Geometry and trigonometry

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6 Geometry

A geometry problem is a chain from facts you are given to a fact you need. Mark the diagram, name each usable relationship, and choose the shortest chain that reaches the target.

Route	Useful when
angle facts	the target is an angle or parallel relationship
congruence or similarity	corresponding lengths or angles must be transferred
transformations	one shape is moved, reflected, rotated or enlarged
vectors	parallelism, ratios or a common point can be expressed algebraically

Start with statements that are guaranteed by labels or the question, not by appearance. Add one justified consequence at a time; if a route stalls, return to the givens and try a different valid route.

Formal theorem proofs are not required, but every step in a solution still needs a valid geometric reason. A diagram that is not accurately drawn is never evidence that lines are equal, parallel or perpendicular.

Write geometry as paired statements: make a claim, then give the fact that guarantees it. This exposes missing assumptions and makes a multi-step argument checkable.

Claim	Suitable reason
$ngle ABC=ngle BCD$	alternate angles, because $AB\parallel CD$
$OA=OB$	radii of the same circle
$riangle ABC\sim riangle DEF$	two corresponding angles are equal
$AC=DF$	corresponding sides of congruent triangles

Name points in matching order, state any intermediate equality you need, and finish by explicitly connecting the established result to the requested conclusion.

Do not write a theorem name without showing that its conditions hold. For example, alternate angles are equal only after the relevant lines have been established as parallel.

Structure	Angle fact
straight line	sum $180^\circ$
around a point	sum $360^\circ$
triangle	sum $180^\circ$
quadrilateral	sum $360^\circ$
interior angles of an $n$ -gon	sum $(n-2)180^\circ$
regular $n$ -gon exterior angle	$360^\circ/n$

With parallel lines, corresponding and alternate angles are equal; co-interior angles sum to $180^\circ$ . Mark the two parallel lines and the transversal before naming the relationship.

\text{anticlockwise turn}>0,\qquad \text{clockwise turn}<0

Use only marked parallel lines or established facts. Do not transfer an angle merely because two lines look parallel, and do not confuse a polygon's interior-angle sum with one interior angle of a regular polygon.

Shape	Guaranteed properties
parallelogram	opposite sides parallel and equal; opposite angles equal; diagonals bisect each other
rectangle	parallelogram with four right angles; diagonals equal
rhombus	parallelogram with four equal sides; diagonals perpendicular
square	rectangle and rhombus properties
trapezium	one pair of opposite sides parallel
kite	two pairs of adjacent equal sides; one diagonal bisects the other at right angles

Classify from the properties that are guaranteed, not from appearance. A square is also a rectangle, rhombus and parallelogram because it satisfies all their defining conditions.

When solving, translate each mark into a property: arrows mean parallel, ticks mean equal lengths, a small square means a right angle. Then select only conclusions licensed by those marks.

Diagonals are not automatically equal, perpendicular or angle bisectors in every quadrilateral. State the specific shape property before using one of these conclusions.

A line of symmetry reflects a shape onto itself. Corresponding points lie on opposite sides of the mirror line, at equal perpendicular distances from it.

Rotational symmetry of order $n$ means the shape matches itself $n$ times in one full turn. The smallest positive matching angle is $360^\circ/n$ ; order 1 means only the full turn works.

Symmetry	Fixed object
about a point	the centre of rotation
about a line	every point on the mirror line
about a plane	every point in the mirror plane

To complete a reflected shape, count perpendicular grid steps from the axis rather than copying a horizontal or vertical offset. Do not count the starting position twice when finding rotational order.

a^2+b^2=c^2\qquad(c\text{ is opposite the right angle})

Find or create a right-angled triangle, label the hypotenuse first, substitute lengths, then take the positive square root. In an acute triangle, an altitude splits the shape into right-angled triangles.

\text{cuboid space diagonal}=\sqrt{l^2+w^2+h^2}

Pythagoras applies only to a right-angled triangle. In 3D, use a visible face diagonal as an intermediate length if necessary; never add unsquared lengths, and do not use the excluded angle-bisector theorems.

