeter of 2D shapes
  3. Circles and sectors
  4. Surface area and volume of 3D solids
  5. Why mensuration matters

What this dot point is asking

OCR references G16, G17 and G18 cover area and perimeter of 2D shapes (including circles and sectors) and surface area and volume of 3D solids (prisms, cylinders, pyramids, cones and spheres). Mensuration is a core, heavily examined topic on every tier. Some formulae must be recalled (rectangle, triangle, circle, prism), while the more complex ones (sphere, cone) are given on the OCR formulae sheet, so knowing which is which, and how to use each, matters.

Area and perimeter of 2D shapes

The standard area formulae cover the common polygons.

The height in the triangle and parallelogram formulae is the perpendicular height, not a slant side. For a composite shape, split it into known shapes, find each area, then add (or subtract a cut-out). Perimeter is found by adding every side; for a composite shape, take care to include only the outer edges.

Circles and sectors

Circles have their own two key formulae.

So a circle of radius 66 has area 36π113.136\pi \approx 113.1 and circumference 12π37.712\pi \approx 37.7. A sector with angle 9090^\circ is a quarter circle, so its area is 14πr2\tfrac{1}{4}\pi r^2. The circle area must be recalled; mixing up the area πr2\pi r^2 with the circumference 2πr2\pi r is one of the most common slips in the whole subject.

Surface area and volume of 3D solids

Volume measures the space inside; surface area measures the total outer area.

Surface area is found by adding the areas of every face. For a cylinder, that is two circles (2πr22\pi r^2) 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, where ll is the slant height). Building the surface area face by face, and labelling each piece, keeps long calculations organised.

Why mensuration matters

Area and volume appear in packaging, construction, capacity and density problems, and OCR links them to compound measures and to ratio (areas scale by the square of a length factor, volumes by the cube). Recalling the core formulae and selecting the given ones correctly, then keeping units consistent (and squared or cubed), is exactly what the marks reward. Leaving an answer "in terms of π\pi" when asked keeps it exact.

Exam-style practice questions

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

OCR 20193 marksA circle has radius 77 cm. Work out its area, giving your answer to 1 decimal place. (Foundation, Paper 1, calculator.)
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The area of a circle is A=πr2A = \pi r^2.

Substitute the radius: A=π×72=49πA = \pi \times 7^2 = 49\pi.

49π153.93849\pi \approx 153.938, so A153.9A \approx 153.9 cm2^2 to 1 decimal place.

Markers award a mark for the correct formula, a mark for 49π49\pi, and a mark for 153.9153.9 cm2^2. Using the diameter instead of the radius, or using 2πr2\pi r (the circumference), are the common errors. The area formula must be recalled, as it is not on the basic list given in the question.

OCR 20214 marksA solid cylinder has radius 55 cm and height 1212 cm. Work out its total surface area in terms of π\pi. (Higher, Paper 5, non-calculator.)
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A cylinder's surface is two circles plus the curved side, which unrolls to a rectangle of width 2πr2\pi r and height hh.

Two circles: 2×πr2=2×π×25=50π2 \times \pi r^2 = 2 \times \pi \times 25 = 50\pi.

Curved surface: 2πrh=2π×5×12=120π2\pi r h = 2\pi \times 5 \times 12 = 120\pi.

Total: 50π+120π=170π50\pi + 120\pi = 170\pi cm2^2.

Markers give a mark for the two circle areas, a mark for the curved surface, a mark for adding, and a mark for 170π170\pi. Forgetting one of the two end circles, or the curved surface, is the usual error.

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