C4.2; E4.2 - Measuring and constructing triangles

C4.2; E4.2 - Measuring and constructing triangles

A careful drawing can solve a geometry problem without guessing its shape. You will use a ruler and protractor accurately, construct triangles from three side lengths with compasses, and combine two triangles to construct a rhombus.

Measuring and drawing accurately

Can three lengths fix a triangle even when no angle is given? Yes: two compass arcs can locate the missing vertex. Before building that triangle, make sure each measurement starts from the correct place.

Place a ruler's zero mark, not its physical end, at the first endpoint of a segment. Read the other endpoint at eye level. Each millimetre is 0.1 cm0.1\,\mathrm{cm}, so 47 mm=4.7 cm47\,\mathrm{mm}=4.7\,\mathrm{cm}. If zero cannot be aligned, subtract the starting reading from the finishing reading. To draw a segment, mark its two endpoints at the required readings and join them with a ruled line.

For an angle, place the protractor's centre exactly on the vertex. Align its baseline with one arm. Read the scale that starts at zero on that arm, then follow it to the second arm. If an arm is too short, extend it lightly with a ruler. Judge acute or obtuse first: an obtuse angle cannot correctly read 34∘34^\circ, even if that number appears on the other scale. To draw an angle, mark its required reading, then rule a ray from the vertex through the mark.

[DIAGRAM: asset_name: Reading ruler and protractor scales; asset_slug: ruler_and_protractor; caption: Start at the correct zero and follow the correct angle scale.; description: A ruler diagram shows endpoints at readings 1.2 and 6.4 centimetres and the difference 5.2 centimetres. Beside it a semicircular protractor has its centre on O, baseline pointing right and an obtuse ray at 128 degrees, with the competing 52 degree reading identified as the wrong scale. The illustration explicitly says screen guide, not physical centimetres.]
Start at the correct zero and follow the correct angle scale.

On your own paper, draw lengths in actual centimetres using your ruler. A displayed image can resize on different screens: do not infer physical centimetres from its displayed width. If a diagram gives no guaranteed print size, use its stated measurements or scale bar instead.

Worked example 1: drawing an acute angle

Question. Draw AB=5.2 cmAB=5.2\,\mathrm{cm}, then draw ray ACAC so that ∠BAC=52∘\angle BAC=52^\circ, above ABAB.

Plan. Draw the baseline accurately, then measure the turn from ABAB at AA.

AB=52 mmconvert to ruler divisions∠BAC=52∘start the protractor at AB\begin{aligned} AB&=52\,\mathrm{mm} && \text{convert to ruler divisions}\\ \angle BAC&=52^\circ && \text{start the protractor at }AB \end{aligned}

Mark AA at zero and BB at 5.2 cm5.2\,\mathrm{cm}; rule ABAB. Centre the protractor at AA, with zero towards BB. Mark 52∘52^\circ above the line and draw ACAC through that mark. CC can be any point farther along that ray.

Answer. A ruled 5.2 cm5.2\,\mathrm{cm} baseline and an acute 52∘52^\circ angle at AA.

Check. Re-measure ABAB from its other end. Then place the protractor on ACAC and measure back towards ABAB; the smaller angle is still 52∘52^\circ.

Keep the same baseline but change only the angle. This makes the choice between the two protractor scales visible.

Worked example 2: drawing an obtuse angle

Question. Draw AB=5.2 cmAB=5.2\,\mathrm{cm}, then draw ray ADAD so that ∠BAD=128∘\angle BAD=128^\circ, above ABAB.

Plan. Start from the same zero direction and continue past the right-angle mark.

AB=5.2 cmkeep the baseline unchanged90∘<128∘<180∘the angle must be obtuse\begin{aligned} AB&=5.2\,\mathrm{cm} && \text{keep the baseline unchanged}\\ 90^\circ&<128^\circ<180^\circ && \text{the angle must be obtuse} \end{aligned}

Draw the baseline. Centre the protractor at AA, with zero towards BB. Follow that scale to 128∘128^\circ, mark the point and rule ADAD. The other scale reads 52∘52^\circ at the same mark; it starts at the wrong end for this question.

Answer. An obtuse 128∘128^\circ angle on a 5.2 cm5.2\,\mathrm{cm} baseline.

Check. The ray points above and to the left of AA, so the marked turn from the rightward baseline is larger than 90∘90^\circ. Re-measure with the protractor correctly centred.

Watch out

“6.4−1.2=6.4 cm6.4-1.2=6.4\,\mathrm{cm}” ignores the starting reading. The length between those ruler marks is 5.2 cm5.2\,\mathrm{cm}. Likewise, a protractor reading is measured from the chosen arm, not whichever zero is nearest the printed number.

Practise on paper so the instruments, rather than the screen size, set the measurements.

Constructing a triangle from three sides

Compasses keep a fixed distance from their centre. If AC=5 cmAC=5\,\mathrm{cm}, point CC must lie on a circle of radius 5 cm5\,\mathrm{cm} centred at AA. A second distance from BB places CC on a second circle. Where the circles meet, both length conditions hold at once.

SSS construction

An SSS construction uses the three side lengths to construct a triangle with a ruler and compasses. Draw one side, then use arcs from its endpoints to locate the third vertex. Keep the construction arcs visible.

[DIAGRAM: asset_name: Triangle from intersecting arcs; asset_slug: triangle_sss_arcs; caption: The arc intersection satisfies both remaining side lengths at once.; description: Two exact SSS constructions share baseline AB equal to 7 centimetres. Both have AC equal to 5 centimetres; the first has BC equal to 4 centimetres and the second BC equal to 6 centimetres. Blue and sage compass arcs cross at each labelled C, and the ruled triangle sides are shown. A note says lengths are labels, not screen measurements.]
The arc intersection satisfies both remaining side lengths at once.

Only short arcs near the intended crossing are needed; full circles are unnecessary. Keep the compass point on the endpoint, and set the opening using the ruler before drawing the arc. Normally there are two intersections, one on each side of the baseline, producing mirror-image triangles with the same side lengths. Use the side the question specifies; otherwise either is valid.

Worked example 3: locating the third vertex

Question. Using a ruler and compasses only, construct triangle ABCABC with AB=7 cmAB=7\,\mathrm{cm}, AC=5 cmAC=5\,\mathrm{cm} and BC=4 cmBC=4\,\mathrm{cm}. Put CC above ABAB.

Plan. Draw ABAB, then intersect an arc of radius 55 from AA with an arc of radius 44 from BB.

AB=7 cmdraw the fixed baselineAC=5 cmarc centred at ABC=4 cmarc centred at B\begin{aligned} AB&=7\,\mathrm{cm} && \text{draw the fixed baseline}\\ AC&=5\,\mathrm{cm} && \text{arc centred at }A\\ BC&=4\,\mathrm{cm} && \text{arc centred at }B \end{aligned}

Mark the upper crossing CC. Rule ACAC and BCBC, and leave both arcs visible.

Answer. The constructed triangle has the three required side lengths.

Check. Use the ruler to re-measure all three sides. In particular, the arc from AA fixes ACAC, while the arc from BB independently fixes BCBC.

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