\text{length factor}=k,\qquad \text{area factor}=k^2,\qquad \text{volume factor}=k^3

Write the direction of comparison before forming a ratio. If shape B is an enlargement of A by $k$ , every corresponding length in B is $k$ times its partner in A.

k=\sqrt{\frac{A_B}{A_A}}\quad\text{or}\quad k=\sqrt[3]{\frac{V_B}{V_A}}

Square or cube the linear factor, not the measured area or volume. Similar figures have equal corresponding angles and proportional corresponding lengths; equal area alone does not prove similarity.

To prove similarity, establish enough corresponding information: AA uses two equal angle pairs; SSS uses three proportional side pairs; SAS uses two proportional side pairs with the included angle equal.

\triangle ABC\sim\triangle DEF\ \Rightarrow\ A\leftrightarrow D,\ B\leftrightarrow E,\ C\leftrightarrow F

After proving similarity, use the same correspondence order to equate angle pairs or form side ratios. Keep one consistent scale-factor direction throughout the calculation.

Equal angles prove the same shape, not the same size. Do not assume similarity because triangles look alike, and do not mix non-corresponding sides in a proportion.

Congruent shapes have exactly the same size and shape. One can be translated, rotated or reflected to fit the other without enlargement.

Match distinctive vertices, equal angles and equal sides, then list vertices in corresponding order. Orientation may reverse after a reflection, but corresponding lengths remain equal.

Relationship	Angles	Corresponding lengths
congruent	equal	equal
similar	equal	proportional

Equal area, equal perimeter or the same general outline alone does not guarantee congruence. A scale factor other than 1 gives similar, not congruent, shapes.

Test	Information required
SSS	all three corresponding sides equal
SAS	two sides and their included angle equal
ASA	two angles and the included side equal
RHS	right triangles with equal hypotenuse and one other side

State the three matching facts with reasons, name the test, then conclude using correctly ordered triangle names. Only after congruence is established may you transfer other corresponding sides or angles.

\triangle ABC\cong\triangle DEF\ \Rightarrow\ AB=DE,\ BC=EF,\ AC=DF

AAA proves similarity, not congruence. SSA is not a general congruence test, and in SAS the known angle must be between the two known sides.

Configuration	Theorem
same chord or arc	angles at the circumference are equal
same arc	angle at centre = twice angle at circumference
angle in a semicircle	$90^\circ$
radius and tangent at contact	perpendicular
tangent and chord	angle equals angle in alternate segment

Intersections	Product relationship
chords intersect inside at $P$	$PA\cdot PB=PC\cdot PD$
two secants from external $P$	$PA\cdot PB=PC\cdot PD$ using whole secants
tangent $PT$ and secant $PAB$	$PT^2=PA\cdot PB$

Identify the chord, arc, centre, tangent or intersection that triggers a theorem. Mark equal radii and add triangle angle facts only after the circle relationship is clear.

A tangent theorem needs the exact point of contact. For an external secant product, multiply each external segment by its whole secant, not by the internal segment alone.

A cyclic quadrilateral has all four vertices on one circle. Its opposite interior angles are supplementary.

\angle A+\angle C=180^\circ,\qquad \angle B+\angle D=180^\circ

An exterior angle of a cyclic quadrilateral equals the interior opposite angle, because the adjacent interior angle and the exterior angle also sum to $180^\circ$ .

The sides need not be equal and the diagonals need not be diameters. Do not use the supplementary-opposite-angle rule unless all four vertices are established on the same circle.

Condition	Locus
fixed distance $r$ from point $A$	circle centre $A$ , radius $r$
equidistant from points $A$ and $B$	perpendicular bisector of $AB$
fixed perpendicular distance from a line	two parallel lines
equidistant from two intersecting lines	both angle bisectors

For 'less than' or 'within', shade the side or interior satisfying the inequality; for 'more than', use the opposite region. Intersections of conditions give the feasible set.

Test one point from each candidate region against the original distance wording. Keep construction arcs and boundaries visible when an exact locus is required.

A locus is the complete set of points satisfying a condition, not one example path. Use geometric construction or measured distance; tracing-paper methods are not acceptable.

