Geometrical Constructions

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Construct a triangle from three given sides

Question: Construct triangle ABCABC with AB=6AB = 6 cm, AC=8AC = 8 cm and BC=10BC = 10 cm, using ruler and compasses only. Measure BAC\angle BAC.

Step 1. Draw the base: rule AB=6AB = 6 cm exactly, and label the ends.

Step 2. The third vertex is fixed by two distances, so use two compass arcs:

set the compasses to 88 cm and draw an arc centred at AA;

set them to 1010 cm and draw an arc centred at BB.

Step 3. The arcs cross at CC (choose either intersection — the two options are mirror images). Join ACAC and BCBC with a ruler.

Step 4. Measure with a protractor:

BAC=90°\angle BAC = 90°

— as it must be, since 62+82=1026^2 + 8^2 = 10^2 makes this a right-angled triangle (a built-in accuracy check for your drawing: within ±1°\pm 1° is fine).

⚠ Watch out: Leave every construction arc visible — rubbed-out arcs lose the method marks; the arcs are the working. Compasses, not protractor, fix CC: measuring angles to place it answers a different question. Label the vertices as the question names them, or the measured angle can't be credited.

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