A construction is an accurate drawing made with only a ruler (for straight lines) and a pair of compasses (for arcs). A protractor is not allowed.
The perpendicular bisector cuts $AB$ exactly in half, at $90^\circ$.
- Open your compasses to more than half the length of $AB$.
- Put the point on $A$ and draw arcs above and below the line.
- Keeping the same radius, put the point on $B$ and draw arcs that cross the first two.
- Join the two crossing points with a straight line.
Two mobile phone masts stand at $A$ and $B$. Show the region of a map that is closer to mast $A$ than to mast $B$.
- Put the compass point on the vertex of the angle and draw an arc crossing both arms.
- Put the compass point on each crossing point in turn and draw two arcs that meet inside the angle.
- Join the vertex to where those two arcs cross.
Perpendicular from a point $P$ not on the line
- With the compass point on $P$, draw an arc that crosses the line twice. Call the crossings $X$ and $Y$.
- Construct the perpendicular bisector of $XY$ in the usual way.
- That bisector passes through $P$ and meets the line at $90^\circ$.
Perpendicular at a point $P$ on the line
- With the compass point on $P$, draw arcs cutting the line either side of $P$ at equal distances. Call them $X$ and $Y$.
- Construct the perpendicular bisector of $XY$ — it will pass through $P$.
| You are given | Method |
|---|---|
| Three sides (SSS) | Draw the longest side. Arc from each end with the other two lengths; they cross at the third vertex. |
| Two sides and the included angle (SAS) | Draw one side, measure the angle with a protractor, draw the second side, join up. |
| Two angles and a side (ASA) | Draw the side, measure both angles at its ends, extend the arms until they meet. |
Construct triangle $ABC$ with $AB = 8$ cm, $BC = 6$ cm and $AC = 5$ cm.
A locus (plural loci) is the set of all points that satisfy a given rule. There are four standard loci to learn.
| The rule | The locus |
|---|---|
| A fixed distance $r$ from a point $P$ | A circle of radius $r$ centred on $P$ |
| A fixed distance $d$ from a line segment $AB$ | A "racetrack": two parallel lines $d$ away, joined by semicircles of radius $d$ at each end |
| Equidistant from two points $A$ and $B$ | The perpendicular bisector of $AB$ |
| Equidistant from two lines | The angle bisector of the angle between them |
A goat is tethered in a rectangular field $ABCD$. It must be within $10$ m of corner $A$ and nearer to side $AB$ than to side $AD$. Shade the region it can reach. (Scale: $1$ cm represents $5$ m.)
"Within" or "less than" → the boundary is not included; draw it dashed if the question asks.
"At most" or "no more than" → the boundary is included; draw it solid.
When two conditions are given, shade only the overlap.
Leave the arcs
Construction arcs earn the marks — never rub them out.
Perp. bisector
Equal arcs from both ends; the line is equidistant from $A$ and $B$.
Angle bisector
Arc from the vertex, then arcs from both crossings; equidistant from both arms.
Perpendicular
Arc from $P$ cutting the line twice, then bisect that chord.
Shortest distance
From a point to a line it is always the perpendicular distance.
$60^\circ$
Two arcs of equal radius build an equilateral triangle; bisect for $30^\circ$.
Locus: point
Fixed distance from a point = a circle.
Locus: segment
Fixed distance from a segment = a racetrack shape.
Two conditions
Shade only the overlap of the two regions.
Describe the steps needed to construct the perpendicular bisector of a line segment $PQ$.
▶ Show solution
1. Open the compasses to more than half of $PQ$.
2. With the point on $P$, draw arcs above and below $PQ$.
3. Keeping the same radius, put the point on $Q$ and draw arcs crossing the first pair.
4. Draw a straight line through the two intersection points.
Leave all arcs showing.
What is the locus of points exactly $4$ cm from a fixed point $O$?
▶ Show solution
A circle of radius $4$ cm with centre $O$.
(If the question said "at most $4$ cm", it would be the circle and everything inside it.)
Two trees stand $12$ m apart at points $A$ and $B$. Describe the locus of points that are the same distance from both trees.
