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Cambridge 0580 IGCSE  ·  Topic C4.2

C4.2 Geometrical Constructions

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Construction questions are marked by measurement, not just by eye — examiners check finished drawings against a tolerance of about ±2 mm\pm 2\text{ mm} and ±2∘\pm 2^\circ. Every mark on the page must come from a ruler, a protractor, or a pair of compasses — never estimated by "it looks about right."

01 — Explanation

How do I measure lines and angles accurately?

Lines are measured with a ruler to the nearest millimetre, reading from the 00 mark, not the edge of the ruler itself. Angles are measured with a protractor, placing its centre-cross exactly on the vertex and its baseline exactly along one arm of the angle, then reading from 0∘0^\circ along that same arm around to the other arm.

Exam tip

protractors have two scales running in opposite directions. Before reading off a value, check which arm of the angle sits on the 0∘0^\circ line of the scale you're using — reading the wrong scale is the single most common way to turn an obtuse angle into (an incorrect) acute one, or vice versa.

02 — Explanation

How do I construct a triangle from three given side lengths?

To construct triangle ABCABC with AB=7 cmAB = 7\text{ cm}, BC=5 cmBC = 5\text{ cm}, CA=6 cmCA = 6\text{ cm}, using only a ruler and compasses:

  1. Draw the longest side first with a ruler: AB=7 cmAB = 7\text{ cm}.
  2. Open the compasses to 6 cm6\text{ cm} (CACA). Put the point on AA and draw an arc above the line.
  3. Open the compasses to 5 cm5\text{ cm} (BCBC). Put the point on BB and draw an arc crossing the first.
  4. Label the crossing point CC, then join ACAC and BCBC with a ruler.

Drawing the longest side first, flat along the bottom, gives the two compass arcs the best chance of crossing cleanly above the line, which keeps the whole construction tidy and easy to check.

Exam tip

never rub out your construction arcs — they are the examiner's only proof that you used the compasses correctly, not a protractor or a ruler estimate. A perfectly accurate triangle with no visible arcs can lose the method mark.

Perpendicular bisectors and angle bisectors are not required at Core tier. The only construction skill needed beyond a single triangle is building a compound shape from two such triangles, e.g. constructing a rhombus by drawing two congruent triangles back-to-back along a shared side — each triangle uses the rhombus's diagonal as one side, and its two equal edges as the other two.

03 — Explanation

How do I draw and use a net of a 3D solid?

A net is a 2D shape that folds up into a 3D solid. To draw one, picture "unfolding" the solid along its edges: a cuboid's net is six rectangles arranged in three matching pairs; a triangular prism's net is two triangular end-faces plus three rectangles, one per side of the triangle, each rectangle's width equal to the prism's length.

Worked example. The net of a closed cuboid box has a 6 cm×4 cm6\text{ cm} \times 4\text{ cm} base, and the box is 3 cm3\text{ cm} tall. Find its total surface area.

A cuboid net unfolds into 3 pairs of identical rectangles:

top & bottom: 6×4=24 cm2 eachfront & back: 6×3=18 cm2 eachsides: 4×3=12 cm2 each\text{top \& bottom: } 6\times4=24\text{ cm}^2 \text{ each} \qquad \text{front \& back: } 6\times3=18\text{ cm}^2 \text{ each} \qquad \text{sides: } 4\times3=12\text{ cm}^2 \text{ each}
SA=2(24)+2(18)+2(12)=48+36+24=108 cm2SA = 2(24) + 2(18) + 2(12) = 48+36+24 = 108\text{ cm}^2

Common mistake

counting only the faces that are easy to see and forgetting the base. List all six faces in matching pairs before adding — a cuboid always has exactly 3 pairs of identical faces, and missing one pair is an easy way to lose an otherwise straightforward accuracy mark.

Worked example. A closed triangular prism has length 10 cm10\text{ cm}. Its triangular cross-section is right-angled with the two shorter sides 3 cm3\text{ cm} and 4 cm4\text{ cm}, and hypotenuse 5 cm5\text{ cm}. Find its total surface area.

Two triangular ends: 2×(12×3×4)=2×6=12 cm22\times\left(\frac{1}{2}\times3\times4\right) = 2\times6 = 12\text{ cm}^2. Three rectangular faces, one per side of the triangle, each 10 cm10\text{ cm} long:

(3×10)+(4×10)+(5×10)=30+40+50=120 cm2(3\times10)+(4\times10)+(5\times10) = 30+40+50 = 120\text{ cm}^2
SA=12+120=132 cm2SA = 12+120 = 132\text{ cm}^2