Euclidean Geometry (Grade 10) — Mathematics Grades 8–11

Prove and apply properties of parallel lines, triangles and quadrilaterals.

Gr 10 Terms 2 and 3 - Euclidean GeometryTerm 2 and Term 3 - 4 weeks (CAPS)

Practise Euclidean Geometry in the app →

What gets asked

Calculate unknown angles using parallel lines, and give a reason for each step.
Prove two triangles congruent and name which condition you used.
State the definition of a kite, parallelogram, rectangle, rhombus or square.
Use the midpoint theorem to find a length or to prove lines parallel.

You must be able to

Give a reason for every angle statement, not just the final answer.
Match corresponding vertices in the same order when naming triangles congruent or similar.
State the definition of a quadrilateral, not just one of its properties.
Explain why a single counter-example disproves a general claim.

Traps that cost marks

Using a parallel-line reason, like co-interior angles, when the lines are not marked parallel. Check the diagram or given information shows the lines are parallel before using that reason.
Using two sides and an angle that is not between them to claim two triangles are congruent. Only four conditions work: SSS, SAS, AAS or RHS. The angle in SAS must be the included one.
Saying a square is not a rectangle, or a rectangle is not a parallelogram. A square meets every rectangle property, so it counts as one. Always give the most specific name.

Worked example

Two parallel lines are cut by a transversal. One co-interior angle is

65°

. Find the other one.

Co-interior angles on parallel lines add up to $180°$ .
$180° - 65° = 115°$ .
Reason: co-interior angles, parallel lines.

Lines, angles and triangles

Every angle you find needs a reason, and the reason must be one you are allowed to use.

transversal: A line that cuts across two or more other lines.
co-interior angles: The two angles inside a pair of lines, on the same side of the transversal.
exterior angle: The angle outside a triangle, made by carrying one side past a corner.
congruent: Two shapes that would fit exactly on top of each other.
included angle: The angle sitting between two named sides.

When a transversal cuts two parallel lines, three angle pairs matter. Corresponding angles are equal. Alternate angles are equal. Co-interior angles add up to $18 0^{\circ}$ , so they are not equal.

Those rules only work on parallel lines. Look for the arrow marks, or a line in the question saying so. No marks, no parallel-line reason.

The three angles of a triangle add up to $18 0^{\circ}$ . Carry one side past a corner and you get an exterior angle. It equals the two inside angles not next to it, added together. Both, not one.

Equal sides come from the markings, never from the picture. A rough sketch can make any triangle look isosceles. Only tick marks, or a proved fact, let you call two sides equal.

Four conditions prove two triangles congruent: SSS, SAS, AAS and RHS. In SAS the angle must be the included angle. Write the corners in matching order: $△ A B C \equiv △ D E F$ says $A$ pairs with $D$ .

Rules to remember

co-interior angles on parallel lines: $x + y = 18 0^{\circ}$
angles of a triangle: $a + b + c = 18 0^{\circ}$
congruency: SSS, SAS, AAS or RHS

Examples

Two parallel lines are cut by a transversal. One co-interior angle is

6 5^{\circ}

. Find the other one.

Worked answer

The lines are parallel, so the co-interior rule is allowed.
Co-interior angles on parallel lines add up to $18 0^{\circ}$ .
$18 0^{\circ} - 6 5^{\circ} = 11 5^{\circ}$ .
Reason: co-interior angles, parallel lines.

Answer: $11 5^{\circ}$

The two angles were not equal. Co-interior is the one pair that adds instead.

△ P QR

\hat{P} = 4 8^{\circ}

and

\hat{Q} = 5 7^{\circ}

. Side

QR

is carried on to

S

. Find

\hat{P R S}

Worked answer

$\hat{P R S}$ is outside the triangle, so it is the exterior angle at $R$ .
The two inside angles not next to it are $\hat{P}$ and $\hat{Q}$ .
Add them: $4 8^{\circ} + 5 7^{\circ} = 10 5^{\circ}$ .
Check the other way: the angle inside at $R$ is $18 0^{\circ} - 10 5^{\circ} = 7 5^{\circ}$ .