Angle bisector: from the vertex draw an arc cutting both arms; from those two cut points draw equal-radius arcs that intersect; join the vertex to that intersection.

Perpendicular bisector of segment $AB$ : with compass radius greater than half $AB$ , draw arcs from $A$ and $B$ above and below the segment; join the two arc intersections.

Construction	Guaranteed result
angle bisector	points on it are equidistant from the two arms
perpendicular bisector	points on it are equidistant from $A$ and $B$ ; it crosses $AB$ at $90^\circ$ and its midpoint

Use only an unmarked ruler and compasses, preserve the construction arcs, and keep the same compass radius for paired arcs. Measuring halfway with ruler marks or estimating an angle is not a valid construction.

7 Mensuration

Length measures one dimension, area measures a surface in square units, and volume measures space in cubic units. Decide which quantity the question asks for before choosing a formula.

Quantity	Example unit	Conversion pattern
length	$\mathrm{cm}$	multiply the linear conversion factor
area	$\mathrm{cm^2}$	square the linear conversion factor
volume	$\mathrm{cm^3}$	cube the linear conversion factor

1\text{ m}=100\text{ cm}\Rightarrow 1\text{ m}^2=10,000\text{ cm}^2,\quad 1\text{ m}^3=1,000,000\text{ cm}^3

Convert all lengths to one unit before substitution and attach the correct power to the final unit. Perimeter is a length, not an area; surface area is not volume.

Shape	Area
rectangle	$lw$
parallelogram	$bh$
triangle	$frac12bh$
trapezium	$frac12(a+b)h$
circle	$\pi r^2$

The height is the perpendicular distance to the chosen base. A sloping side is not the height unless it is perpendicular to that base.

For a composite region, draw or imagine cuts into known shapes, calculate each part, then add included pieces or subtract holes. Check that the pieces cover the target once with no gaps or overlap.

Use radius, not diameter, in $\pi r^2$ . Keep $\pi$ exact until the final rounding step, and do not add areas expressed in different square units.

Solid	Volume
prism	cross-sectional area $imes$ length
cuboid	$lwh$
cylinder	$\pi r^2h$
pyramid	$frac13 imes$ base area $imes h$
cone	$frac13\pi r^2h$
sphere	$frac43\pi r^3$

For a cone or pyramid, $h$ is the perpendicular height from the base to the apex, not a sloping edge. For a prism, its length runs perpendicular to the constant cross-section.

Split a compound solid into standard solids or subtract a drilled-out part. A hemisphere has half the volume of a sphere; match shared radii and heights before combining expressions.

Do not confuse curved surface features with volume. Convert every length first, cube the resulting unit, and delay decimal rounding until the complete volume has been evaluated.

\text{arc length}=\frac{\theta}{360^\circ}\times2\pi r

\text{sector area}=\frac{\theta}{360^\circ}\times\pi r^2

Treat $heta/360^\circ$ as the fraction of a full circle. For a perimeter, include any straight radii as well as the arc; for a shaded region, subtract inner sectors when necessary.

The syllabus uses degrees and excludes radian measure. Arc length has linear units, sector area has square units, and the major sector uses $360^\circ- heta$ when $heta$ labels the minor angle.

8 Vectors and transformation geometry

A scalar has magnitude only. A vector has both magnitude and direction; in this syllabus vectors are two-dimensional.

Scalar	Vector
mass, time, temperature	displacement, velocity, force
distance	displacement
speed	velocity

Ask whether changing direction while keeping the numerical size would change the quantity. If yes, the quantity is vector-valued.

A negative scalar is still a scalar. A vector is not identified merely by having two numbers: the numbers must encode directed components.

\overrightarrow{OA}=\mathbf a=\begin{pmatrix}a_x\a_y\end{pmatrix}

The top component is horizontal: positive right, negative left. The bottom component is vertical: positive up, negative down.

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf b-\mathbf a

Order matters: $\overrightarrow{AB}=-\overrightarrow{BA}$ . Do not swap the two components or omit the direction arrow when naming a vector between points.

A directed line segment shows a vector by its length and arrow direction. Its starting position is irrelevant: equal vectors may be drawn in different places.

