Geometry

Geometry

:::info Board Coverage AQA Paper 1 & 2 | Edexcel Paper 1 & 2 | OCR Paper 2 & 3 | WJEC Unit 2 :::

1. Angle Properties

1.1 Basic Angle Facts

Property	Statement
Angles on a straight line	Sum to $180^{\circ}$
Angles at a point	Sum to $360^{\circ}$
Vertically opposite angles	Are equal
Angles in a triangle	Sum to $180^{\circ}$
Angles in a quadrilateral	Sum to $360^{\circ}$

Theorem (Angle sum of a triangle). The interior angles of any triangle sum to $180^{\circ}$.

Proof. Let triangle $\triangle ABC$ have vertices $A$, $B$, $C$. Draw a line through $C$ Parallel to $AB$. Label the intersection points of this line with the exterior of the triangle as $D$ and $E$ (so $D$ is on the side of $B$ and $E$ is on the side of $A$).

By alternate angles: $\angle B = \angle BCD$ and $\angle A = \angle ACE$.

Since $D$, $C$, $E$ lie on a straight line: $\angle BCD + \angle ACB + \angle ACE = 180^{\circ}$.

Therefore $\angle A + \angle B + \angle C = 180^{\circ}$. $\blacksquare$

Theorem (Angle sum of a quadrilateral). The interior angles of any quadrilateral sum to $360^{\circ}$.

Proof. Draw a diagonal, dividing the quadrilateral into two triangles. Each triangle has angles Summing to $180^{\circ}$So the total is $360^{\circ}$. $\blacksquare$

Corollary. The interior angles of any $n$-sided polygon sum to $180(n - 2)^{\circ}$.

Proof by induction. A triangle ($n = 3$) has sum $180^{\circ}$. Adding a vertex to an $(n-1)$-gon creates a triangle, adding $180^{\circ}$. So an $n$-gon has sum $180 + 180(n - 3) = 180(n - 2)^{\circ}$. $\blacksquare$

1.2 Angles in Parallel Lines

Type	Description
Corresponding angles (F-angles)	Equal
Alternate angles (Z-angles)	Equal
Co-interior (allied) angles (U-angles)	Sum to $180^{\circ}$

Worked Example. In the diagram, line $AB$ is parallel to line $CD$. A transversal intersects $AB$ at $E$ and $CD$ at $F$. If $\angle AEF = 65^{\circ}$Find all the other angles at $E$ and $F$.

$\angle AEF = \angle EFD = 65^{\circ}$ (alternate angles).

$\angle BEF = 180^{\circ} - 65^{\circ} = 115^{\circ}$ (angles on a straight line).

$\angle EFC = 115^{\circ}$ (corresponding to $\angle BEF$).

$\angle CFD = 65^{\circ}$ (vertically opposite $\angle EFD$).

$\angle EFB = \angle EFD = 65^{\circ}$ (vertically opposite).

Worked Example (Higher Tier). Two parallel lines are cut by a transversal. One interior angle is $3x + 10^{\circ}$ and the co-interior angle is $2x + 20^{\circ}$. Find the value of $x$.

Since co-interior angles sum to $180^{\circ}$:

$(3x + 10) + (2x + 20) = 180$

$5x + 30 = 180$

$5x = 150$

$x = 30$

1.3 Angles in Polygons

The sum of interior angles of an $n$-sided polygon:

$S = 180(n - 2)^{\circ}$

The interior angle of a regular $n$-sided polygon:

\mathrm{Each interior angle = \frac{180(n - 2)}{n}^{\circ}

The exterior angle of a regular polygon:

\mathrm{Each exterior angle = \frac{360}{n}^{\circ}

Worked Example. Find the interior angle of a regular pentagon.

\mathrm{Interior angle = \frac{180(5 - 2)}{5} = \frac{540}{5} = 108^{\circ}

Worked Example (Higher Tier). A regular polygon has an interior angle of $150^{\circ}$. How many Sides does it have?

$\frac{180(n - 2)}{n} = 150$ $180n - 360 = 150n$ $30n = 360$ $n = 12$

It is a regular dodecagon (12 sides).

