IGCSE Mathematics: Geometry

Worked Example 1.1 — Angles on a Straight Line

Two angles on a straight line are $(3x + 10)°$ and $(2x - 5)°$. Find the value of $x$ and hence each angle.

Step 1 Use the angle rule: angles on a straight line sum to 180°.

$$(3x + 10) + (2x - 5) = 180$$

Step 2 Simplify and solve.

$$5x + 5 = 180 \implies 5x = 175 \implies x = 35$$

Step 3 Find each angle.

First angle: $3(35) + 10 = 115°$. Second angle: $2(35) - 5 = 65°$.

Check: $115° + 65° = 180°$ ✓

Worked Example 1.2 — Parallel Lines

Two parallel lines are cut by a transversal. One of the co-interior angles is $112°$. Find the other co-interior angle.

Step 1 State the theorem: co-interior angles sum to 180°.

Step 2 Other angle $= 180° - 112° = 68°$.

Worked Example 2.1 — Exterior Angle Theorem

In triangle $PQR$, angle $P = 48°$ and angle $Q = 65°$. An exterior angle is formed at $R$ by extending side $QR$. Find the exterior angle at $R$.

Step 1 The exterior angle at $R$ equals the sum of the two remote interior angles.

$$\text{Exterior angle} = \angle P + \angle Q = 48° + 65° = 113°$$

Verify Interior angle $R = 180° - 48° - 65° = 67°$. Exterior angle $= 180° - 67° = 113°$ ✓

Worked Example 2.2 — Parallelogram

In parallelogram $ABCD$, angle $A = (2x + 10)°$ and angle $B = (3x - 5)°$. Find all four angles.

Step 1 In a parallelogram, co-interior angles (between parallel sides) sum to 180°.

$$\angle A + \angle B = 180° \implies (2x + 10) + (3x - 5) = 180$$

$$5x + 5 = 180 \implies x = 35$$

Step 2 $\angle A = 80°$, $\angle B = 100°$. Since opposite angles are equal: $\angle C = 80°$, $\angle D = 100°$.

Worked Example 3.1 — Interior Angle of a Regular Hexagon

Find the interior angle of a regular hexagon.

Step 1 A hexagon has $n = 6$ sides. Sum of interior angles $= (6-2) \times 180° = 4 \times 180° = 720°$.

Step 2 Each interior angle $= 720° \div 6 = 120°$.

Worked Example 3.2 — Finding the Number of Sides

Each exterior angle of a regular polygon is 24°. How many sides does it have?

Step 1 Sum of exterior angles $= 360°$.

$$n = \frac{360°}{24°} = 15 \text{ sides}$$

Worked Example 3.3 — Irregular Polygon

A pentagon has interior angles $110°, 95°, 130°, 118°,$ and $x°$. Find $x$.

Step 1 Sum of interior angles of pentagon $= (5-2) \times 180° = 540°$.

Step 2 $x = 540° - 110° - 95° - 130° - 118° = 87°$.

Worked Example 4.1 — Angle at the Centre

O is the centre of a circle. Angle $AOB = 134°$ where $A$ and $B$ are points on the circumference. $C$ is another point on the major arc $AB$. Find angle $ACB$.

Step 1 Angle at the centre is twice the angle at the circumference.

$$\angle ACB = \frac{134°}{2} = 67°$$

Worked Example 4.2 — Cyclic Quadrilateral

$ABCD$ is a cyclic quadrilateral. $\angle DAB = 85°$ and $\angle ABC = 110°$. Find $\angle BCD$ and $\angle CDA$.

Step 1 Opposite angles sum to 180°.

$$\angle BCD = 180° - \angle DAB = 180° - 85° = 95°$$

$$\angle CDA = 180° - \angle ABC = 180° - 110° = 70°$$

Check $85° + 95° = 180°$ ✓ and $110° + 70° = 180°$ ✓

Worked Example 4.3 — Tangent and Radius

$TA$ is a tangent to a circle with centre $O$ at point $A$. $OB$ is a radius and $\angle BOA = 52°$. Find $\angle TAB$.

Step 1 $OA$ is a radius and $TA$ is a tangent at $A$, so $\angle OAT = 90°$.

Step 2 Triangle $OAB$ is isosceles (OA = OB, both radii), so $\angle OAB = \angle OBA = \dfrac{180° - 52°}{2} = 64°$.

Step 3 $\angle TAB = \angle OAT - \angle OAB = 90° - 64° = 26°$.

Worked Example 5.1 — Proving Congruence

In the diagram, $ABCD$ is a rectangle. Prove that triangle $ABD$ is congruent to triangle $CDB$.

Step 1 List corresponding equal parts.

• $AB = CD$ (opposite sides of a rectangle are equal)
• $BD = BD$ (common side)
• $\angle ABD = \angle CDB$ (alternate angles, $AB \parallel DC$)

Step 2 Two sides and the included angle are equal, so triangles $ABD$ and $CDB$ are congruent (SAS).

Worked Example 5.2 — Using Scale Factor

Triangles $PQR$ and $XYZ$ are similar. In triangle $PQR$: $PQ = 6$ cm, $QR = 9$ cm, $PR = 12$ cm. In triangle $XYZ$: $XY = 4$ cm. Find $YZ$ and $XZ$.

Step 1 Find the scale factor from $PQR$ to $XYZ$.

$$k = \frac{XY}{PQ} = \frac{4}{6} = \frac{2}{3}$$

Step 2 Apply to corresponding sides.

$$YZ = QR \times \frac{2}{3} = 9 \times \frac{2}{3} = 6 \text{ cm}$$

$$XZ = PR \times \frac{2}{3} = 12 \times \frac{2}{3} = 8 \text{ cm}$$

Worked Example 6.1 — Locus Problem

Points $A$ and $B$ are 8 cm apart. Describe and sketch the locus of all points that are equidistant from $A$ and $B$, and also 5 cm from $A$.

Locus 1 Equidistant from $A$ and $B$: the perpendicular bisector of $AB$. This passes through the midpoint of $AB$ (at 4 cm from each) and is perpendicular to $AB$.

Locus 2 5 cm from $A$: a circle of radius 5 cm centred at $A$.

Solution The required points are the intersections of the perpendicular bisector and the circle. Since the perpendicular bisector is 4 cm from $A$ (horizontally) and the radius is 5 cm, by Pythagoras the intersection points are $\sqrt{5^2 - 4^2} = 3$ cm above and below $AB$ on the perpendicular bisector.

Worked Example 6.2 — Constructing a Perpendicular Bisector

Describe carefully the steps to construct the perpendicular bisector of a line segment $PQ$.

1. Open the compass to a radius greater than half the length of $PQ$.
2. Place the compass at $P$ and draw arcs above and below the line.
3. Without changing the radius, place the compass at $Q$ and draw arcs above and below the line, crossing the first arcs at points $X$ and $Y$.
4. Draw a straight line through $X$ and $Y$. This line is the perpendicular bisector of $PQ$.