A(x_1,y_1),\ B(x_2,y_2)\Rightarrow\overrightarrow{AB}=\begin{pmatrix}x_2-x_1\y_2-y_1\end{pmatrix}

From any chosen start, move by the horizontal component, then the vertical component, and place the arrowhead at the endpoint.

A segment without an arrow has no specified direction. Parallel segments of different lengths are not equal vectors, and reversing an arrow changes the sign.

Type	Test
parallel vectors	one is a scalar multiple of the other
unit vector	magnitude is 1
position vector of $P$	vector $\overrightarrow{OP}$ from the origin

\mathbf b=k\mathbf a\Rightarrow \mathbf a\parallel\mathbf b\quad(k<0\text{ gives opposite directions})

\widehat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}\quad(\mathbf a\ne\mathbf0)

Parallel does not mean equal: the multiplier may change length or reverse direction. The zero vector has no unit direction and cannot be normalised.

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

For addition, place vectors head-to-tail; the resultant joins the first tail to the last head. Subtracting $\mathbf b$ means adding the reversed vector $-\mathbf b$ .

\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}

Only combine matching horizontal and vertical components. Vector subtraction is not commutative, and a valid path must respect each arrow's direction.

\left|\begin{pmatrix}x\y\end{pmatrix}\right|=\sqrt{x^2+y^2}

Magnitude is the non-negative length of the vector. It follows from Pythagoras applied to the horizontal and vertical components.

\left|\begin{pmatrix}-6\8\end{pmatrix}\right|=\sqrt{36+64}=10

Square each signed component before adding and give a non-negative result. $|\mathbf a|$ is a scalar, not a column vector.

k\begin{pmatrix}a\b\end{pmatrix}=\begin{pmatrix}ka\kb\end{pmatrix}

Multiplier $k$	Effect
$k>1$	same direction, longer
$0<k<1$	same direction, shorter
$k<0$	reverse direction, length scaled by $\|k\|$
$k=0$	zero vector

|k\mathbf a|=|k|,|\mathbf a|

Multiply every component. A negative multiplier reverses direction; it does not create a negative magnitude.

The resultant is the single vector with the same overall effect as all the given vectors applied in sequence.

\mathbf r=\mathbf v_1+\mathbf v_2+\cdots+\mathbf v_n

Choose a consistent positive direction, convert every vector to components, add components, then state the result as a directed vector. A closed head-to-tail path has resultant $\mathbf0$ .

Do not add magnitudes unless all vectors lie on the same line in the same direction. Opposing components must carry opposite signs.

Goal	Vector evidence
collinear points	connecting vectors are scalar multiples
parallel lines	direction vectors are scalar multiples
same midpoint	position-vector averages are equal
concurrency	independently derived position vectors give the same point

\overrightarrow{OP}=\overrightarrow{OA}+\lambda\overrightarrow{AB}

Express every route from a common origin, simplify to the same vector basis, and compare coefficients or scalar multiples. State the geometric conclusion after the algebra.

Equal-looking coefficients are meaningful only in a consistent vector basis. A scalar-multiple result proves parallelism; extra point-sharing or position information is needed to conclude collinearity.

Transformation	Complete description
reflection	mirror line
rotation	centre, angle, direction
translation	column vector
enlargement	centre and scale factor

Transform every vertex, preserve vertex order, and join the images. Use perpendicular equal distances for reflections, constant centre-distance for rotations, and centre-to-point rays for enlargements.

\begin{pmatrix}x\y\end{pmatrix}\mapsto\begin{pmatrix}x+a\y+b\end{pmatrix}

A transformation is not fully described by its type alone. For negative enlargement factors the image lies on the opposite ray from the centre; do not infer a centre or mirror line by appearance alone.

Apply transformations in the stated order and use the first image as the input to the second. The order can change the final result.

P\xmapsto{T_1}P'\xmapsto{T_2}P''

Track one labelled vertex at a time, record intermediate coordinates, then transform the remaining vertices consistently. To describe a combined result, compare the original and final shapes only after the chain is complete.

Do not collapse two transformations by guessing from the picture. Rotations, reflections and translations generally do not commute, so reversing the order may produce another image.

