Chapter – 6

Triangles

In this post we have given the detailed notes of class 10 Maths Chapter 6 (Triangles) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 6: Triangles

What is a Triangle?

A plane figure bounded by three line segments is called a triangle. A triangle is denoted by the symbol Δ. Every triangle has three sides, three angles, and three vertices. The sum of all three angles of a triangle is always 180°.

Classification of Triangles

Triangles can be classified on two bases:

• (i) On the basis of sides
• (ii) On the basis of angles

Triangles Based on Sides

On the basis of sides, triangles are of three types:

• Scalene triangle — all three sides are of different lengths.
• Isosceles triangle — any two sides are equal in length.
• Equilateral triangle — all three sides are equal in length (and all angles are 60° each).

Triangles Based on Angles

On the basis of angles, triangles are of three types:

• Acute-angled triangle — all three angles are acute (less than 90°).
• Obtuse-angled triangle — one angle is obtuse (greater than 90°).
• Right-angled triangle — one angle is exactly 90° (a right angle).

Congruent and Similar Figures

Congruent Triangles

When all the sides and all the angles of two triangles are equal (i.e. the two triangles are of the exact same shape and size), the triangles are said to be congruent.

Similar Triangles

Two figures having the same shape but not necessarily the same size are called similar figures.

All congruent figures are similar, but all similar figures need not be congruent.

Two triangles are similar if:

• (i) their corresponding angles are equal, and
• (ii) their corresponding sides are in the same ratio (proportional).

Criteria for Similarity of Triangles

1. AAA (Angle-Angle-Angle) Similarity Criterion

If in two triangles, corresponding angles are equal, then the triangles are similar. As a corollary, if two angles of one triangle are respectively equal to two angles of another triangle, the two triangles are similar (this is called the AA similarity criterion, since the third angle is automatically equal by the angle-sum property).

2. SSS (Side-Side-Side) Similarity Criterion

If in two triangles, the corresponding sides are in the same ratio (proportional), then their corresponding angles are equal, and hence the two triangles are similar.

3. SAS (Side-Angle-Side) Similarity Criterion

If one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are in the same ratio (proportional), then the two triangles are similar.

Basic Proportionality Theorem (Thales’ Theorem)

The Basic Proportionality Theorem, also known as Thales’ Theorem, states:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

That is, in a triangle ABC, if DE ∥ BC and DE intersects AB at D and AC at E, then:

AD/DB = AE/EC

Converse of the Basic Proportionality Theorem

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Worked Example — Finding EC and AD

Question: In a triangle, DE ∥ BC. In figure (i), AD = 1.5 cm, DB = 3 cm, AE = 1 cm; find EC. In figure (ii), DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm; find AD.

Solution (i): By the Basic Proportionality Theorem:

AD/DB = AE/EC
⇒ 1.5/3 = 1/EC
⇒ EC = (1 × 3)/1.5 = 2 cm

Solution (ii): By the Basic Proportionality Theorem:

AD/DB = AE/EC
⇒ AD/7.2 = 1.8/5.4
⇒ AD = (1.8 × 7.2)/5.4 = 2.4 cm

So, EC = 2 cm in figure (i), and AD = 2.4 cm in figure (ii).

Worked Example — Checking whether EF ∥ QR

Question: E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF ∥ QR:

• (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
• (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
• (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

Solution: By the converse of the Basic Proportionality Theorem, EF ∥ QR if PE/EQ = PF/FR.

(i) PE/EQ = 3.9/3 = 1.3, and PF/FR = 3.6/2.4 = 1.5. Since 1.3 ≠ 1.5, PE/EQ ≠ PF/FR, so EF is not parallel to QR.

(ii) PE/QE = 4/4.5 = 8/9, and PF/RF = 8/9. Since PE/QE = PF/RF, EF is parallel to QR.

(iii) EQ = PQ − PE = 1.28 − 0.18 = 1.10 cm, and FR = PR − PF = 2.56 − 0.36 = 2.20 cm.
PE/EQ = 0.18/1.10 ≈ 0.164, and PF/FR = 0.36/2.20 ≈ 0.164. Since PE/EQ = PF/FR, EF is parallel to QR.

Worked Example — Proving a Ratio Using BPT

Question: In the figure, if LM ∥ CB and LN ∥ CD, prove that AM/AB = AN/AD.

Solution: Given LM ∥ CB.

So in triangle ABC, applying the Basic Proportionality Theorem:
AM/AB = AL/AC … (i)

Given LN ∥ CD.

So in triangle ACD, applying the Basic Proportionality Theorem:
AN/AD = AL/AC … (ii)

From equations (i) and (ii):
AM/AB = AN/AD (Proved)

Worked Example — Proving Two Ratios Equal

Question: In a triangle ABC, DE ∥ AC and DF ∥ AE, with D on AB, E on BC and F on BE. Prove that BF/FE = BE/EC.

Solution: Given DE ∥ AC.

So in triangle ABC, by the Basic Proportionality Theorem:
BD/DA = BE/EC … (i)

Given DF ∥ AE.

So in triangle ABE, by the Basic Proportionality Theorem:
BD/DA = BF/FE … (ii)

From equations (i) and (ii):
BF/FE = BE/EC (Proved)

Worked Example — Proving EF ∥ QR

Question: In the figure, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR.

Solution: Given DE ∥ OQ.

So in triangle POQ, by the Basic Proportionality Theorem:
PE/EQ = PD/DO … (i)

Given DF ∥ OR.

So in triangle POR, by the Basic Proportionality Theorem:
PF/FR = PD/DO … (ii)

From (i) and (ii): PE/EQ = PF/FR.

Hence, by the converse of the Basic Proportionality Theorem applied in triangle PQR, EF ∥ QR (Proved).