Worked Example (Higher Tier). Find the sum of the interior angles of a polygon with 15 sides.

$S = 180(15 - 2) = 180 \times 13 = 2340^{\circ}$

Worked Example (Higher Tier). Find the exterior angle of a regular decagon, and hence find the Interior angle.

\mathrm{Exterior angle = \frac{360}{10} = 36^{\circ}

\mathrm{Interior angle = 180 - 36 = 144^{\circ}

1.4 Bearings

A bearing is an angle measured clockwise from north, always given as a three-figure number (e.g. $045^{\circ}$Not $45^{\circ}$).

Worked Example. A ship sails from port $A$ on a bearing of $070^{\circ}$ for 80 km to point $B$. It then sails on a bearing of $150^{\circ}$ for 60 km to point $C$. Find the bearing of $C$ from $A$.

The angle $\angle NAB = 70^{\circ}$ and $\angle NBC = 150^{\circ}$.

The angle between $AB$ and north at $B$ is $180^{\circ} - 70^{\circ} = 110^{\circ}$ (back bearing).

The angle $\angle ABC = 110^{\circ} + 150^{\circ} = 260^{\circ}$… This is the external angle. The Internal angle is $360^{\circ} - 260^{\circ} = 100^{\circ}$.

Using the cosine rule on $\triangle ABC$:

$AC^2 = 80^2 + 60^2 - 2 \times 80 \times 60 \times \cos(100^{\circ})$ $AC^2 = 6400 + 3600 - 9600 \times (-0.1736\ldots)$ $AC^2 = 10000 + 1667.1\ldots = 11667.1\ldots$ AC \approx 108.0 \mathrm{ km

Using the sine rule to find $\angle BAC$:

$\frac{\sin \angle BAC}{60} = \frac{\sin 100^{\circ}}{108.0}$ $\sin \angle BAC = \frac{60 \times 0.9848}{108.0} = 0.5471\ldots$ $\angle BAC = 33.2^{\circ}$

Bearing of $C$ from $A = 70^{\circ} + 33.2^{\circ} = 103.2^{\circ} \approx 103^{\circ}$.

2. Triangles and Trigonometry

2.1 Pythagoras’ Theorem

Theorem. In a right-angled triangle with hypotenuse $c$ and legs $a$ and $b$:

$a^2 + b^2 = c^2$

Proof (area-based). Consider a square of side $(a + b)$. Place four identical right-angled Triangles inside, each with legs $a$ and $b$Arranged so that their hypotenuses form a smaller Square of side $c$ in the centre.

The area of the large square equals the area of the four triangles plus the area of the inner Square:

$(a + b)^2 = 4 \times \frac{1}{2}ab + c^2$ $a^2 + 2ab + b^2 = 2ab + c^2$ $a^2 + b^2 = c^2 \quad \blacksquare$

Converse of Pythagoras’ Theorem. If $a^2 + b^2 = c^2$ for a triangle with sides $a, b, c$ where $c$ is the longest side, then the triangle is right-angled.

Worked Example. Is a triangle with sides 5 cm, 12 cm, and 13 cm right-angled?

$5^2 + 12^2 = 25 + 144 = 169 = 13^2$. Yes, it is right-angled.

Worked Example. A ladder of length 10 m leans against a wall with its foot 6 m from the base of The wall. How high up the wall does it reach?

$h^2 + 6^2 = 10^2$ $h^2 = 100 - 36 = 64$ h = 8 \mathrm{ m

Worked Example (Higher Tier). Is a triangle with sides 7 cm, 11 cm, and 13 cm acute, Right-angled, or obtuse?

$7^2 + 11^2 = 49 + 121 = 170$. Since $170 \gt 13^2 = 169$The triangle is acute (the angle opposite The longest side is less than $90^{\circ}$).