\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}ax+by\cx+dy\end{pmatrix}

The columns are the images of the basis vectors: transform $(1,0)^T$ to get column 1 and $(0,1)^T$ to get column 2.

M\begin{pmatrix}1\0\end{pmatrix}=\begin{pmatrix}a\c\end{pmatrix},\qquad M\begin{pmatrix}0\1\end{pmatrix}=\begin{pmatrix}b\d\end{pmatrix}

A $2 imes2$ matrix here represents only transformations that leave the origin fixed. Ordinary translations cannot be encoded by such a matrix; keep point vectors as $(x,y)^T$ .

9 Trigonometry

In a right-angled triangle, name the sides relative to the marked angle: opposite is across from it, adjacent touches it but is not the hypotenuse, and the hypotenuse is opposite the right angle.

Known or wanted sides	Ratio
opposite and hypotenuse	$\sin\theta=\frac{O}{H}$
adjacent and hypotenuse	$\cos\theta=\frac{A}{H}$
opposite and adjacent	$\tan\theta=\frac{O}{A}$

Mark the right angle and target angle, label O/A/H, choose the ratio containing the known and unknown quantities, substitute, solve, then round only at the end. Use inverse sine, cosine or tangent when the angle is unknown.

\theta=\sin^{-1}!\left(\frac{O}{H}\right),\quad\cos^{-1}!\left(\frac{A}{H}\right),\quad\tan^{-1}!\left(\frac{O}{A}\right)

Use degree mode and give angles in degrees or decimals of a degree. These right-triangle ratios choose sides relative to the target angle; do not reuse O and A labels unchanged when the target angle changes. The syllabus considers angles up to 180°.

Information pattern	Tool
opposite side-angle pair plus another side or angle	sine rule
three sides, or two sides and included angle	cosine rule
two sides and included angle, area wanted	$\tfrac12ab\sin C$

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C},\qquad c^2=a^2+b^2-2ab\cos C

\text{Area}=\frac12ab\sin C

For a 3D problem, redraw only the triangle that contains the requested length or angle. Establish its sides from earlier right-triangle, Pythagoras, sine-rule or cosine-rule calculations, keep unrounded values, then solve the final triangle.

Match each side with its opposite angle. For the sine rule, an unknown angle may have a second solution $180°- heta$ ; test it against the diagram, side ordering and angle sum. In the cosine rule, isolate the cosine before applying $\cos^{-1}$ .

Problems may be solved by calculation or accurate drawing. Latitude and longitude will not be set, nor will direct calculations of the angle between two planes or between a line and a plane. Do not treat a perspective sketch as a scale drawing.

An angle of elevation is measured upward from an observer's horizontal line of sight. An angle of depression is measured downward from that horizontal.

Draw a horizontal through the observer, a vertical height, and the line of sight. Mark the angle at the observer—not at the object—and use the right angle between horizontal and vertical to expose a right triangle.

\text{horizontal sight lines are parallel};\Rightarrow;\text{alternate elevation/depression angles are equal}

Translate the context into a right triangle, include any observer height or different ground levels, choose sine/cosine/tangent from the labelled sides, and report the requested height or distance with units.

Angles are in degrees or decimals of a degree. Depression is not measured from the vertical, and a person's eye height cannot be ignored when the question distinguishes it from ground level.

A bearing is measured clockwise from north at the starting point and written with three figures, such as $047°$ , $120°$ or $305°$ .

Draw a north line at every relevant point. Put the protractor centre at the journey's start, begin at north, turn clockwise, then draw the route. The bearing of B from A starts at A; the bearing of A from B starts at B.

\text{reverse bearing}=\begin{cases}b+180°,&b<180°\b-180°,&b\ge180°\end{cases}

Use parallel north lines to transfer angles into the route triangle, then apply angle facts, the sine rule or cosine rule as appropriate. Convert the final direction back to a clockwise angle from north and pad it to three figures.

Do not measure anticlockwise, from east, or from the arrival point unless that is the named start. A triangle's interior angle is not automatically the bearing; relate it explicitly to a north line first.