Sec 1 Polygons: Angles, Symmetry and Constructions

Secondary 1 Mathematics · G3 focus · Study guide

Understand triangle and quadrilateral properties, interior and exterior angles, symmetry and ruler-and-compass constructions.

Concept → worked examples → practice20 min read · practice linked below

Use triangle properties, not the appearance of the sketch

A polygon is a closed plane figure made from straight sides. A triangle has three sides, and its interior angles add to $180^\circ$ . Extend one side: the exterior angle equals the sum of the two opposite interior angles because both are supplementary to the adjacent interior angle.

An equilateral triangle has three equal sides and three $60^\circ$ angles.
An isosceles triangle has two equal sides; the angles opposite those sides are equal. Match the equal-side marks to the opposite angles.
A scalene triangle has three unequal sides. A right-angled triangle has one $90^\circ$ angle; its other two angles total $90^\circ$ .

Triangle side lengths must satisfy the triangle inequality: the sum of any two sides must exceed the third. For positive lengths, it is enough to check that the two shorter sides add to more than the longest side. Lengths 2, 3 and 5 cannot form a triangle: they lie flat.

Worked example 1: an isosceles triangle and an exterior angle

In triangle $ABC$ , $AB=AC$ and $\angle BAC=44^\circ$ . Extend $BC$ beyond $C$ to $D$ . Find $\angle ABC$ , $\angle ACB$ and $\angle ACD$ .

The equal base angles are opposite the equal sides. The exterior angle is outside the triangle at

C

Since $AB=AC$ , the base angles at $B$ and $C$ are equal. Call each $b$ .
Use the triangle sum: $44+2b=180,\qquad 2b=136,\qquad b=\boxed{68^\circ}.$
$B,C,D$ are collinear, so $\angle ACD=180^\circ-68^\circ=\boxed{112^\circ}.$

Check: the exterior angle is also $44^\circ+68^\circ=112^\circ$ . All three interior angles total $180^\circ$ .

Classify quadrilaterals by their properties

A diagonal splits any convex quadrilateral into two triangles, so its interior angles total $360^\circ$ . A convex polygon has every interior angle less than $180^\circ$ . The special quadrilaterals below have additional properties.

Properties of familiar quadrilaterals
Shape	Sides and angles	Diagonals and symmetry
Parallelogram	Both pairs of opposite sides parallel and equal; opposite angles equal; adjacent angles supplementary	Diagonals bisect each other; rotational symmetry of order 2
Rectangle	A parallelogram with four right angles	Equal diagonals that bisect each other; 2 symmetry lines unless it is a square
Rhombus	A parallelogram with four equal sides	Diagonals bisect each other at right angles and bisect opposite angles; 2 symmetry lines unless it is a square
Square	Four equal sides and four right angles; both a rectangle and a rhombus	Equal perpendicular diagonals that bisect each other; 4 symmetry lines and rotational order 4
Trapezium	One pair of opposite sides parallel	No symmetry required; an isosceles trapezium has equal non-parallel sides, equal base angles and 1 symmetry line
Kite	Two pairs of equal adjacent sides	The diagonal through the equal-side meeting vertices bisects the other at right angles; an ordinary kite has 1 symmetry line

Here “trapezium” means exactly one parallel pair. Some books use “at least one”; follow the stated definition when classifying. A square belongs to several families at once. A typical parallelogram has no line of symmetry even though it matches itself after a half-turn.

Worked example 2: angles in a parallelogram

The adjacent angles $A$ and $B$ of parallelogram $ABCD$ are $(3x+10)^\circ$ and $(2x+20)^\circ$ . Find all four angles.

Adjacent angles of a parallelogram total $180^\circ$ , because its opposite sides are parallel: $(3x+10)+(2x+20)=180.$
Then $5x+30=180$ , so $x=30$ .
$A=3(30)+10=100^\circ$ and $B=2(30)+20=80^\circ$ . Opposite angles are equal, so $\boxed{A=C=100^\circ,\quad B=D=80^\circ}.$

Check: $100+80+100+80=360$ . Equal opposite sides do not make all four angles equal.

Interior sums and exterior turns

From one vertex of a convex $n$ -sided polygon, draw diagonals to the non-adjacent vertices. They split the shape into $n-2$ triangles. Therefore

\text{sum of interior angles}=(n-2)180^\circ.

Walking around a convex polygon and turning through one exterior angle at each vertex brings you back facing your starting direction: the exterior turns total $360^\circ$ . At each vertex, the adjacent interior and exterior angles total $180^\circ$ .

A regular polygon has all sides equal and all interior angles equal. Only then can we divide the total equally among all $n$ vertices:

\text{each exterior angle}=\frac{360^\circ}{n},\qquad \text{each interior angle}=180^\circ-\frac{360^\circ}{n}.

Count the triangles inside a polygon

A pentagon can be split from one vertex into 3 triangles: its interior sum is 3 × 180° = 540°. If the pentagon is regular, each interior angle is 108° and each exterior turn is 72°.

All triangles meet at A. Count triangles, not vertices, when finding the interior-angle sum.

5 sides → 3 triangles → 540° in total. Regular pentagon: each interior angle 108°, each exterior angle 72°.

Regular polygons: angles and symmetry
Polygon	Sides	Interior angle	Exterior angle	Symmetry
Pentagon	5	108°	72°	5 lines; rotational order 5
Hexagon	6	120°	60°	6 lines; rotational order 6
Octagon	8	135°	45°	8 lines; rotational order 8
Decagon	10	144°	36°	10 lines; rotational order 10

An equilateral triangle has 3 symmetry lines and rotational order 3; a non-equilateral isosceles triangle has 1 symmetry line, and a scalene triangle has none. A line of symmetry folds a shape onto itself. Rotational order counts how many times a shape matches itself during one full turn, including the final position. A regular $n$ -gon has $n$ lines of symmetry and rotational order $n$ .