Test for triangle type:

Condition	Type
$a^2 + b^2 = c^2$	Right-angled
$a^2 + b^2 \gt c^2$	Acute
$a^2 + b^2 \lt c^2$	Obtuse

2.2 SOH CAH TOA

For a right-angled triangle with angle $\theta$:

\sin \theta = \frac{\mathrm{opposite}{\mathrm{hypotenuse}, \quad \cos \theta = \frac{\mathrm{adjacent}{\mathrm{hypotenuse}, \quad \tan \theta = \frac{\mathrm{opposite}{\mathrm{adjacent}

Proof that $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

\tan \theta = \frac{\mathrm{opp}{\mathrm{adj} = \frac{\mathrm{opp/\mathrm{hyp}{\mathrm{adj/\mathrm{hyp} = \frac{\sin \theta}{\cos \theta} \quad \blacksquare

Proof that $\sin^2\theta + \cos^2\theta = 1$.

By Pythagoras: \mathrm{opp^2 + \mathrm{adj^2 = \mathrm{hyp^2.

Dividing by \mathrm{hyp^2:

\frac{\mathrm{opp^2}{\mathrm{hyp^2} + \frac{\mathrm{adj^2}{\mathrm{hyp^2} = 1

$\sin^2\theta + \cos^2\theta = 1 \quad \blacksquare$

Worked Example. Find the length of the hypotenuse in a right-angled triangle where the opposite Side is 5 cm and the angle is $35^{\circ}$.

$\sin 35^{\circ} = \frac{5}{h}$ h = \frac{5}{\sin 35^{\circ}} = \frac{5}{0.5736\ldots} = 8.72 \mathrm{ cm (to 3 s.f.)

Worked Example. Find the angle $\theta$ in a right-angled triangle where the adjacent side is 8 Cm and the hypotenuse is 15 cm.

$\cos \theta = \frac{8}{15}$ $\theta = \cos^{-1}\!\left(\frac{8}{15}\right) \approx 57.8^{\circ}$

2.3 Sine Rule

For any triangle $\triangle ABC$:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Used when you know: an angle and its opposite side, plus one other angle or side.

Proof sketch. Drop an altitude from $C$ to $AB$. In the two right-angled triangles formed, Express the height as both $b \sin A$ and $a \sin B$. Setting equal: $b \sin A = a \sin B$Giving $\frac{a}{\sin A} = \frac{b}{\sin B}$. $\blacksquare$

Worked Example (Higher Tier). In $\triangle ABC$, $a = 10$ cm, $A = 45^{\circ}$ $B = 70^{\circ}$. Find $b$.

$\frac{b}{\sin 70^{\circ}} = \frac{10}{\sin 45^{\circ}}$ b = \frac{10 \sin 70^{\circ}}{\sin 45^{\circ}} = \frac{10 \times 0.9397}{0.7071} \approx 13.29 \mathrm{ cm

2.4 Cosine Rule

For any triangle $\triangle ABC$:

$a^2 = b^2 + c^2 - 2bc \cos A$

Rearranged to find an angle:

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$

Used when you know: two sides and the included angle (to find the third side), or all three sides (to find an angle).

Worked Example. In $\triangle ABC$, $a = 8$ cm, $b = 5$ cm, $c = 7$ cm. Find angle $A$.

$\cos A = \frac{25 + 49 - 64}{2 \times 5 \times 7} = \frac{10}{70} = \frac{1}{7}$ A = \cos^{-1}\!\left(\frac{1}{7}\right) = 81.8^{\circ} \mathrm{ (to 1 d.p.)

Worked Example (Higher Tier). In $\triangle ABC$, $a = 12$ cm, $b = 8$ cm, $C = 60^{\circ}$. Find $c$.

$c^2 = 144 + 64 - 2 \times 12 \times 8 \times \cos 60^{\circ}$ $c^2 = 208 - 96 = 112$ c = \sqrt{112} = 4\sqrt{7} \approx 10.58 \mathrm{ cm

2.5 Area of a Triangle

\mathrm{Area = \frac{1}{2}ab \sin C

Where $a$ and $b$ are two sides and $C$ is the included angle.

Worked Example. Find the area of $\triangle ABC$ where $a = 10$ cm, $b = 8$ cm, and $C = 45^{\circ}$.

