Triangles: Class 10 Maths · Chapter 6

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

75 min
1 simulation
38 practice questions
14 MCQs

What you will learn in Triangles

Tell similar figures from congruent ones, and test two polygons for similarity using matching angles and matching side ratios.
Use a scale factor to find lengths in enlargements, reductions, maps and plans.
State and prove the Basic Proportionality Theorem and its converse, and use them to find lengths and to show that lines are parallel.
Write similarity statements with the vertices in matching order.
Prove that two triangles are similar using the AA, SSS or SAS criterion.
Use similar triangles to find unknown sides, angles and heights, including shadow problems and ratios of altitudes, medians and angle bisectors.

Original companion text aligned to NCERT Mathematics Class 10, Chapter 6, pp. 73–98. Not affiliated with or endorsed by NCERT.

6.1 Introduction

Pinch a photo on your phone to zoom in. The picture grows, yet nothing in it gets squashed: faces, windows and kites keep their shape. In Class 9 you studied congruent figures, which match in shape and size. This chapter is about figures that match in shape alone. They are called similar figures (समरूप आकृतियाँ).

Similarity is also a measuring tool. Nobody climbs the school flagpole with a measuring tape. Instead, on a sunny morning, compare the flagpole with a metre stick: both cast shadows at the same moment, and the shadows reveal the height (Fig. 6.1).

3/3The two right triangles have the same angles, so height ÷ shadow is the same for both: $h = 7.5 m$ .

Fig. 6.1Parallel sun rays make two right triangles of the same shape.

A flat ground line with a metre stick on the left and a tall flagpole with a tricolour flag. In the first step only the shadows are shown: 1.6 m for the stick and 12 m for the flagpole. In the second step dashed sun rays are drawn from the top of each object to the tip of its shadow; the two rays are parallel. In the third step the two right triangles formed by each object, its shadow and the ray are shaded, and the flagpole height is given as h = 7.5 m.

In each triangle, height ÷ shadow is the same number. For the stick it is $\frac{1}{1.6}$ , so the flagpole is $\frac{12}{1.6} = 7.5 m$ tall. Surveyors find the heights of hills and the widths of rivers with the same idea, and you will meet it again in trigonometry (Chapters 8 and 9). In this chapter you will learn exactly when two triangles are similar and how to prove it.

6.2 Similar Figures

Some families of shapes are always similar. Any two circles are similar: a ₹1 coin and a dinner plate differ only in size. Any two squares are similar, and so are any two equilateral triangles. Each of these families has just one shape; only the size can change.

Congruent figures match in shape and in size, so every pair of congruent figures is also similar. The reverse fails: a 2 cm square and a 5 cm square are similar but not congruent. Some pairs are never similar, whatever their sizes. A circle and a square, for example: no enlargement turns a smooth curve into four corners.

2/3Every length is multiplied by 1.6 and the shape is kept: the copy is similar but not congruent.

Fig. 6.2Same size, even enlargement, or stretching in one direction only.NCERT Fig. 6.1 · p. 74

A small house shape (a square with a triangular roof, a door and a window) stands on the left, labelled original. Beside it is a second house. In the first step the second house is an exact copy of the same size, labelled congruent. In the second step it grows evenly so that every length is 1.6 times as long, labelled similar. In the third step it is instead stretched only sideways, to 2.2 times the width but the same height, so the roof becomes flatter; it is labelled not similar.

With circles and squares the eye is enough. With two quadrilaterals or two pentagons it is not: two four-sided shapes can look alike and still differ a little. We need a test that we can check.

Think about printing a photo. A shop enlarges a 9 cm × 12 cm print to 15 cm × 20 cm. Every length in the picture is multiplied by the same number, $\frac{15}{9} = \frac{20}{12} = \frac{5}{3} .$ Every angle stays as it was: the corner of a window is still a right angle. Now open the photo in an editing app and drag only its width, from 9 cm to 18 cm. Faces look wide and every slanting angle changes, so the stretched photo is not similar to the original.