Answer: $10 5^{\circ}$

Using only $\hat{P}$ would give $4 8^{\circ}$ , and the check would not close on $18 0^{\circ}$ .

△ A B C

and

△ D E F

A B = D E

B C = E F

, and

\hat{B} = \hat{E}

. Are the triangles congruent?

Worked answer

Two pairs of sides are equal, and one pair of angles.
The angle at $B$ lies between $A B$ and $B C$ , so it is included.
The same is true at $E$ , between $D E$ and $E F$ .
That is SAS, so the triangles are congruent.

Answer: yes, by SAS

The angle had to sit between the two sides. Anywhere else proves nothing.

Traps

Using a parallel-line reason, like co-interior angles, when the lines are not marked parallel. Check the diagram or given information shows the lines are parallel before using that reason.
Calling co-interior angles equal, the way corresponding and alternate angles are. On parallel lines they add up to $18 0^{\circ}$ . They are only equal when both are $9 0^{\circ}$ .
Setting the exterior angle equal to one of the two inside angles opposite it. It equals their sum. With $4 8^{\circ}$ and $5 7^{\circ}$ inside, the exterior angle is $10 5^{\circ}$ .
Using two sides and an angle that is not between them to claim two triangles are congruent. Only four conditions work: SSS, SAS, AAS or RHS. The angle in SAS must be the included one.

Learn and practise “Lines, angles and triangles” in the app →

Similarity and midpoint theorem

Same shape, different size: what stays equal, what stays in proportion, and what a midpoint hands you.

similar: Two triangles whose angles are equal and whose sides are in the same ratio.
corresponding sides: Sides in matching places in the two triangles, each facing a pair of equal angles.
midpoint: The point exactly halfway along a line, cutting it into two equal pieces.
midline: The line joining the midpoints of two sides of a triangle.
converse: A theorem read backwards. What was given and what was proved swap places.

Congruent triangles are the same size. Similar triangles need not be. One is a clean enlargement of the other: every angle unchanged, every side grown by the same number of times.

Two pairs of equal angles is enough. The third pair must then be equal too, because all three add up to $18 0^{\circ}$ . Equal angles first, ratios second, and never the other way round.

Corresponding sides are paired by the angles they face, not by how long they look. Write the triangles in matching order first. Then first letter goes with first letter.

The midpoint theorem says two things about the midline, and marks go to both. It is parallel to the third side, AND half as long. Say only the parallel part and you have said half of it.

The converse runs the other way. A line through the midpoint of one side, parallel to another side, must cut the third side at its midpoint. One midpoint alone tells you nothing.

Rules to remember

similar triangles: equal angles, and sides in the same ratio
midpoint theorem: the midline is parallel to the third side and half its length

Examples

△ A B C

and

△ D E F

have

\hat{A} = \hat{D}

and

\hat{B} = \hat{E}

A B = 6

cm,

B C = 9

cm and

D E = 4

cm. Find

E F

Worked answer

Two pairs of equal angles, so the triangles are similar.
$A B$ and $D E$ correspond. $B C$ and $E F$ correspond.
Let $E F = x$ . Then $\frac{6}{4} = \frac{9}{x}$ .
Cross-multiply: $6 x = 36$ , so $x = 6$ .

Answer: $6$ cm

The order $A B C$ to $D E F$ chose the pairs. Guessing them would have paired $6$ with $9$ .

△ P QR

S

is the midpoint of

P Q

and

T

is the midpoint of

P R

QR = 18

cm. Find

S T

Worked answer

Both $S$ and $T$ are midpoints, so $S T$ is the midline.
The midpoint theorem says $S T$ is parallel to $QR$ and half as long.
$S T = \frac{1}{2} \times 18 = 9$ .