\mathrm{Area = \frac{1}{2} \times 10 \times 8 \times \sin 45^{\circ} = 40 \times \frac{\sqrt{2}}{2} = 20\sqrt{2} \approx 28.3 \mathrm{ cm^2

2.6 The Ambiguous Case of the Sine Rule (Higher Tier)

When using the sine rule to find an angle, there may be two possible solutions: $\theta$ and $180^{\circ} - \theta$.

Worked Example. In $\triangle ABC$, $a = 8$ cm, $b = 10$ cm, $A = 40^{\circ}$. Find angle $B$.

$\frac{\sin B}{10} = \frac{\sin 40^{\circ}}{8}$ $\sin B = \frac{10 \sin 40^{\circ}}{8} = \frac{10 \times 0.6428}{8} = 0.8035$

$B = \sin^{-1}(0.8035) \approx 53.5^{\circ}$ or $B = 180^{\circ} - 53.5^{\circ} = 126.5^{\circ}$.

Both are valid since $53.5 + 40 = 93.5 \lt 180$ and $126.5 + 40 = 166.5 \lt 180$.

When to check for the ambiguous case: Only when finding an angle using the sine rule. If Finding a side, there is at most one solution.

3. Circle Theorems

3.1 Key Theorems

Theorem	Statement
Centre and chord	The perpendicular from the centre to a chord bisects the chord
Tangent and radius	A tangent is perpendicular to the radius at the point of contact
Two tangents	Two tangents from an external point are equal in length
Angle at centre	The angle at the centre is twice the angle at the circumference subtended by the same arc
Angle in a semicircle	The angle in a semicircle is a right angle
Cyclic quadrilateral	Opposite angles sum to $180^{\circ}$
Same segment	Angles in the same segment are equal

3.2 Proof: Angle at Centre is Twice Angle at Circumference

Proof. Let $O$ be the centre of a circle. Let arc $AB$ subtend angle $\angle AOB = 2\theta$ at The centre and angle $\angle ACB = \theta$ at a point $C$ on the circumference.

Draw the radius $OC$. Since $OA = OC$ (radii), $\triangle OAC$ is isosceles, so $\angle OAC = \angle OCA$.

Similarly, $OB = OC$So $\triangle OBC$ is isosceles, and $\angle OBC = \angle OCB$.

The exterior angle at $O$ for $\triangle OAC$: $2\angle OCA = \angle AOC$.

The exterior angle at $O$ for $\triangle OBC$: $2\angle OCB = \angle BOC$.

Adding: $\angle AOC + \angle BOC = 2\angle OCA + 2\angle OCB$

$\angle AOB = 2(\angle OCA + \angle OCB) = 2\angle ACB \quad \blacksquare$

3.3 Proof: Angle in a Semicircle is a Right Angle

Proof. Let $AB$ be a diameter of a circle with centre $O$. Let $C$ be any point on the Circumference.

Since $AB$ is a diameter, $\angle AOB = 180^{\circ}$.

By the angle at centre theorem: $\angle AOB = 2\angle ACB$.

Therefore $\angle ACB = 90^{\circ}$. $\blacksquare$

3.4 Proof: Opposite Angles of a Cyclic Quadrilateral Sum to $180^{\circ}$

Proof. Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$. The arc $ABC$ subtends angle $\angle ADC$ at the circumference and angle $\angle AOC$ at the centre. Similarly, arc $ADC$ Subtends $\angle ABC$ and $\angle AOD$.

The arcs $ABC$ and $ADC$ together make the full circle, so $\angle AOC + \angle AOD = 360^{\circ}$.

By the angle at centre theorem: $\angle ADC = \frac{1}{2}\angle AOC$ and $\angle ABC = \frac{1}{2}\angle AOD$.

Adding: $\angle ABC + \angle ADC = \frac{1}{2} \times 360^{\circ} = 180^{\circ}$. $\blacksquare$

3.5 Worked Examples with Circle Theorems

Worked Example. $A$, $B$, $C$And $D$ lie on a circle. $\angle ABC = 75^{\circ}$ and $\angle CAD = 40^{\circ}$. Find $\angle ABD$.

$\angle ABC$ and $\angle ADC$ are opposite angles of cyclic quadrilateral $ABCD$: $\angle ADC = 180^{\circ} - 75^{\circ} = 105^{\circ}$.

