G1, G2 - Geometric terms and constructions | AQA GCSE Mathematics

Use G1's geometric language to read, label and draw shapes accurately. Then use G2's ruler-and-compass constructions to make exact midpoints, right angles and angle bisectors, including an angle of $60^\circ$ .

Points, lines and shapes

Can you find the exact middle of a line without reading the halfway mark on a ruler? Two pairs of equal compass arcs can do it. First, you need a language for saying precisely what you have made.

A point marks a position and has no size. A line is straight and extends indefinitely in both directions; a line segment has two endpoints. A plane is a flat surface extending in all directions. A vertex is a corner; vertices is its plural. In a solid, an edge is where two faces meet. A cube has eight vertices and twelve edges, even when a drawing hides some of them.

[DIAGRAM: asset_name: Geometric language; asset_slug: aqa8300_g01_language; caption: The middle letter identifies the vertex of an angle; description: A labelled triangle ABC shows side AB and angle ABC at B. Beside it a shaded flat patch represents part of a plane.]

Parallel and perpendicular

Parallel lines lie in the same plane and never meet, however far they are extended. Perpendicular lines meet at a right angle, $90^\circ$ . For example, opposite edges of a rectangular page are parallel; adjacent edges are perpendicular. Two lines which cross at $70^\circ$ are not perpendicular.

An arrow mark on each of two lines means that they are parallel. Matching short ticks mean equal lengths; matching angle arcs mean equal angles. A small square means a right angle. These marks carry information: the apparent shape of a sketch does not.

A polygon is a closed plane shape made from straight line segments. A circle is not a polygon. A regular polygon has all sides equal and all interior angles equal. An equilateral triangle and a square are regular; a rectangle with unequal neighbouring sides is not.

A line of reflection symmetry folds a shape onto itself. Rotational symmetry means it fits itself after a turn smaller than a full turn. The order counts the matching positions in one full turn, including the starting position. A non-square rectangle has two reflection lines and rotation order two. A sloping parallelogram can have rotation order two but no reflection lines. Symmetry and regularity are different tests.

Key idea

A geometric statement comes from its definition, measurements or given marks. A shape looking regular or a line looking perpendicular is not enough.

A tilted square keeps all its properties. Turning the page changes your view, not the mathematics.

Read labels and draw from words

In angle $ABC$ , the middle letter $B$ is the vertex. The arms point from $B$ towards $A$ and $C$ . Angle $CBA$ names the same angle, but angle $BAC$ is at a different vertex.

The side opposite vertex $A$ is $BC$ . When lower-case side letters are used, $a$ is opposite angle $A$ , $b$ opposite $B$ , and $c$ opposite $C$ . A diagonal joins non-neighbouring vertices of a polygon. Label a polygon's vertices in order around its boundary so that $ABCD$ names its four sides unambiguously.

Worked example 1: turn a description into a sketch

Question. Sketch triangle $ABC$ with $AB=5\,\mathrm{cm}$ , $BC=3\,\mathrm{cm}$ , and a right angle at $B$ . Label the side opposite $B$ .

Plan. Put the stated right angle at the middle letter, then attach the given lengths to its two arms.

Draw a horizontal segment $AB$ , with $A$ left of $B$ : this is the first named side.
Draw $BC$ vertically upwards from $B$ : this makes $AB$ perpendicular to $BC$ .
Join $A$ to $C$ , label the two lengths, and add a right-angle square at $B$ : all the given information is now recorded.

Why this step? Both named sides contain $B$ , so they must meet at the stated right angle.

Answer. The side opposite $B$ is $AC$ , which may also be labelled $b$ .

Check. $AC$ does not touch $B$ ; the right-angle square is at $B$ , not $A$ .

The second example changes the vertex named in the description. Keep following the letters rather than copying the orientation.

Worked example 2: move the right angle

Question. Sketch triangle $PQR$ , with $PQ=4\,\mathrm{cm}$ , $QR=6\,\mathrm{cm}$ , and angle $PQR=90^\circ$ . Name the side opposite $Q$ .

Plan. Make $Q$ the corner where the two given sides meet.

Draw $PQ$ : its endpoints fix the first side.
At $Q$ , draw $QR$ perpendicular to $PQ$ : the middle letter of $PQR$ identifies the right-angle vertex.
Join $P$ to $R$ and label all given facts: this closes the triangle.

Why this step? The length $QR$ belongs on a side touching $Q$ , not on the opposite side.