Definition

Similar polygons

Two polygons with the same number of sides are similar when both of these hold. (i) Each angle of one equals the matching angle of the other. (ii) Each side of one is the same multiple of the matching side of the other.

ExampleA 9 cm × 12 cm photo and its 15 cm × 20 cm enlargement: all angles are $90^{\circ}$ and every side is multiplied by $\frac{5}{3} .$

That common multiple is the scale factor, also called the representative fraction. Maps and building plans are drawn to a scale. On a map with scale 1 : 50 000, 2 cm on paper stands for 100 000 cm on the ground, which is 1 km.

Activity 6.1NCERT Activity 1 · p. 76

The shadow of a paper kite

You’ll need

a phone torch
a kite shape cut from thick card
a plain wall or a sheet of chart paper taped to it
a ruler, a protractor and a pencil

Predict first

You hold the card parallel to the wall, exactly halfway between the torch and the wall. How long will each edge of the shadow be?

the same as the edge of the cardtwice the edge of the cardhalf the edge of the cardlonger for some edges only

In a dark room, rest the torch on a table so that it shines straight at the wall. Call the bulb $O$ .
Hold the card parallel to the wall, halfway between the torch and the wall. Name its corners $A$ , $B$ , $C$ , $D$ .
Ask a friend to trace the shadow on the chart paper. Name the matching corners $A^{'}$ , $B^{'}$ , $C^{'}$ , $D^{'}$ .
Measure the sides and two angles of the card and of the shadow, and fill in the table.

Table 6.1

Parts	Card	Shadow	Ratio
$A B$ and $A^{'} B^{'}$
$B C$ and $B^{'} C^{'}$
$C D$ and $C^{'} D^{'}$
$∠ A$ and $∠ A^{'}$
$∠ C$ and $∠ C^{'}$

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 6.4

3/3Each corner of the shadow is twice as far from $O$ : $O A^{'} = 2 \times O A$ . Every side doubles and every angle stays.

Fig. 6.3An enlargement from a point: every distance from

O

doubles.NCERT Fig. 6.4 · p. 76

On the left is a point O marked as the torch. A small kite ABCD sits a little to its right. In the first step only the torch and the small kite are shown. In the second step dashed rays run from O through each corner of the kite and continue to the right. In the third step a larger kite with corners A′, B′, C′ and D′ appears on these rays, each corner twice as far from O as the matching corner of the small kite, with the notes OA′ = 2 × OA, A′B′ = 2 × AB and angles kept.

Light travels in straight lines, so $A^{'}$ lies on ray $O A$ , $B^{'}$ on ray $O B$ , and so on. The corners pair up: $A$ with $A^{'}$ , $B$ with $B^{'}$ . We write $A \leftrightarrow A^{'}$ and call such pairs corresponding vertices. Sides and angles built from corresponding vertices are corresponding parts: $A B$ corresponds to $A^{'} B^{'}$ and $∠ B$ to $∠ B^{'} .$

SimulationMake a similar copy

A square grid from 0 to 11 on both axes. Fixed points A (7, 2), B (10, 2) and C (7, 4) form a right triangle whose sides AB = 3, BC = root 13 and CA = 2 are drawn and measured. Fixed points P (1, 6) and Q (7, 6) are 6 apart, twice AB. Point R starts at (5, 9) and can be dragged to whole-number positions; the length QR is measured as you move it, so you can look for the place where every side of PQR is twice the matching side of ABC.

Fig. 6.4Triangle

A B C

is fixed, and

P Q = 2 \times A B

. Drag

R

so that

P Q R

is an enlargement of

A B C

with scale factor 2.

Try this

Find the place for $R$ where $Q R = 2 \sqrt{13}$ and $P R = 4$ . What are its coordinates?
Is there a second place for $R$ below $P Q$ that also works? Is that triangle still similar to $A B C$ ?
At the start, $Q R$ equals $B C$ . Why does that not make $P Q R$ similar to $A B C$ ?

One more fact is easy to check. If a first polygon is similar to a second, and the second to a third, then the first is similar to the third. (Multiply the two scale factors.)