$\angle CAD$ and $\angle CBD$ are in the same segment (subtended by arc $CD$), so $\angle CAD = \angle CBD = 40^{\circ}$.

Therefore $\angle ABD = \angle ABC - \angle CBD = 75^{\circ} - 40^{\circ} = 35^{\circ}$.

Worked Example (Higher Tier). $AB$ is a diameter of a circle with centre $O$. $C$ is a point on The circle such that $\angle BAC = 32^{\circ}$. Find $\angle OCA$.

Since $AB$ is a diameter, $\angle ACB = 90^{\circ}$.

$\angle ABC = 90^{\circ} - 32^{\circ} = 58^{\circ}$.

Since $OA = OC$ (radii), $\triangle OAC$ is isosceles: $\angle OCA = \angle OAC = 32^{\circ}$.

Worked Example (Higher Tier). $AB$ and $AC$ are tangents to a circle at points $B$ and $C$ Respectively. Prove that $AB = AC$.

Join $O$ to $A$, $B$And $C$. Since $OB$ and $OC$ are radii, and tangents are perpendicular to Radii at the point of contact, $\angle OBA = \angle OCA = 90^{\circ}$.

$OA$ is common, and $OB = OC$ (radii). By RHS (right angle, hypotenuse, side), $\triangle OBA \cong \triangle OCA$.

Therefore $AB = AC$. $\blacksquare$

4. Area and Perimeter

4.1 2D Shapes

Shape	Area	Perimeter
Rectangle	$l \times w$	$2(l + w)$
Triangle	$\frac{1}{2}bh$	Sum of sides
Parallelogram	$bh$	Sum of sides
Trapezium	$\frac{1}{2}(a + b)h$	Sum of sides
Circle	$\pi r^2$	$2\pi r$ (circumference)
Sector	$\frac{\theta}{360} \pi r^2$	Arc: $\frac{\theta}{360} \times 2\pi r$

Proof of the area of a trapezium. A trapezium with parallel sides $a$ and $b$ and height $h$ can Be divided into a rectangle and two triangles. The rectangle has area $ah$ and the two triangles Have total area $\frac{1}{2}(b-a)h + \frac{1}{2}(b-a)h = (b-a)h$. Total: $ah + (b-a)h = bh - ah + ah = (a+b)h/2$. $\blacksquare$

4.2 3D Shapes

Shape	Volume	Surface Area
Cuboid	$lwh$	$2(lw + lh + wh)$
Cylinder	$\pi r^2 h$	$2\pi r^2 + 2\pi rh$
Sphere	$\frac{4}{3}\pi r^3$	$4\pi r^2$
Cone	$\frac{1}{3}\pi r^2 h$	$\pi r l + \pi r^2$ (where $l$ = slant height)
Pyramid	\frac{1}{3} \times \mathrm{base area \times h	Base area + triangular faces

Worked Example. A cylinder has radius 5 cm and height 12 cm. Find its volume and total surface Area.

V = \pi \times 5^2 \times 12 = 300\pi \approx 942 \mathrm{ cm^3

\mathrm{SA = 2\pi \times 25 + 2\pi \times 5 \times 12 = 50\pi + 120\pi = 170\pi \approx 534 \mathrm{ cm^2

Worked Example (Higher Tier). A cone has base radius 6 cm and slant height 10 cm. Find its Volume.

The height $h$: $h^2 + 6^2 = 10^2$So $h = 8$ cm.

V = \frac{1}{3}\pi \times 36 \times 8 = 96\pi \approx 301.6 \mathrm{ cm^3

Worked Example (Higher Tier). A solid hemisphere has radius 7 cm. Find its total surface area.

Curved surface area: $2\pi r^2 = 2\pi \times 49 = 98\pi$.

Flat face: $\pi r^2 = 49\pi$.

Total: 147\pi \approx 461.8 \mathrm{ cm^2.

Worked Example (Higher Tier). A frustum is formed by removing a small cone of height 4 cm from The top of a cone of height 12 cm. Both cones have the same base radius 5 cm. Find the volume of the Frustum.