Worked example 3: recover the number of sides

Each interior angle of a regular polygon is $156^\circ$ . Find its number of sides.

Find one exterior angle: $180^\circ-156^\circ=24^\circ$ .
The exterior angles total $360^\circ$ and are equal, so $n=\frac{360}{24}=\boxed{15}.$
Check using the interior sum: $(15-2)180=2340$ degrees; $2340\div15=156$ degrees.

A number of sides must be a whole number at least 3. An interior angle of $130^\circ$ , for example, would give $360\div50=7.2$ , so no regular polygon has that interior angle.

Worked example 4: an irregular polygon

Four interior angles of a convex pentagon are $98^\circ$ , $115^\circ$ , $123^\circ$ and $104^\circ$ . Find the fifth angle and its adjacent exterior angle.

The pentagon has 5 sides, so its interior sum is $(5-2)180=540^\circ$ .
The four known angles total $98+115+123+104=440^\circ$ . The fifth is $540-440=\boxed{100^\circ}$ .
Its adjacent exterior angle is $180-100=\boxed{80^\circ}$ .

Check: all five exterior angles are $82^\circ,65^\circ,57^\circ,76^\circ,80^\circ$ ; they add to $360^\circ$ . Dividing $540$ by 5 would incorrectly assume the pentagon was regular.

Construct accurately from given measurements

A construction builds a figure using the stated lengths and angles. Use a sharp pencil, a ruler for lengths and straight lines, a protractor for specified angles, and compasses for equal distances. Keep construction arcs visible. When using a protractor, place its centre on the vertex, align its baseline with one ray, and read from the zero on that ray.

For two sides and their included angle, draw one side, construct the angle at the correct endpoint, and mark the second side along the ray. For a side and the two angles at its endpoints, draw the side, draw both angle rays on the same side of it, and join their intersection. Check that the two supplied angles total less than $180^\circ$ . A set square can draw a perpendicular; sliding it along a fixed ruler keeps its edge parallel.

Worked example 5: construct a triangle from three sides

Construct triangle $ABC$ with $AB=6$ cm, $AC=5$ cm and $BC=4$ cm.

The compass setting represents a distance from its centre. Use the given centimetre lengths on paper; the screen drawing is not a measuring scale.

Check the lengths can form a triangle: $4+5>6$ .
Draw $AB=6$ cm with a ruler.
Set the compasses to 5 cm. With centre $A$ , draw an arc above $AB$ .
Set the compasses to 4 cm. With centre $B$ , draw an arc crossing the first. Label the intersection $C$ .
Join $AC$ and $BC$ . Check all three lengths with the ruler; do not change the compass setting while drawing an arc.

Why it works: every point on the first arc is 5 cm from $A$ ; every point on the second is 4 cm from $B$ . Their intersection satisfies both distances. The reflected intersection below $AB$ gives an equally valid triangle unless the question specifies a side.

Worked example 6: construct a perpendicular bisector

Construct the perpendicular bisector of a 6 cm segment $AB$ , and explain why any point on it is equally far from $A$ and $B$ .

A perpendicular bisector both halves a segment and meets it at

90^\circ

Open the compasses to a radius greater than 3 cm, for example 4 cm. Half the segment would make the arcs touch at only one point.
With centre $A$ , draw arcs above and below the segment. Without changing the radius, repeat with centre $B$ . Label their intersections $P$ and $Q$ .
Draw the straight line $PQ$ ; it crosses $AB$ at $M$ .
Check $AM=MB=3$ cm and that the crossing is a right angle. The line $PQ$ is the required perpendicular bisector.

Why it works: both $P$ and $Q$ have equal distances from $A$ and $B$ . The line through them is the line of points equally far from those endpoints. “Bisect” means divide into two equal parts.

Bisect an angle: equal angles, not equal side lengths

For $\angle AOB$ , draw an arc centred at $O$ that crosses its two rays at $X$ and $Y$ . With equal compass radii and centres $X$ and $Y$ , draw two arcs that intersect inside the angle at $P$ . Join $OP$ . The ray $OP$ divides $\angle AOB$ into two equal angles: a $70^\circ$ angle becomes two $35^\circ$ angles. A point on this bisector has equal perpendicular distances from the two sides of the angle.

The first arc makes

OX=OY

; the next pair makes

PX=PY

. Equal-distance constructions produce the equal angles.

Common mistakes and checks

Confusing total and individual angles: $(n-2)180^\circ$ is a sum. Divide by $n$ only for a regular polygon.
Using an interior angle in $360\div\text{angle}$ : find its supplementary exterior angle first.
Assuming a parallelogram has equal diagonals: a rectangle does; an ordinary parallelogram need not.
Giving a fractional number of sides: it means the proposed regular polygon is impossible or an earlier calculation is wrong.
Mixing up bisectors: a perpendicular bisector halves a segment at right angles; an angle bisector halves an angle.
Erasing construction arcs: leave the small intersecting arcs to show how the vertex or bisector was located.

Practise reasoning and construction

Use our Sec 1 triangles, quadrilaterals and polygons question bank for angle reasoning, regular polygons and construction questions. Draw constructions with the stated instruments on paper. Review angles and parallel lines whenever you need to justify an equality or a sum.