For triangles, as you will see in Section 6.4, one condition always brings the other with it. For other polygons it does not: you must check both conditions. Fig. 6.5 shows why.

2/3A rhombus of side 3 cm: every side ratio is 1.5, but its angles are $60^{\circ}$ and $120^{\circ}$ . Not similar.

Fig. 6.5Equal angles alone, or proportional sides alone, are not enough for quadrilaterals.NCERT Figs. 6.6–6.7 · p. 77

On the left is a square of side 2 cm with right-angle marks. On the right, one shape at a time is compared with it. First a rectangle 3 cm by 2 cm: all its angles are right angles, but the side ratios are 1.5 and 1, so it is not similar. Second a rhombus of side 3 cm with a 60 degree angle: every side ratio is 1.5 but the angles differ, so it is not similar. Third a square of side 3 cm: all ratios are 1.5 and all angles are 90 degrees, so it is similar.

Quick check

1 of 3

Which pair of figures is always similar?

two rectanglestwo rhombusestwo regular hexagonstwo isosceles triangles

Show answer

Answer (C) two regular hexagons

Every regular hexagon has six equal sides and six

120^{\circ}

angles, so one is an enlargement of another. Rectangles can be long or squat, and rhombuses thin or fat, so their shapes vary.

Practice · 4 questions

Practice 6.1: questions with answers

1
Quick answerEasyNCERT Exercise 6.1, Q1
Complete each statement.
1. Two circles of radii 3 cm and 5 cm are but not congruent.
  Show answer
  Answer similar
  Same shape, different sizes.
2. Two squares have equal diagonals. The most exact word for them is:
  similar onlycongruentneither
  Show answer
  Answer (B) congruent
  Equal diagonals give equal sides, so the squares are congruent (and therefore also similar).
3. Which triangles are always similar to each other?
  isoscelesright-angledequilateralscalene
  Show answer
  Answer (C) equilateral
  Every equilateral triangle has three $60^{\circ}$ angles and equal sides. The other kinds come in many shapes.
4. For two pentagons to be similar, each pair of matching angles must be and each pair of matching sides must be .
  Show answer
  Answer (1) equal, (2) proportional
  Both conditions are needed for polygons.
Short answerEasyNCERT Exercise 6.1, Q2
Give two pairs of everyday objects that are similar and two pairs that are not. Give a reason for each.
Show solution
Similar: a ₹1 coin and a ₹5 coin (both circles); an A4 sheet and an A5 sheet (A5 is A4 folded in half, and both have sides in the ratio about 1 : 1.41).
Not similar: the two set squares in a geometry box (one has angles $30^{\circ}, 60^{\circ}, 90^{\circ}$ , the other $45^{\circ}, 45^{\circ}, 90^{\circ}$ ); a cricket pitch and a carrom board (a long rectangle and a square have different side ratios).
3
Short answerEasyNCERT Exercise 6.1, Q3
Decide whether the two quadrilaterals in each pair are similar. Give a reason.
1. A square of side 2 cm and a rhombus of side 5 cm with one angle of $70^{\circ}$ .
  Show solution
  Not similar. The sides are in one ratio (2.5), but the square has $90^{\circ}$ angles and the rhombus has $70^{\circ}$ and $110^{\circ}$ angles.
2. A 3 cm × 2 cm rectangle and a 7.5 cm × 5 cm rectangle.
  Show solution
  Similar. All angles are $90^{\circ}$ , and $\frac{7.5}{3} = \frac{5}{2} = 2.5$ , so the sides are in one ratio.
Quick answerEasy
A 10 cm × 15 cm photo is enlarged so that its shorter side becomes 25 cm. How long is the longer side, in cm?
cm
Show answer
Answer 37.5 cm
Scale factor $\frac{25}{10} = 2.5$ , so the longer side is $15 \times 2.5 = 37.5 c m$ .

6.3 Similarity of Triangles

A triangle is a polygon, so the test does not change: matching angles equal, and matching sides all in one ratio. Triangles whose matching angles are equal are called equiangular triangles.

