Triangle: Grade 9 Math – Science Playground

How tall is the tree in your school ground? You can’t climb it with a tape, and you have no instrument for angles. All you have is a small mirror, a measuring tape and a friend.

Lay the mirror flat on the ground, 9 m from the foot of the tree. Stand close to the mirror, then walk slowly backwards, away from it, until the very top of the tree shows in it. Your friend measures two things: how far your feet are from the mirror, and how high your eyes are. From those two numbers and the 9 m, you can work out the height of the tree. Try it below: step back and forward until the top of the tree shows in the mirror.

Move the slider to walk towards or away from the mirror. The yellow line is the light from the tree to your eye; it leaves the mirror at the same angle as it arrives.

You stand from the mirror, d: 3.00 m

Your eyes: 1.3 m up
In the mirror you see: 3.9 m up

In the mirror you see the tree 3.9 m up, not its top. Step closer to the mirror.

You see only sky above the tree. Step back from the mirror.

You see the very top. The two green triangles have the same angles, so they are the same shape: the tree is to 9 m as your eyes’ 1.3 m are to your 3.00 m. Height = 9 × 1.3 ÷ 3.00 ≈ 3.9 m.

Tree: W. Heubach (d. 1923) in O. Schmeil, Lehrbuch der Botanik (1911), No restrictions, via Wikimedia Commons

Why does it work? Because two triangles, yours and the tree’s, have the same angles, and triangles with the same angles are the same shape, only different in size. By Back to the tree you will be able to prove it and find the height in one line. On the way, you will meet the rules that every triangle obeys: what its outside angles are, why two equal sides mean two equal angles, and why one side can never be longer than the other two together.

For the exam

The question: Grade 9’s final exam follows CDC’s specification grid for Compulsory Mathematics, Grades 9–10, where Geometry is 13 marks in three questions. In the grid’s model paper, question 11 is this unit: a. verify experimentally, with two triangles of different sizes, that an exterior angle equals the sum of the two opposite interior angles (3 marks, application); b. in an equilateral triangle ABC, a point D is taken so that CD = CA, and A is joined to D, making triangle ACD (drawn with BC produced to D): find the ratio ∠ABC:∠CAD (2 marks, higher ability).
What it needs:

Exterior angle = sum of the two opposite interior angles; the three angles of a triangle add up to 180°.
Equal sides face equal angles (isosceles triangle); in an equilateral triangle every angle is 60°.
The bisector of the vertical angle of an isosceles triangle is perpendicular to the base and bisects it.
The longest side faces the largest angle; any two sides together are longer than the third.
A line parallel to one side cuts the other two in the same ratio; similar triangles have their sides in the same ratio.
An experimental verification: two figures, a table of measurements, a conclusion.

How the exam asks it has a full question marked step by step; Exam practice has every exercise question to mark yourself.

13.0 Review

Draw two triangles on a sheet of paper. Measure all their sides and angles and put the measurements in a table. Discuss with your partner:

a.What type is each triangle, by its sides and by its angles?
b.How can you tell whether two triangles are congruent, similar, or neither?

Present your pair’s work on newsprint, with the figures, in your class.

Figure 13.1 Triangles by their sides: equilateral (all three sides equal), isosceles (two equal), scalene (none equal). Sides marked with the same number of small strokes are equal.

Figure 13.2 Triangles by their angles: acute-angled (every angle less than 90), right-angled (one angle 90, the small square), obtuse-angled (one angle more than 90, here 115).

Every triangle has a name of each kind: an equilateral triangle is always acute-angled, while an isosceles triangle can be acute, right or obtuse. You also know from Grade 8 that the three angles of a triangle add up to 180°. That one fact is behind almost everything in this unit.

A triangle has sides 5 cm, 5 cm and 8 cm, and its largest angle is 106°. What is it?

13.1 Properties of triangles and their verification

Work in groups. Each group draws a pair of triangles of each kind: equilateral, isosceles, right-angled and scalene. You will use them to test the facts below.