The large cone has volume $\frac{1}{3}\pi \times 25 \times 12 = 100\pi$.

The small cone has height 4 cm. By similar triangles, the radius ratio is $4/12 = 1/3$So the small Cone has radius $5/3$ cm.

Volume of small cone: $\frac{1}{3}\pi \times \frac{25}{9} \times 4 = \frac{100\pi}{27}$.

Volume of frustum: 100\pi - \frac{100\pi}{27} = \frac{2600\pi}{27} \approx 302.3 \mathrm{ cm^3.

5. Transformations

5.1 Types of Transformation

Transformation	Description
Translation	Movement by a vector $\begin{pmatrix} x \\ y \end{pmatrix}$
Reflection	Mirror image across a line of reflection
Rotation	Turned about a centre by an angle and direction
Enlargement	Scaled from a centre by a scale factor

Worked Example. Describe fully the transformation that maps $\triangle ABC$ with vertices at (1, 2)$$(3, 5)$$(5, 2) to $\triangle A'B'C'$ with vertices at (-1, -2)$$(-3, -5) $(-5, -2)$.

$(1, 2) \to (-1, -2)$: the $x$-coordinate is negated and the $y$-coordinate is negated. This is a Reflection in the origin, which is equivalent to a rotation of $180^{\circ}$ about the origin.

5.2 Vectors

A vector has both magnitude and direction. We write vectors as column vectors or using bold letters.

Addition: $\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}$

Scalar multiplication: $k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}$

Magnitude: $\left|\begin{pmatrix} a \\ b \end{pmatrix}\right| = \sqrt{a^2 + b^2}$

Parallel vectors: $\mathbf{a}$ and $\mathbf{b}$ are parallel if $\mathbf{a} = k\mathbf{b}$ for Some scalar $k$.

Worked Example. Points A$$BAnd $C$ have position vectors \begin{pmatrix} 2 \\ 3 \end{pmatrix}$$\begin{pmatrix} 8 \\ 7 \end{pmatrix}And $\begin{pmatrix} 14 \\ 11 \end{pmatrix}$. Show that A$$BAnd $C$ are collinear.

$\overrightarrow{AB} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}, \qquad \overrightarrow{BC} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$

Since $\overrightarrow{AB} = \overrightarrow{BC}$The vectors are parallel and share point $B$So $A$, $B$, $C$ are collinear. $\blacksquare$

5.3 Column Vectors in Geometry

The position vector of a point $P$ relative to an origin $O$ is $\overrightarrow{OP}$.

The vector from $A$ to $B$ is $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$.

The midpoint $M$ of $AB$ has position vector $\frac{1}{2}(\mathbf{a} + \mathbf{b})$.

Worked Example (Higher Tier). Point $P$ divides the line segment $AB$ in the ratio $2 : 3$. If $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 11 \\ 9 \end{pmatrix}$Find $\overrightarrow{OP}$.

$\overrightarrow{OP} = \overrightarrow{OA} + \frac{2}{5}\overrightarrow{AB} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} + \frac{2}{5}\begin{pmatrix} 10 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}$

6. Similarity and Congruence

6.1 Congruent Triangles

Triangles are congruent if they are identical in shape and size. The conditions are:

Condition	Abbreviation
Three sides equal	SSS
Two sides and included angle	SAS
Two angles and a corresponding side	AAS
Right angle, hypotenuse, one side	RHS

:::caution ASS (two sides and a non-included angle) is NOT a valid congruence condition. It can Produce two different triangles.

6.2 Similar Triangles

Triangles are similar if they have the same shape but different sizes. Corresponding angles are Equal, and corresponding sides are in the same ratio.

Area scale factor = (length scale factor)$^2$.

Volume scale factor = (length scale factor)$^3$.

Worked Example. Two similar solids have volumes of 27 cm$^3$ and 125 cm$^3$. The surface area of The smaller solid is 54 cm$^2$. Find the surface area of the larger solid.

Length scale factor $= \sqrt[3]{\frac{125}{27}} = \frac{5}{3}$.

Area scale factor $= \left(\frac{5}{3}\right)^2 = \frac{25}{9}$.