Thales of Miletus noticed something special about equiangular triangles: the ratio of a pair of matching sides is the same for every pair. So for triangles, equal angles already force proportional sides. The key is a theorem about a line parallel to one side of a triangle.

Activity 6.2NCERT Activity 2 · p. 79

Parallel lines on a notebook page

You’ll need

a page of a ruled notebook
a sharp pencil
a ruler with a millimetre scale

Predict first

A ruled line crosses a triangle 4 lines below its top corner and 3 lines above its base. In what ratio does it cut each slanting side?

3 : 44 : 34 : 71 : 1

Mark a point $A$ on one ruled line. Seven lines further down, mark two points $B$ and $C$ far apart. Join $A B$ , $B C$ and $C A$ .
The fourth ruled line below $A$ crosses $A B$ at $D$ and $A C$ at $E$ . Ink in $D E$ : it is parallel to $B C$ , because ruled lines are parallel.
Measure $A D$ , $D B$ , $A E$ and $E C$ to the nearest millimetre and fill in the table.
Repeat with a new triangle, using the second ruled line below $A$ for $D E$ .

Table 6.2

Length or ratio	First triangle	Second triangle
$A D$
$D B$
$A D \div D B$
$A E$
$E C$
$A E \div E C$

Your entries are saved automatically.

3/3Each slanting side has 4 equal gaps above $D E$ and 3 below: $A D : D B = A E : E C = 4 : 3.$

Fig. 6.6Equally spaced parallel lines cut both sides into equal pieces.NCERT Fig. 6.9 · p. 79

Eight parallel ruled lines run across the page. Triangle ABC has A on the top line and B and C on the eighth line. In the first step only the lines and the triangle are shown. In the second step the fourth line below A is highlighted between D on AB and E on AC. In the third step dots mark where every ruled line crosses the two sides, showing 4 equal gaps above DE and 3 below on each side, with the notes AD : DB = 4 : 3 and AE : EC = 4 : 3.

This fact is the Basic Proportionality Theorem (आधारभूत समानुपातिकता प्रमेय). Its proof compares areas of triangles; Fig. 6.7 shows the three pairs of triangles it uses.

2/4 $△ A D E$ and $△ B D E$ share the height $E N$ , so their areas are in the ratio $A D : D B .$

Fig. 6.7Three comparisons of area prove the theorem.NCERT Fig. 6.10 · p. 80

Triangle ABC with D on AB and E on AC, and DE parallel to BC, marked with arrowheads. In the first step only this is shown. In the second step BE is joined, triangles ADE and BDE are shaded and the perpendicular EN from E to AB is drawn: they share the height EN. In the third step CD is joined, triangles ADE and CDE are shaded and the perpendicular DM from D to AC is drawn: they share the height DM. In the fourth step triangles BDE and CDE are shaded together: they have the same base DE and lie between the same parallels, so their areas are equal.

Theorem 6.1Basic Proportionality Theorem (Thales’ theorem)

If a line parallel to one side of a triangle cuts the other two sides at distinct points, it divides them in the same ratio.

Given: in $△ A B C$ , a line meets $A B$ at $D$ and $A C$ at $E$ , and $D E ∥ B C .$

To prove: $\frac{A D}{D B} = \frac{A E}{E C} .$

Construction: join $B E$ and $C D$ . Draw $E N ⊥ A B$ and $D M ⊥ A C .$

Proof: area of a triangle $= \frac{1}{2} \times$ base $\times$ height. Triangles $A D E$ and $B D E$ have bases $A D$ and $D B$ on line $A B$ and the same height $E N$ . So $ar (A D E) = \frac{1}{2} A D \cdot E N$ and $ar (B D E) = \frac{1}{2} D B \cdot E N .$ Dividing, the factor $\frac{1}{2} E N$ cancels:

\frac{ar (A D E)}{ar (B D E)} = \frac{A D}{D B}

(6.1)

In the same way, triangles $A D E$ and $C D E$ have bases $A E$ and $E C$ on line $A C$ and the same height $D M$ :

\frac{ar (A D E)}{ar (C D E)} = \frac{A E}{E C}

(6.2)

Triangles $B D E$ and $C D E$ stand on the same base $D E$ and lie between the same parallels $D E$ and $B C$ . From Class 9, such triangles have equal areas:

\begin{aligned} ar (B D E) \\ = ar (C D E) \end{aligned}

(6.3)

By (6.3), the left sides of (6.1) and (6.2) are the same fraction. Hence $\frac{A D}{D B} = \frac{A E}{E C} .$ ∎

Is the converse true? Suppose a line cuts two sides of a triangle proportionally. Must it then be parallel to the third side? Try it.