When one side of a triangle is produced (made longer past a corner), it makes an angle outside the triangle with the next side. This is an exterior angle. The interior angle beside it, at the same corner, is its adjacent interior angle; the other two interior angles are the opposite (or non-adjacent) interior angles. An exterior angle and its adjacent interior angle lie on a straight line, so together they make 180°.

Activity

1: verify the exterior angle property

Verify experimentally that if any side of a triangle is produced, the exterior angle so formed is equal to the sum of the two opposite interior angles.

I. Paper cutting. Draw a triangle ABC on paper and produce BC to D. Shade the corners at A and B, cut them off, and lay the two pieces side by side inside the exterior angle ACD, their points at C. Do they fill it? Try it with your own triangle, or in the drawing below.

Drag the yellow corner A to change the triangle; press the button to lay the corners A and B inside the exterior angle.

∠A: 68°
∠B: 62°
∠A + ∠B: 130°
Exterior ∠ACD: 130°

Drag A anywhere. Whatever the triangle, the exterior angle ACD is 130°, and ∠A + ∠B is 68° + 62° = 130°. Tear off the corners to see why.

The blue corner fits against CD and the red one against CA, with nothing over and no gap: together they make the exterior angle exactly. The dashed line CE is parallel to BA.

II. Measuring angles. Draw the two figures below in your copy and measure the angles with a protractor. Put them in a table like this one.

Figure 13.3 Figure I: BC produced to D, making the exterior angle ACD.

Figure 13.4 Figure II: BA produced to D, making the exterior angle CAD.

Figure	Opposite interior angles	Exterior angle	Sum of opposite interior angles	Result
I	∠BAC = ?, ∠ABC = ?	∠ACD = ?	?	?
II	∠ABC = ?, ∠ACB = ?	∠CAD = ?	?	?

What you should find

An experiment shows the rule holds for the triangles we drew. Here is why it holds for every triangle. In Tear off the corners, the dashed line CE, parallel to BA, splits the exterior angle into two parts: one equal to ∠B (corresponding angles) and one equal to ∠A (alternate angles). Or, shorter, from the angle sum:

	Statements	Reasons
1.	$\angle ACB + \angle ACD = 180^\circ$	BCD is a straight line (linear pair)
2.	$\angle BAC + \angle ABC + \angle ACB = 180^\circ$	the angles of a triangle add up to 180°
3.	∠ACD = ∠BAC + ∠ABC	from 1 and 2, take ∠ACB from both

What it means: you never need to find the third angle first. The outside angle at one corner is just the other two inside angles added.

In triangle PQR, $\angle P = 48^\circ$ and $\angle Q = 75^\circ$ . QR is produced to S. What is the exterior angle ∠PRS?

Exercise 13.1

Question 1

Write whether each statement is true or false, with a reason.

There are two right angles in a triangle.

Answer

All the angles of a triangle are acute.

Answer

If an exterior angle of a triangle is 130°, its adjacent interior angle is also obtuse.

Answer

If an interior angle is acute, its adjacent exterior angle is obtuse.

Answer

Question 2

Find the values of x and y.

Figure 13.8

Answer

Figure 13.9

Answer

Figure 13.10

Answer

Figure 13.11

Answer

Question 3

Find the values of x, y, z and a.

Figure 13.12

Answer

Figure 13.13

Answer

Figure 13.14

Answer

Figure 13.15

Answer

Figure 13.16

Answer

Figure 13.17

Answer

Question 4

The exterior angle of a triangle is 120° and the two opposite interior angles are in the ratio 1 : 2. Find all the interior angles of the triangle.

Answer

Question 5

In the figure, ∠BAC:∠ABC:∠ACB = 2:1:1 and ∠ABC = ∠ECD. Find ∠ACE.

Figure 13.18

Answer

13.2 Verification of properties of isosceles triangle

Work in groups. Each group draws a pair of isosceles triangles, measures their sides and angles with a ruler and protractor, and looks for a relation between the equal sides and the angles. Write your conclusion and present it in class.