Answer: $9$ cm

Doubling instead of halving would give $36$ cm, longer than the side it sits inside.

△ K L M

N

is the midpoint of

K L

. The line through

N

parallel to

L M

meets

K M

P

. If

K M = 14

cm, find

K P

Worked answer

Only one midpoint is given, so the theorem itself does not apply.
But $N P$ is parallel to $L M$ , so the converse does.
The converse makes $P$ the midpoint of $K M$ .
So $K P = \frac{1}{2} \times 14 = 7$ .

Answer: $7$ cm

The parallel line was the second piece of evidence. One midpoint alone proves nothing.

Traps

Writing the ratio with sides that are not corresponding, like the shortest with the middle one. Name the triangles in matching order first. Then pair the sides that face equal angles.
Calling two triangles similar because they look alike on the page. Similarity must be earned. Show two pairs of equal angles before writing any ratio.
Stating the midpoint theorem as the parallel part only. There are two conclusions: parallel to the third side, and half its length. State both.
Using the midpoint theorem when only one midpoint has been given. The theorem needs two midpoints. With one, you also need a parallel line: that is the converse.

Learn and practise “Similarity and midpoint theorem” in the app →

Special quadrilaterals

The shortest true description of each shape, and why noticing something is not proving it.

diagonal: A straight line joining two corners that are not next to each other.
kite: A quadrilateral with two pairs of equal sides, the equal ones next to each other.
trapezium: A quadrilateral with at least one pair of parallel sides, so a parallelogram is one. Exam papers often mean the shape with only one pair.
rhombus: A parallelogram with all four sides equal.
conjecture: Something you have noticed and believe, but have not yet proved.
counter-example: One case that fits the setup but breaks the claim. That alone sinks it.

A definition is the shortest list that pins a shape down. A parallelogram has both pairs of opposite sides parallel. Everything else, like its equal opposite angles, is a property you prove.

A rectangle is a parallelogram with a right angle. A square is a rhombus with one too. Equal diagonals belong to a rectangle as a result, not as its definition.

The families sit inside one another. Every parallelogram is a trapezium, and a square is a rhombus and a rectangle. Name a shape as fully as you can.

Name from the markings, not from the drawing. A parallelogram's diagonals cut each other in half, but do not meet at right angles. A rectangle's are equal, but do not halve its corners.

Measuring shapes is where a conjecture comes from, not a proof. A hundred kites that agree leave every untested kite open. One counter-example ends the claim.

Rules to remember

every square is a rhombus and a rectangle, and all three are parallelograms

Examples

A quadrilateral has both pairs of opposite sides parallel, and its diagonals meet at right angles. Name it as fully as possible.

Worked answer

Both pairs of opposite sides parallel makes it a parallelogram.
A parallelogram whose diagonals meet at right angles has all four sides equal, so it is a rhombus.
Nothing forces a right angle at a corner, so it need not be a square.

Answer: a rhombus

Rhombus is the most detailed name the given facts reach.

P QR S

is a rectangle with

P Q = 8

cm and

QR = 6

cm. Its diagonals meet at

M

. Find

QM

Worked answer

In a rectangle the diagonals are equal and cut each other in half.
$△ P QR$ has a right angle at $Q$ , so use the Theorem of Pythagoras.
$P R^{2} = 8^{2} + 6^{2} = 64 + 36 = 100$ , so $P R = 10$ cm.
The other diagonal is equal: $QS = 10$ cm.
$M$ halves it: $QM = \frac{1}{2} \times 10 = 5$ .

Answer: $5$ cm

Two rectangle properties were needed: equal diagonals, each cut in half.

A learner draws ten kites and finds that in every one the diagonals meet at right angles. Has she proved it?

Worked answer

Ten kites that agree are good evidence, and evidence is all it is.
The claim covers every kite, and most are untested.
So it is still a conjecture, not a proved result.
A proof splits any kite into congruent triangles, covering all at once.