Surface area = 54 \times \frac{25}{9} = 150 \mathrm{ cm^2.

Worked Example (Higher Tier). Two similar triangles have areas in the ratio $16 : 49$. The Perimeter of the smaller triangle is 24 cm. Find the perimeter of the larger triangle.

Length scale factor $= \sqrt{\frac{49}{16}} = \frac{7}{4}$.

Perimeter of larger $= 24 \times \frac{7}{4} = 42$ cm.

7. Construction and Loci

7.1 Standard Constructions

• Perpendicular bisector of a line segment: using compasses, draw arcs from each endpoint, then join the intersection points.
• Angle bisector: using compasses, draw arcs from the vertex, then from the intersection points with each arm.
• Perpendicular from a point to a line: using compasses centered at the point, find two equidistant points on the line, then construct the perpendicular bisector.
• Regular polygons: constructed by dividing a circle into equal arcs.

7.2 Loci

A locus is the set of all points satisfying a given condition.

Locus	Description
Fixed distance from a point	Circle
Fixed distance from a line	Two parallel lines
Equidistant from two points	Perpendicular bisector
Equidistant from two lines	Angle bisector

7.3 Regions (Higher Tier)

Loci problems often require shading the region satisfying multiple conditions simultaneously.

Worked Example. A goat is tethered to a corner of a rectangular field measuring 20 m by 15 m by A rope of length 8 m. Shade the region the goat can graze.

The region is a quarter circle of radius 8 m centred at the corner.

Worked Example (Higher Tier). Point $A$ is at $(2, 3)$ and point $B$ is at $(8, 7)$. Shade the Region of points that are within 5 units of $A$ and closer to $A$ than to $B$.

The first condition is a circle of radius 5 centred at $A$. The second condition is the half-plane On $A$‘s side of the perpendicular bisector of $AB$. The shaded region is the intersection.

8. 3D Geometry (Higher Tier)

8.1 3D Pythagoras and Trigonometry

For a cuboid with dimensions $a, b, c$The longest diagonal is:

$d = \sqrt{a^2 + b^2 + c^2}$

Worked Example. A cuboid has dimensions 5 cm, 12 cm, and 8 cm. Find the length of the longest Diagonal.

d = \sqrt{25 + 144 + 64} = \sqrt{233} \approx 15.26 \mathrm{ cm

Worked Example. A cone has base radius 3 cm and height 4 cm. Find the angle between the slant Height and the base.

Slant height $l = \sqrt{9 + 16} = 5$ cm.

$\cos \theta = \frac{3}{5}$ $\theta = \cos^{-1}(0.6) \approx 53.1^{\circ}$

Worked Example (Higher Tier). A pyramid has a square base of side 6 cm and all its triangular Faces are equilateral. Find the height of the pyramid.

The slant height equals the side length: $l = 6$ cm.

The distance from the centre of the base to a vertex: $\frac{6\sqrt{2}}{2} = 3\sqrt{2}$ cm.

Height: $h = \sqrt{6^2 - (3\sqrt{2})^2} = \sqrt{36 - 18} = \sqrt{18} = 3\sqrt{2}$ cm.

Common Pitfalls

• Using degrees when your calculator is in radians mode (or vice versa). Always check.
• Misidentifying which sides are opposite/adjacent in trigonometry. Draw and label the triangle.
• Using Pythagoras for non-right-angled triangles. Use the sine or cosine rule instead.
• Confusing arc length and sector area formulas. Arc length is a fraction of $2\pi r$; sector area is a fraction of $\pi r^2$.
• Forgetting that the angle in the cosine rule must be the included angle (between the two known sides).
• Mixing up similarity and congruence. Congruent shapes are also similar, but similar shapes are not necessarily congruent.
• The ambiguous case of the sine rule. When finding an angle, always check whether the supplementary angle is also valid.
• Forgetting the perpendicular bisector theorem. Points on the perpendicular bisector are equidistant from both endpoints.
• Using the wrong scale factor for area or volume. Area uses the square of the length scale factor; volume uses the cube.
• Calculating the exterior angle incorrectly. The exterior angle is $\frac{360}{n}$Not $\frac{180}{n}$.