In an isosceles triangle ABC with AB = AC, the angle between the two equal sides, ∠A, is the vertical angle, the third side BC is the base, and ∠B and ∠C are the base angles.

Activity

2: the bisector of the vertical angle

(a) Verify experimentally that the bisector of the vertical angle of an isosceles triangle is perpendicular to the base and bisects it.

I. Folding paper. Take a rectangular sheet of paper. Fold one corner over so that a short side lies along a long side, and cut off the extra strip: you have a square. Fold the square along a diagonal and cut along the fold: each half is an isosceles triangle. Name it ABC, with A at the right angle. Now fold it so that AB lies exactly on AC, and open it. The crease meets BC at D. Compare a. BD and CD; b. ∠BDA and ∠CDA.

II. Measuring sides and angles. Draw two isosceles triangles ABC of different sizes. Bisect the angle at A with a compass or protractor; the bisector meets BC at D. Measure ∠ADB, ∠ADC, BD and DC, and tabulate them.

Figure 13.19 Two isosceles triangles with AB = AC, and the bisector AD of the vertical angle.

Figure	∠ADB	∠ADC	BD	DC	Result
I	?	?	?	?	?
II	?	?	?	?	?

What you should find

Activity

3: the perpendicular bisector of the base

(b) Verify experimentally that the perpendicular bisector of the base of an isosceles triangle bisects the vertical angle.

Draw two isosceles triangles ABC of different sizes. With a compass, draw the perpendicular bisector of the base BC; it cuts BC at D and passes through A. Measure ∠BAD and ∠CAD, and tabulate them.

Figure 13.20 Two isosceles triangles with the perpendicular bisector of the base drawn from D, the middle of BC.

Figure	∠BAD	∠CAD	Result
I	?	?	?
II	?	?	?

What you should find

The fold explains everything at once. An isosceles triangle is symmetric about the line AD: fold it there and one half lies exactly on the other. So, together:

the base angles are equal: ∠B lands on ∠C (and if two angles of a triangle are equal, the sides facing them are equal: the converse);
the bisector of the vertical angle, the perpendicular bisector of the base, and the line from A to the middle of BC are one and the same line.

The same thing, as a proof:

	Statements	Reasons
1.	In △ABD and △ACD: i. AB = AC (S) ii. ∠BAD = ∠CAD (A) iii. AD = AD (S)	i.given ii. AD bisects ∠A iii. common side
2.	$\triangle ABD \cong \triangle ACD$	by S.A.S.
3.	BD = DC, ∠ABD = ∠ACD, ∠ADB = ∠ADC	corresponding parts of congruent triangles
4.	$\angle ADB = \angle ADC = 90^\circ$	from 3, and together they make 180°

Proved: the base angles are equal, and AD is perpendicular to BC and bisects it.

In triangle XYZ, XY = XZ and $\angle Y = 52^\circ$ . What is ∠X?

Exercise 13.2

Question 1

Find the value of x.

Figure 13.24

Answer

Figure 13.25

Answer

Figure 13.26

Answer

Question 2

Find the values of x, y and z.

Figure 13.27

Answer

Figure 13.28

Answer

Question 3

Find the values of x and y.

Figure 13.29

Answer

Figure 13.30

Answer

Figure 13.31

Answer

Figure 13.32

Answer

Figure 13.33

Answer

Figure 13.34

Answer

Question 4

In the figure, XY = 3y, XZ = 7x, XP = 9x and XQ = 13 + 2y. Find the values of x and y.
The textbook draws P and Q between Y and Z. With its numbers, XP = 27 comes out longer than XY = 21, but no point between Y and Z is farther from X than Y and Z are, so the book’s figure doesn’t fit its numbers. With P and Q on YZ produced, as drawn here, they fit.

Figure 13.35

Answer

Question 5

Triangles ABC and BCD are isosceles. Find ∠BDE.