▶ Show solution
The perpendicular bisector of $AB$.
It is a straight line crossing $AB$ at its midpoint, $6$ m from each tree, at $90^\circ$ to $AB$.
Explain why a triangle with sides $3$ cm, $4$ cm and $9$ cm cannot be constructed.
▶ Show solution
The two shorter sides must add to more than the longest side.
$3 + 4 = 7$, which is less than $9$.
Arcs of radius $3$ cm and $4$ cm from the ends of a $9$ cm line would never meet, so no triangle exists.
Describe the locus of all points exactly $3$ cm from a straight line segment $XY$ of length $10$ cm.
▶ Show solution
Two straight lines of length $10$ cm, each parallel to $XY$ and $3$ cm from it (one either side).
These are joined at each end by a semicircle of radius $3$ cm centred on $X$ and on $Y$.
The whole shape looks like a running track.
How would you construct an angle of exactly $30^\circ$ using only a ruler and compasses?
▶ Show solution
1. Draw a line and mark a point $A$ on it.
2. With the point on $A$, draw an arc crossing the line at $B$.
3. With the same radius and the point on $B$, draw an arc crossing the first at $C$.
4. Join $AC$: triangle $ABC$ is equilateral, so angle $CAB = 60^\circ$.
5. Bisect angle $CAB$ to get $30^\circ$.
A rectangular garden $ABCD$ measures $20$ m by $12$ m. A path must be built that is always the same distance from $AB$ as from $AD$. Describe and locate the path.
▶ Show solution
Points equidistant from two lines lie on the bisector of the angle between them.
$AB$ and $AD$ meet at $A$ at $90^\circ$, so the path is the bisector of that right angle.
It starts at corner $A$ and runs at $45^\circ$ into the garden.
Since the garden is $20$ m by $12$ m, the path reaches the far side $BC$ after $12$ m across and $12$ m up, at the point on $BC$ that is $12$ m from $B$.
A point $P$ lies $5$ cm from a line $L$. Explain why $5$ cm is the shortest distance from $P$ to $L$, and describe how you would construct that shortest path.
▶ Show solution
The shortest distance is the perpendicular distance. Any other line from $P$ to $L$ forms a right-angled triangle in which that line is the hypotenuse, and the hypotenuse is always longer than the perpendicular leg.
Construction: with the compass point on $P$, draw an arc cutting $L$ at two points $X$ and $Y$; then construct the perpendicular bisector of $XY$. It passes through $P$ and meets $L$ at $90^\circ$.
On a map with scale $1$ cm : $50$ m, a radio transmitter at $T$ has a range of $325$ m. A second transmitter at $S$ has a range of $200$ m.
(a) What radii would you draw on the map? (b) Describe the region that receives both signals.
▶ Show solution
(a) $T$: $325 \div 50 = 6.5$ cm. $S$: $200 \div 50 = 4$ cm.
(b) Draw a circle of radius $6.5$ cm about $T$ and a circle of radius $4$ cm about $S$. The region receiving both signals is the overlap of the two circles — a lens-shaped area. (If the circles do not overlap on the map, no point receives both.)
Triangle $ABC$ has $AB = 9$ cm, $BC = 7$ cm and $AC = 6$ cm. A treasure is buried inside the triangle so that it is:
• nearer to $AB$ than to $AC$, and
• less than $4$ cm from $B$.
Describe fully how to find and shade the possible region.
▶ Show solution
Step 1 — Construct the triangle (SSS). Draw $AB = 9$ cm. Arc $6$ cm from $A$ and $7$ cm from $B$; they cross at $C$. Join $AC$ and $BC$.
Step 2 — First condition. "Nearer to $AB$ than to $AC$" — these two lines meet at $A$, so construct the bisector of angle $BAC$. The required side is the one containing $AB$.
Step 3 — Second condition. "Less than $4$ cm from $B$" — draw an arc of radius $4$ cm centred on $B$. The required region is inside the arc, so draw it dashed.
Step 4 — Shade. Shade only the part that is inside the triangle, on the $AB$ side of the angle bisector, and within $4$ cm of $B$.
Leave every construction arc visible.