Practice Questions

1. A regular hexagon and a regular octagon share a common side. Find the size of the angle between them.

2. In $\triangle ABC$, $a = 12$ cm, $b = 9$ cm, and $B = 40^{\circ}$. Find angle $A$.

3. Prove that the exterior angle of a triangle equals the sum of the two interior opposite angles.

4. A sector has radius 8 cm and angle $135^{\circ}$. Find its perimeter and area.

5. Point $P$ divides the line segment $AB$ in the ratio $2 : 3$. If $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 11 \\ 9 \end{pmatrix}$Find $\overrightarrow{OP}$.

6. Two similar cones have heights in the ratio $3 : 5$. The volume of the smaller cone is 108 cm$^3$. Find the volume of the larger cone.

7. $A$, $B$And $C$ are points on a circle with centre $O$. Angle $ABC = 55^{\circ}$. Find angle $AOC$.

8. A triangle has sides 7 cm, 8 cm, and 10 cm. Determine whether it is acute, right-angled, or obtuse.

9. Describe fully the single transformation that maps $\triangle ABC$ with vertices at $(1, 2)$ (3, 5)$$(5, 2) to $\triangle A'B'C'$ with vertices at (-1, -2)$$(-3, -5)$$(-5, -2).

10. Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm.

11. Prove that the angle between a tangent and a chord equals the angle in the alternate segment.

12. A frustum is formed by removing a cone of height 4 cm from the top of a cone of height 10 cm. Both cones have base radius 6 cm. Find the volume of the frustum.

13. In \triangle ABC$$\angle A = 30^{\circ}$$b = 8 cm, $c = 5$ cm. Find the two possible values of $a$.

14. Points $P$ and $Q$ have position vectors $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 7 \\ 5 \end{pmatrix}$. Find the position vector of the midpoint of $PQ$ and the magnitude of $\overrightarrow{PQ}$.

15. A cylinder and a cone have the same base radius and the same volume. If the cylinder has height 9 cm, find the height of the cone.

16. Prove that the angle at the centre of a circle is twice the angle at the circumference.

17. A regular polygon has each exterior angle of $24^{\circ}$. How many sides does it have? Find the sum of its interior angles.

18. Find the shortest distance from the point $(1, 2, 3)$ to the $xy$-plane.

19. Prove that the sum of the interior angles of a pentagon is $540^{\circ}$.

20. A cone has slant height 10 cm and total surface area $165\pi$ cm$^2$. Find its radius and volume.

21. Prove that the angle between a tangent and a chord equals the angle in the alternate segment.

22. A regular hexagon is inscribed in a circle of radius 8 cm. Find the perimeter and area of the hexagon.

23. Two circles have radii 5 cm and 3 cm, and their centres are 10 cm apart. Determine whether the circles intersect, are tangent, or are separate.

Extended Practice (Higher Tier)

24. A cylinder and a cone have the same base radius and the same height. Prove that the volume of the cylinder is three times the volume of the cone.

25. Triangle $\triangle ABC$ has vertices at (2, 3)$$(8, 7)And $(6, 1)$. Find: (a) the length of side $AB$(b) the area of the triangle, (c) the equation of the line through $C$ perpendicular to $AB$.

26. A sector of a circle has radius 12 cm and angle $75^\circ$. Find its perimeter and area.

27. Prove that if two chords of a circle are equal in length, they are equidistant from the centre.

28. The points A(1, 2)$$B(5, 6)And $C(3, k)$ are collinear. Find $k$.

29. A sphere has surface area $144\pi$ cm$^2$. Find its volume.

30. In \triangle ABC$$AB = 8 cm, $BC = 6$ cm, and $\angle ABC = 120^\circ$. Find the area of the triangle.

Worked Examples

Example 1:

A typical exam question on Geometry requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Geometry often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

Summary

This topic covers the mathematical techniques and concepts related to geometry, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• sine, cosine, and tangent functions
• trigonometric identities
• solving trigonometric equations
• the sine and cosine rules
• radian measure and arc length

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

:::