Figure 13.36

Answer

Question 6

Two wires of equal length are tied from the top of a pole to the ground. Each wire is 13 m long and the pole is 12 m high.

a.Find the distance on the ground between the foot of the pole and each wire.
b.Are the areas of △ABC and △ABD equal? Justify.

Figure 13.37

Answer

13.3 Relation among sides and angles of triangle

Activity

4: three sticks

In groups, collect sticks of different lengths. Each group picks three and tries to join their ends into a triangle. Does every set of three make a triangle? If it does, what type is it? If not, why not? Try it with the sliders below as well.

Set the three sticks’ lengths. Each side has the same color as the angle facing it.

Stick BC: 9 cm Stick CA: 7 cm Stick AB: 5 cm

Two shorter sticks together: 7 + 5 = 12 cm > 9 cm
Largest angle: ∠A = 96°, facing BC

All three sticks are the same length: an equilateral triangle, with every angle 60°.

The two shorter sticks together are longer than the longest, so they meet: a triangle. The longest side, BC, faces the largest angle (same color).

The two shorter sticks together are shorter than BC: even swung down flat they can’t reach each other. No triangle.

The two shorter sticks together are exactly as long as BC: they meet only lying flat along it, a straight line, not a triangle.

What you should find

a. The sum of any two sides and the third side

Activity

5: two sides against the third

Experimentally verify that the sum of any two sides of a triangle is greater than the third side.
a. Work in groups. b. Each student draws a triangle of a different shape and size. c. Measure all the sides and tabulate the sums of each two, as below. d. Discuss and write your conclusion.

Figure 13.38 Figure I and Figure II, to measure.

Figure	AB	BC	CA	AB + BC	BC + CA	CA + AB	Result
I	?	?	?	?	?	?	?
II	?	?	?	?	?	?	?

What you should find

If the sum of two sides is equal to or less than the third side, there is no triangle: as the sticks showed, the two shorter sides lie flat along the longest one, or can’t reach each other at all. This is called the triangle inequality. It is also why a straight road is the shortest way: going from B to C by way of any point A off the road is always longer than going straight.

Which three lengths can be the sides of a triangle?

b. The sides and the angles facing them

Activity

6: the longer side and the larger angle

Experimentally verify that the angle opposite the longer side is greater than the angle opposite the shorter side.

I. Using Meccano strips. Each group makes a triangle ABC from Meccano strips (metal strips with holes, bolted together; strips of cardboard pinned at the ends work too), with AB and AC fixed and C free to move. Move C down as in Figure II, then up as in Figure III. Each time, measure ∠CAB and the side BC, and tabulate them.

Figure 13.39 Figures I, II and III: the strips AB and AC stay the same length; only the angle at A and the side BC change.

II. Measuring sides and angles. Draw three triangles: acute-angled, right-angled and obtuse-angled. Measure every angle and the side opposite it, and tabulate them.

Figure 13.40 Figure I acute-angled, Figure II right-angled at C, Figure III obtuse-angled at C.

What you should find

What it means: in a triangle, sides and angles come in the same order. Know the order of the angles and you know the order of the sides, without measuring. In a right-angled triangle the right angle is the largest, so the side facing it, the hypotenuse, is always the longest side; in an obtuse-angled triangle the side facing the obtuse angle is the longest.

The Eiffel Tower seen from just below its first floor, rising against a blue evening sky: its iron lattice is crossed again and again by diagonal girders into small triangles, up to the top — **Figure 13.42** The Eiffel Tower, Paris (1889): its iron lattice is braced into triangles all the way up, so it keeps its shape in the wind. Photo: Arnaud Ligny from Paris, France, CC BY-SA 2.0

Exercise 13.3

Question 1

Measure all the sides of triangle ABC, and find its largest and smallest angles from the lengths of the sides.

Figure 13.43

Answer

Question 2

Measure all the angles of triangle ABC, and find its longest and shortest sides from the sizes of the angles.

Figure 13.44

Answer

Question 3

The sides of a triangle are 5 cm, 6 cm and x cm, and x is the longest side. Which of these is correct?