Answer: no, it is still a conjecture

Examples never finish a general claim. One counter-example could end it.

Traps

Giving a rectangle's equal diagonals as its definition. That is a property you prove. The definition is a parallelogram with a right angle.
Saying a square is not a rectangle, or a rectangle is not a parallelogram. A square meets every rectangle property, so it counts as one. Always give the most specific name.
Claiming the diagonals of every parallelogram meet at right angles. They only do in a rhombus and a square. Otherwise they just cut each other in half.
Treating a pile of supporting examples as a proof. Examples support a conjecture. A proof must cover every case, tested or not.

Learn and practise “Special quadrilaterals” in the app →

Areas and mixed problems

Turning what you know about a shape into its area, and using only the facts you were actually given.

area: The flat space inside a shape, counted in squares. Its unit always carries a square.
perpendicular height: The distance from the base to the opposite side, measured at a right angle to the base.

A parallelogram's area is base times perpendicular height, exactly like a rectangle's. The slanted side is longer than the height, so multiplying two side lengths always gives too much.

A trapezium has a pair of parallel sides. Add them, halve the total, then multiply by the perpendicular height between them. Halving the sum is what makes the trapezium formula different.

A kite and a rhombus are handled by their diagonals, which cross at right angles. Multiply the two diagonals and halve. The half is the part learners drop most often.

In a mixed problem, use only what is marked or stated. A shape that looks like a parallelogram is not one until the question says so. Then its parallel sides let you use angle rules like co-interior angles.

Rules to remember

parallelogram: $A = bh$
trapezium: $A = \frac{1}{2} (a + b) h$
kite or rhombus: $A = \frac{1}{2} pq$ , with $p$ and $q$ the diagonals

Examples

A parallelogram has a base of

12

cm and a slanted side of

7

cm. Its perpendicular height is

5

cm. Find its area.

Worked answer

Area needs the base and the perpendicular height.
The $7$ cm side is slanted, so it is not the height.
$A = 12 \times 5 = 60$ .

Answer: $60$ cm $^{2}$

The $7$ cm was there to tempt you. It never enters the area at all.

A vegetable plot is a trapezium. Its parallel sides are

9

m and

5

m, and they are

4

m apart. Find its area.

Worked answer

Add the parallel sides: $9 + 5 = 14$ .
Halve that: $\frac{1}{2} \times 14 = 7$ .
Multiply by the perpendicular height: $7 \times 4 = 28$ .

Answer: $28$ m $^{2}$

Halving the sum averages the two parallel sides into one usable width.

A B C D

is a parallelogram and

\hat{A} = 7 0^{\circ}

. Find

\hat{B}

Worked answer

$A B C D$ is given as a parallelogram, so $A D$ is parallel to $B C$ .
$A B$ cuts across them, so $\hat{A}$ and $\hat{B}$ are co-interior angles.
$\hat{B} = 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ .

Answer: $11 0^{\circ}$

The word parallelogram supplied the parallel lines. Without it, no angle rule applies.

Traps

Using the slanted side of a parallelogram or trapezium as its height. The height meets the base at a right angle. A slanted side is longer, so the area comes out too big.
Giving a kite's area as the two diagonals multiplied together. Halve it. Diagonals of $10$ cm and $6$ cm give $30$ cm squared, not $60$ .
Reading equal sides or right angles off the picture instead of off the markings. A diagram is a sketch, not a measurement. Use only what is marked or stated in words.

Learn and practise “Areas and mixed problems” in the app →

Grade 10 Euclidean Geometry

What gets asked

You must be able to

Traps that cost marks

Worked example

Lines, angles and triangles

Rules to remember

Examples

Traps

Similarity and midpoint theorem

Rules to remember

Examples

Traps

Special quadrilaterals

Rules to remember

Examples

Traps

Areas and mixed problems

Rules to remember

Examples

Traps

About this material