Worked Example — Proving BC ∥ QR

Question: A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

Solution: Given AB ∥ PQ.

So in triangle OPQ, by the Basic Proportionality Theorem:
OA/AP = OB/BQ … (i)

Given AC ∥ PR.

So in triangle OPR, by the Basic Proportionality Theorem:
OA/AP = OC/CR … (ii)

From (i) and (ii): OB/BQ = OC/CR.

Hence, by the converse of the Basic Proportionality Theorem applied in triangle OQR, BC ∥ QR (Proved).

Worked Example — Midpoint Theorem Using BPT

Question: Using the Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (You have already proved this in Class IX.)

Solution: Let PQR be a triangle, and let E be the mid-point of side PQ, so PE = EQ. Let a line EF be drawn parallel to side QR, meeting PR at F.

Since EF ∥ QR, by the Basic Proportionality Theorem in triangle PQR:
PE/EQ = PF/FR

Since PE = EQ, we get PE/EQ = 1, so PF/FR = 1, which gives PF = FR.

Hence, F is the mid-point of PR, i.e. the line EF bisects the third side PR. (Proved)

Worked Example — Identifying Similar Triangle Pairs

Question: State which pairs of triangles are similar in the given figures. Write the similarity criterion used, and express the similarity in symbolic form.

Solution (representative cases from the textbook figures):

Where all three pairs of corresponding sides are proportional, the triangles are similar by the SSS similarity criterion.

Where one angle of a triangle equals one angle of another and the sides including those angles are proportional, the triangles are similar by the SAS similarity criterion.

Where all three pairs of corresponding angles are equal, the triangles are similar by the AAA similarity criterion.

Where the given sides are neither equal nor in the same ratio, and no equal included angle is given, the triangles are not similar.

Worked Example — Similarity with Diagonals of a Trapezium

Question: In a trapezium ABCD, AB ∥ DC, and its diagonals intersect each other at point O. Show that AO/BO = CO/DO.

Solution: Let ABCD be the given trapezium with AB ∥ DC, and let the diagonals AC and BD intersect at O.

In triangles AOB and COD:

∠AOB = ∠COD (vertically opposite angles)

∠OAB = ∠OCD (alternate interior angles, since AB ∥ CD)

So by the AAA (AA) similarity criterion, ΔAOB ~ ΔCOD.

Since corresponding sides of similar triangles are proportional:

AO/CO = BO/DO, i.e. AO/BO = CO/DO (Proved)

Areas of Similar Triangles

Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If ΔABC ~ ΔDEF, then:

ar(ABC)/ar(DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)²

Worked Example — Finding a Side Using Areas

Question: Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Solution: Since ΔABC ~ ΔDEF, we know:

ar(ABC)/ar(DEF) = (BC/EF)²

64/121 = (BC/15.4)²

Taking the square root of both sides:

8/11 = BC/15.4

BC = (8 × 15.4)/11 = 11.2 cm

Worked Example — Similar and Equal Area Implies Congruent

Question: If the areas of two similar triangles are equal, prove that they are congruent.

Solution: Let ΔABC ~ ΔPQR.

We know that ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²

Given: ar(ABC) = ar(PQR), so ar(ABC)/ar(PQR) = 1.

Therefore, (AB/PQ)² = (BC/QR)² = (CA/RP)² = 1

This gives AB = PQ, BC = QR, and CA = RP.

So by the SSS (Side-Side-Side) congruence criterion, ΔABC ≅ ΔPQR.

Hence, two similar triangles with equal areas are congruent. (Proved)

Pythagoras Theorem

Statement: In a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

If ABC is a right triangle, right-angled at B, then:

AC² = AB² + BC²

The Pythagoras Theorem can be proved using the concept of similar triangles: if a perpendicular is drawn from the vertex of the right angle to the hypotenuse, the two smaller triangles formed are each similar to the original triangle and to each other. Using the property that the ratio of areas equals the square of the ratio of corresponding sides in these similar triangles leads to the result AC² = AB² + BC².

Converse of the Pythagoras Theorem

Statement: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

That is, in a triangle ABC, if AC² = AB² + BC², then ∠B = 90°.

Worked Example — Applying Pythagoras Theorem

Question: A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution: Let the ladder be AC, the wall be AB (height 8 m), and the base BC be the unknown distance.

Since the wall is vertical, ∠B = 90°, so by the Pythagoras Theorem:

AC² = AB² + BC²
⇒ 10² = 8² + BC²
⇒ 100 = 64 + BC²
⇒ BC² = 36
⇒ BC = 6 m

So the foot of the ladder is 6 m away from the base of the wall.

Key Points to Remember

• Two figures with the same shape (but not necessarily the same size) are called similar figures. All congruent figures are similar, but similar figures need not be congruent.
• Basic Proportionality Theorem (Thales’ Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, the other two sides are divided in the same ratio: AD/DB = AE/EC.
• Converse of BPT: If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
• AAA/AA Similarity: If corresponding angles of two triangles are equal, the triangles are similar.
• SSS Similarity: If corresponding sides of two triangles are proportional, the triangles are similar.
• SAS Similarity: If one angle of a triangle equals one angle of another, and the sides including these angles are proportional, the triangles are similar.
• Areas of Similar Triangles: ar(ABC)/ar(DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)².
• Similar triangles with equal areas are congruent (proved using the SSS congruence criterion).
• Pythagoras Theorem: In a right triangle, (hypotenuse)² = (base)² + (perpendicular)², i.e. AC² = AB² + BC² when ∠B = 90°.
• Converse of Pythagoras Theorem: If AC² = AB² + BC² in a triangle ABC, then ∠B = 90°.

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