Answer

Question 4

The base of an isosceles triangle is 24 cm and its area is 60 cm². Find the length of each of its other sides.

Answer

Question 5

In the figure, if ∠ABC > ∠DBC, prove that AC > BC.

Figure 13.46

Answer

Question 6

Prove that PQ + QR + RS > SP. (Hint: join P and R.)

Figure 13.47

Answer

13.4 Similar triangles

Figures with the same shape but not necessarily the same size are called similar figures. A photo and its enlargement, a map and the land, a model bus and the bus are all similar.

If the corresponding angles of two triangles are equal, the triangles are similar. We write $\triangle ABC \sim \triangle DEF$ ( $\sim$ is read “is similar to”), listing the corners in matching order. In the figure, the arcs show which angles are equal.

Figure 13.51

\triangle ABC \sim \triangle DEF

: ∠A = ∠D (one arc), ∠B = ∠E (two arcs), ∠C = ∠F (three arcs).

The sides facing equal angles are the corresponding sides: BC and EF (facing ∠A and ∠D), CA and FD, AB and DE.

Since the angles of a triangle add up to 180°, two pairs of equal angles are enough: the third pair must then be equal too.

So for $\triangle ABC \sim \triangle AEF$ :
$\frac{AE}{AB} = \frac{EF}{BC} = \frac{AF}{AC}, \quad\text{or}\quad \frac{AB}{AE} = \frac{BC}{EF} = \frac{AC}{AF}$
Write each ratio with the sides of the same triangle on the same level, top or bottom, and match each side with the side facing the equal angle.

Theorem 1

A line drawn parallel to one side of a triangle cuts the other two sides in the same ratio.

Figure 13.53

XY \parallel BC

, cutting AB at E and AC at F.

Given: in △ABC, the line XY is parallel to BC and meets AB at E and AC at F.

To prove: $\dfrac{AE}{EB} = \dfrac{AF}{FC}$ .

	Statements	Reasons
1.	In △ABC and △AEF: i. ∠ABC = ∠AEF (A) ii. ∠ACB = ∠AFE (A) iii. ∠BAC = ∠EAF (A)	i.corresponding angles, $XY \parallel BC$ ii. corresponding angles, $XY \parallel BC$ iii. common angle
2.	$\triangle ABC \sim \triangle AEF$	all the corresponding angles are equal
3.	$\dfrac{AB}{AE} = \dfrac{AC}{AF}$	corresponding sides of similar triangles are in the same ratio
4.	$\dfrac{AE + EB}{AE} = \dfrac{AF + FC}{AF}$	AB = AE + EB, AC = AF + FC
5.	$1 + \dfrac{EB}{AE} = 1 + \dfrac{FC}{AF}$	dividing each top by its bottom
6.	$\dfrac{EB}{AE} = \dfrac{FC}{AF}$	taking 1 from both sides
7.	$\dfrac{AE}{EB} = \dfrac{AF}{FC}$	turning both fractions upside down

Proved.

This theorem is how you divide a line segment into any ratio with only a ruler, compass and set square: draw a slanting line from one end, mark equal steps along it, and draw parallels. Practice it in Construct it.

In △ABC, $DE \parallel BC$ with D on AB and E on AC. AD = 4 cm, DB = 6 cm and AE = 5 cm. What is EC?

Back to the tree

Now look again at the mirror. Call the foot of the tree T and its top P, the point of the mirror M, your feet F and your eyes E.

The tree and you both stand upright: $\angle PTM = \angle EFM = 90^\circ$ .
Light leaves a mirror at the same angle as it arrives: ∠PMT = ∠EMF.

Two pairs of equal angles, so $\triangle PTM \sim \triangle EFM$ , and their sides are in the same ratio:
$\frac{PT}{EF} = \frac{TM}{FM} \quad\Rightarrow\quad PT = \frac{EF \times TM}{FM}$
In the widget, TM = 9 m, your eyes are EF = 1.3 m up, and you see the top when you stand FM = 1.56 m from the mirror:
$PT = \frac{1.3 \times 9}{1.56} = 7.5 \text{ m}$
The tree is 7.5 m tall. What it means: you measured three short lengths on the ground, and the angles, which you never measured, did the rest. Surveyors, mapmakers and astronomers have measured heights and distances this way for thousands of years.