Answer. $PR$ is opposite $Q$ .

Check. $PR$ joins the other two vertices. There is exactly one right-angle square, at $Q$ .

A sketch records relationships. An accurate drawing also uses the stated lengths and measured or constructed angles. If a question says “Not drawn accurately”, use its labels in calculations instead of measuring it.

Watch out

Wrong: put angle $PQR$ at $P$ because $P$ is written first. Right: put it at $Q$ , the middle letter.

Picture the corner before naming the opposite side.

Construct a perpendicular bisector

Open your compasses more than halfway along a segment. Equal arcs from its two ends meet above and below it. Joining those crossings gives an exact middle line.

Perpendicular bisector

A line which cuts a line segment into two equal lengths and meets it at $90^\circ$ . For example, the perpendicular bisector of an $8\,\mathrm{cm}$ segment passes through its $4\,\mathrm{cm}$ midpoint. A line through the midpoint at another angle bisects the segment but is not its perpendicular bisector.

Every crossing of the equal-radius arcs is equally far from both endpoints. The two crossings sit on the line midway between them. These construction facts are not on the formulae sheet: learn them.

[DIAGRAM: asset_name: Perpendicular bisector construction; asset_slug: aqa8300_g01_bisector; caption: Equal compass settings from A and B locate two points on the perpendicular bisector; description: Segment AB is crossed by equal-radius blue arcs above and below. Their crossings P and Q are joined by a sage perpendicular line through midpoint M.]

Worked example 3: bisect a segment

Question. Use a ruler and compasses to construct the perpendicular bisector of $AB=6\,\mathrm{cm}$ .

Plan. Make two pairs of equal-radius arcs, then join their crossings.

Draw $AB=6\,\mathrm{cm}$ : the segment is the starting object.
Set the compasses to $4\,\mathrm{cm}$ , which is more than half of $6\,\mathrm{cm}$ : the arcs will cross.
From $A$ , draw arcs above and below $AB$ . Without changing the setting, repeat from $B$ : matching radii give equal distances.
Name the crossings $P$ and $Q$ , and draw the straight line $PQ$ : it crosses $AB$ at midpoint $M$ .

Why this step? Keeping the radius the same makes both crossings equally far from $A$ and $B$ .

Answer. $PQ$ is the perpendicular bisector, with $AM=MB=3\,\mathrm{cm}$ .

Check. $3+3=6$ , and the constructed line meets $AB$ at a right angle. Leave the construction arcs visible.

The segment in the second example is longer. The method stays the same; only the minimum useful compass opening changes.

Worked example 4: choose a suitable opening

Question. Construct the perpendicular bisector of $CD=10\,\mathrm{cm}$ . A student suggests an opening of $4\,\mathrm{cm}$ . Explain the problem and complete the construction.

Plan. Compare the opening with half the segment, then use a larger equal setting from both endpoints.

10\div2=5\,\mathrm{cm}\qquad\text{half the segment}

An opening of $4\,\mathrm{cm}$ is too small: the two arcs cannot reach each other.
Use $6\,\mathrm{cm}$ from both $C$ and $D$ : this is greater than $5\,\mathrm{cm}$ .
Join the crossings above and below the segment: their line is its perpendicular bisector.

Why this step? A radius exactly equal to half the segment gives only one touching point. Two crossings are needed to define the new line.

Answer. Use $6\,\mathrm{cm}$ , for example. Each half of $CD$ is $5\,\mathrm{cm}$ .

Check. Both endpoint settings are equal and $5+5=10$ .

The ruler joins points; it does not replace the compass arcs with a measured guess.

Exam tip

When asked to use ruler and compasses, leave the arcs visible. A line drawn through a measured midpoint does not show the required construction.

Now judge an opening for a different segment.

Construct a perpendicular from or at a point

A perpendicular bisector also builds a right angle through a chosen point. First create two points on the given line which are equally far from the chosen point.

[DIAGRAM: asset_name: Perpendicular from and at a point; asset_slug: aqa8300_g01_perpendiculars; caption: Equal distances PU and PV make P lie on the perpendicular bisector of UV; description: Two panels show a point P above a line and a point P on a line. An arc from P marks U and V equally far away on the line. Equal arcs from U and V meet at Q, and PQ is perpendicular to the line.]

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G1, G2 - Geometric terms and constructions

Points, lines and shapes

Read labels and draw from words

Construct a perpendicular bisector

Construct a perpendicular from or at a point

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