Constructing triangles and quadrilaterals · Construction & loci

Most construction questions ask you to draw a triangle or a quadrilateral from some lengths and angles, add a line or two, and then measure something. The constructions are the ones from the last lesson; what’s new is putting them in the right order.

Sketch the shape roughly and mark every given length and angle on it. Decide where each corner goes.
Draw a side that has a constructed angle at one end. Construct the angle, then mark the next corner with the compasses set to the given length.
Use the shape’s properties: a trapezium needs a parallel side; a parallelogram’s diagonals bisect each other; a rhombus has four equal sides.
Add the extra lines asked for (a perpendicular, a bisector, a parallel), then measure.

Construct a triangleBase 8 cm, 60° at B, then BA = 6 cm with the compasses

Try it

Construct a triangleStep through with the buttons

1 of 4stepThe basenow

Sketch the triangle first. Then draw the base BC = 8 cm with a ruler.

Step through the triangle: the angle is built first, then the compasses mark the second side. Then look at the parallel line again: it’s how you draw the second parallel side of a trapezium or parallelogram.

Check your measurements by calculation

Your measured answers should agree with trigonometry to within a millimetre or a degree. A quick calculation catches a slip: a height is the slanting side times the sine of its angle with the base.

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q9

Using a ruler and a pair of compasses only, construct: (i) a trapezium $WXYZ$ such that $|WX| = 10.2\text{ cm}$ , $|XY| = 5.6\text{ cm}$ , $|YZ| = 5.8\text{ cm}$ , $\angle WXY = 60^\circ$ and $\overline{WX}$ is parallel to $\overline{YZ}$ ; (ii) a perpendicular from $Z$ to meet $\overline{WX}$ at $N$ .

Measure: (i) $|WZ|$ ; (ii) $|ZN|$ .

Sketch and draw WX
$\angle WXY = 60^\circ$ is at $X$ , so draw $WX = 10.2$ cm first.
Think first. Which side has a given angle at its end?
60° at X, then Y
Construct $60^\circ$ at $X$ and, with the compasses set to 5.6 cm, cut the arm at $Y$ .
Think first. How do you mark XY = 5.6 cm?
The parallel side YZ
Co-interior angles add to $180^\circ$ , so construct $120^\circ$ at $Y$ (two $60^\circ$ angles), on the side towards $W$ . Mark $YZ = 5.8$ cm along it, and join $WZ$ .
Think first. YZ is parallel to WX. What angle does it make with XY at Y?
(ii) The perpendicular from Z
With centre $Z$ , draw an arc cutting $WX$ twice; from those two points draw equal arcs crossing below $WX$ ; join $Z$ to the crossing. It meets $WX$ at $N$ .
Think first. Which construction drops a perpendicular from a point to a line?
(b) Measure, and check
Measure $|WZ| \approx 5.1$ cm and $|ZN| \approx 4.8$ cm. Check: $|ZN| = 5.6 \sin 60^\circ \approx 4.85$ cm, and $|WN| = 10.2 - 2.8 - 5.8 = 1.6$ cm, so $|WZ| = \sqrt{1.6^2 + 4.85^2} \approx 5.1$ cm.
Think first. What should ZN be? It's the height of the trapezium.

A rectangle equal in area to a parallelogram

A parallelogram’s area is base × height. A rectangle on the same base, between the same parallel lines, has the same height, so it has the same area. To construct it, drop perpendiculars from the two ends of the base to the opposite side (produced if needed).

Parallelogram

A = bh

Rectangle

A = lw

: with

l = b

and

w = h

, the same area

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q8

Using a ruler and a pair of compasses only, construct: (i) a parallelogram $PQRS$ with $RS$ as the base such that $|PQ| = 7.8\text{ cm}$ , $|QR| = 5.6\text{ cm}$ and $\angle QRS = 120^\circ$ ; (ii) a rectangle $ABRS$ equal in area to the parallelogram $PQRS$ .

Measure: (i) $|AP|$ ; (ii) $|AS|$ .

The base and the 120° angle
Draw $RS = 7.8$ cm. Construct $120^\circ$ at $R$ and mark $RQ = 5.6$ cm on the arm.
Think first. The base is RS. Where is the given angle?
Complete the parallelogram
With centre $Q$ and radius 7.8 cm, and with centre $S$ and radius 5.6 cm, draw arcs meeting at $P$ . Join $PQ$ and $PS$ .
Think first. PQ is parallel and equal to SR. How do you find P?
(ii) The rectangle
Produce $PQ$ and drop perpendiculars to it from $R$ and $S$ , meeting it at $B$ and $A$ . $ABRS$ is the rectangle: same base $RS$ , same height.
Think first. Where do the rectangle's other two corners go?
(b) Measure, and check
Measure $|AP| = 2.8$ cm and $|AS| \approx 4.9$ cm. Check: $|AS| = 5.6 \sin 60^\circ \approx 4.85$ cm and $|AP| = 5.6 \cos 60^\circ = 2.8$ cm.
Think first. AS is the height. What should it be?

Your turn

WAEC 2022 · Paper 2 · Q12

Construction & loci

(a)
Using a ruler and a pair of compasses only: (i) construct $\triangle XYZ$ such that $|XY| = 7.2\text{ cm}$ , $|YZ| = 8.4\text{ cm}$ and $\angle XYZ = 60^\circ$ ; (ii) locate, by construction, a point $M$ on $\overline{XY}$ such that $|XM| = |MY|$ ; (iii) construct $\overline{MN} \parallel \overline{YZ}$ such that $MNZY$ is a parallelogram.
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $XY = 7.2$ cm, construct $60^\circ$ at $Y$ and mark $Z$ with $YZ = 8.4$ cm; join $XZ$ . Bisect $XY$ perpendicularly to find its midpoint $M$ . Through $M$ draw the line parallel to $YZ$ , and through $Z$ the line parallel to $XY$ . They meet at $N$ , completing parallelogram $MNZY$ . Measured: $|XZ| \approx 7.9$ cm and $\angle NZX \approx 68^\circ$ .
(b)
Measure: (i) $|XZ|$ ; (ii) $\angle NZX$ .
Show the answer
$|XZ| \approx 7.9\text{ cm}$ ; $\angle NZX \approx 68^\circ$

Try it on a graph

The accurate construction: X(0, 0), Y(7.2, 0), Z(3, 7.27), M(3.6, 0), N(−0.6, 7.27).

Worked solution (try it first)

(a)(i)

Draw $YZ = 8.4$ cm, construct $60^\circ$ at $Y$ and mark $YX = 7.2$ cm on the arm.
Join $XZ$ .

(ii)

Construct the perpendicular bisector of $XY$ .
It cuts $XY$ at its midpoint $M$ .

(iii)

With centre $M$ and radius $|YZ| = 8.4$ cm, and with centre $Z$ and radius $|YM| = 3.6$ cm, draw arcs meeting at $N$ .
Join $MN$ and $NZ$ : $MNZY$ is a parallelogram, with $MN \parallel YZ$ .

(b)

Measure: (i) $|XZ| \approx 7.9$ cm.

(ii)

$\angle NZX \approx 68^\circ$ .
Check: by the cosine rule $|XZ|^2 = 7.2^2 + 8.4^2 - 2(7.2)(8.4)\cos 60^\circ$
$= 61.92$ , so $|XZ| \approx 7.87$ cm.
$ZN \parallel XY$ , so $\angle NZX = \angle ZXY$ , and the sine rule gives $\sin\angle ZXY = \frac{8.4\sin 60^\circ}{7.87}$ , about $68^\circ$ .

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