Construct it: Dividing a line segment

How the exam asks it

Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Geometry is 13 marks in three questions. In the grid’s model paper, this unit is question 11:

Part	Level	Marks	What it asks
a.	application	3	verify a property experimentally with two triangles of different sizes
b.	higher ability	2	a ratio of two angles in a figure with an isosceles or equilateral triangle

An experimental verification has three parts to show: two figures of different sizes, a table of measurements, and a conclusion in words. Part b needs two steps: find each angle with a reason, then write the ratio in its lowest terms. Here is one of our own in that shape.

Exercise 13.4

Question 1

In triangle ABC, $DE \parallel BC$ , AD:DB = 2:3 and EC = 12 cm. Find AE.

Figure 13.58

Answer

Question 2

In triangle XYZ, $YZ \parallel PQ$ , XQ:QZ = 3:5 and PY = 7.5 cm. Find XY.

Figure 13.59

Answer

Question 3

Find the value of x.

Figure 13.60 AB = x, CE = 1.8 cm, BC = 24 cm and CD = 12 cm, drawn to scale.

Answer

Question 4

Two ladders lean at the same angle, as in the figure. The longer ladder is 60 ft long and reaches 45 ft up its wall; the shorter one reaches 18 ft up. Find the length of the shorter ladder.

Figure 13.61 The arcs at the foot of the ladders mark the equal angles they make with the ground.

Answer

Question 5

Find the height of the tree from the figure.

Tree: W. Heubach (d. 1923) in O. Schmeil, Lehrbuch der Botanik (1911), No restrictions, via Wikimedia Commons

Figure 13.62

Answer

A pole 20 m high casts a shadow 30 m long. How high is another pole whose shadow, at the same time, is 7.5 m long?

Answer

Question 6

Mingma puts a mirror flat on the ground between himself and a tree in the school ground. He moves back and forth until he sees the top of the tree in the mirror. How can you find the height of the tree? Discuss with your friends.

Answer

Quiz yourself on this unit →

Practice every question in the book, and keep track of the ones you miss →

Where these facts come from

The mirror and the tree (opening, MirrorTree, Back to the tree, Exercise 13.4 Q6): the law of reflection, the angle of reflection equals the angle of incidence (any physics textbook; Grade 9 Science). The numbers are our own example.
Euclid’s Elements, Book I: Propositions 5 (base angles of an isosceles triangle), 19 (greater angle, greater side), 20 (triangle inequality) and 32 (exterior angle, angle sum); D. E. Joyce’s online edition (Clark University), read November 2026. Wikipedia, “Pons asinorum” (the name and its medieval use), read November 2026.
Proclus and the Epicureans’ donkey: Wikipedia, “Triangle inequality” (history), and Proclus, A Commentary on the First Book of Euclid’s Elements, on Proposition 20.
Trusses and the rigidity of triangles: Wikipedia, “Truss”, read November 2026. The Eiffel Tower, completed 1889: Wikipedia, “Eiffel Tower”.
Thales and the pyramid: Wikipedia, “Thales of Miletus” (Diogenes Laertius and Plutarch’s accounts of measuring the pyramids by their shadows), read November 2026. The Nyatapola temple, Bhaktapur, about 30 m: Wikipedia, “Nyatapola”.
The exam’s question (For the exam, How the exam asks it): CDC, Specification Grid, Compulsory Mathematics, Grades 9–10 (Geometry 13 marks in three questions) and its model question 11 (3A, 2HA); our question is our own.
Every answer: worked out here; every figure drawn from its stated or computed measures (triangle-figures.ts).

A companion to your textbook, not a replacement for it: study from the book your school uses, and where the two differ, follow your teacher. Found a mistake? Please tell us.