Activity 6.3NCERT Activity 3 · p. 81

Equal fractions along both arms

You’ll need

a ruler
two set squares
a sharp pencil

Predict first

On one arm of an angle you mark points every 1.5 cm; on the other arm, every 2 cm. You join the first mark on each arm, and also the fourth mark on each arm. Will the two joining lines be parallel?

yes, for any angleonly if the angle is

90^{\circ}

no, because the spacings differonly if the arms are equal

Draw an angle $X A Y$ of any size.
On $A X$ mark $B_{1}, B_{2}, B_{3}, B$ at 1.5, 3, 4.5 and 6 cm from $A$ . On $A Y$ mark $C_{1}, C_{2}, C_{3}, C$ at 2, 4, 6 and 8 cm from $A$ .
Join $B C$ , then $B_{1} C_{1}$ , $B_{2} C_{2}$ and $B_{3} C_{3}$ .
Test each join against $B C$ : lay one set square along $B C$ , hold a second set square against it, and slide the first along the second.
Finally join $B_{1} C_{2}$ and test it in the same way.

Table 6.3

Join	Ratio on $A X$	Ratio on $A Y$	Parallel?
$B_{1} C_{1}$
$B_{2} C_{2}$
$B_{3} C_{3}$
$B_{1} C_{2}$

Your entries are saved automatically.

Theorem 6.2Converse of the Basic Proportionality Theorem

A line that divides two sides of a triangle proportionally is parallel to the remaining side.

Given: $D$ on $A B$ and $E$ on $A C$ with $\frac{A D}{D B} = \frac{A E}{E C} .$ To prove: $D E ∥ B C .$

Proof (by contradiction): suppose $D E$ is not parallel to $B C$ . Through $D$ draw the line parallel to $B C$ ; it meets $A C$ at a point $E^{'}$ different from $E$ .

Theorem 6.1 in $△ A B C$ gives $\frac{A D}{D B} = \frac{A E^{'}}{E^{'} C} .$ With the given ratio, $\frac{A E}{E C} = \frac{A E^{'}}{E^{'} C} .$

Add 1 to both sides: $\frac{A E + E C}{E C} = \frac{A E^{'} + E^{'} C}{E^{'} C},$ that is, $\frac{A C}{E C} = \frac{A C}{E^{'} C} .$ So $E C = E^{'} C$ .

Both $E$ and $E^{'}$ lie on segment $A C$ at the same distance from $C$ , so they are the same point. This contradicts our supposition. Hence $D E ∥ B C$ . ∎

2/3Draw $D E^{'} ∥ B C$ . By Theorem 6.1, $E^{'}$ divides $A C$ in the ratio $A D : D B$ , just as $E$ does.

Fig. 6.8If

D

and

E

cut the sides in the same ratio,

E

has nowhere to go but onto the parallel.NCERT Fig. 6.12 · p. 82

Triangle ABC with D the midpoint region of AB. In the first step a dashed line joins D to a point E on AC drawn as if DE were not parallel to BC. In the second step the line through D parallel to BC is drawn, meeting AC at a second point E′ above E. In the third step the point E slides up onto E′ and the dashed line disappears, leaving DE parallel to BC, with the note that E′ and E are one point.

InteractiveBPT lab

A triangle ABC with a line DE across it, and slider controls. In Parallel line mode, one slider moves D along AB and E moves with it so that DE stays parallel to BC; a table shows AD, DB, AE and EC in centimetres and the ratios AD/DB and AE/EC, which are always equal. In Equal ratios mode, two sliders set D on AB and E on AC separately; DE turns green with parallel marks only when the two ratios are equal, and a second table compares angle ADE with angle ABC. A New triangle button changes the shape of the triangle.

Fig. 6.9Slide

D

(and

E

) along the sides and compare

\frac{A D}{D B}

with

\frac{A E}{E C} .

Example 6.1NCERT Example 1 · p. 82

Another form of the theorem

In $△ A B C$ , $D E ∥ B C$ with $D$ on $A B$ and $E$ on $A C$ . (a) Show that $\frac{A D}{A B} = \frac{A E}{A C} .$ (b) If $A D = 3$ cm, $A B = 7.5$ cm and $A C = 10$ cm, find $A E$ .

Try it yourself first. Then reveal the solution one step at a time.

Table 6.4Three forms of Theorem 6.1, all true when

D E ∥ B C

Form	It compares
$\frac{A D}{D B} = \frac{A E}{E C}$	the two pieces of each side
$\frac{A D}{A B} = \frac{A E}{A C}$	the top piece with the whole side
$\frac{D B}{A B} = \frac{E C}{A C}$	the bottom piece with the whole side

Never mix the forms: $\frac{A D}{A B}$ is not equal to $\frac{A E}{E C} .$ Pick one form and keep the same pieces on both sides.

2/3Draw the diagonal $P R$ . In $△ P S R$ , $X Z ∥ S R$ .

Fig. 6.10A diagonal splits the trapezium into two triangles.NCERT Figs. 6.14–6.15 · p. 83

A trapezium PQRS with the long side PQ at the bottom and the shorter parallel side SR at the top. A fence XY runs across it, parallel to both, from X on PS to Y on QR. PX is 12 m, XS is 8 m, QY is 15 m and YR is unknown. In the first step only the field and fence are shown. In the second step the diagonal PR is drawn, meeting the fence at Z, and triangle PSR is shaded. In the third step triangle RPQ is shaded instead.

Example 6.2NCERT Example 2 · p. 83

A fence across a trapezium-shaped field

A field $P Q R S$ is a trapezium with $P Q ∥ S R$ (Fig. 6.10). A straight fence $X Y$ , parallel to $P Q$ , runs from $X$ on $P S$ to $Y$ on $Q R$ . $P X = 12$ m, $X S = 8$ m and $Q Y = 15$ m. Find $Y R$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6.3NCERT Example 3 · p. 83

Equal ratios, then an isosceles triangle

In $△ K L M$ , $X$ lies on $K L$ and $Y$ on $K M$ with $K X : X L = K Y : Y M = 3 : 4$ . Also $∠ K X Y = 68^{\circ}$ and $∠ K = 44^{\circ}$ . Find $∠ K M L$ and show that $△ K L M$ is isosceles.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 10 questions

Practice 6.2: questions with answers

1
NumericalEasyNCERT Exercise 6.2, Q1
In both triangles of Fig. 6.11, $D E ∥ B C$ .
Fig. 6.11Lengths in cm; $D E ∥ B C$ in both.NCERT Fig. 6.17 · p. 84
Panel (a): triangle ABC with D on AB and E on AC, DE parallel to BC; AD is 2.4, DB is 3.6, AE is 2 and EC is marked with a question mark. Panel (b): another triangle ABC with DE parallel to BC; AD is marked with a question mark, AE is 2.8, EC is 4.2 and the note says AB = 9. All lengths are in centimetres.
1. Find $E C$ .
  cm
  Show solution
  $\frac{A D}{D B} = \frac{A E}{E C}$ gives $\frac{2.4}{3.6} = \frac{2}{E C},$ so $E C = 2 \times 3.6 \div 2.4 = 3$ cm.
2. Find $A D$ .
  cm
  Show solution
  Use $\frac{A D}{A B} = \frac{A E}{A C}$ with $A C = 2.8 + 4.2 = 7$ : $A D = 9 \times \frac{2.8}{7} = 3.6$ cm.
2
Quick answerEasyNCERT Exercise 6.2, Q2
$E$ and $F$ lie on sides $P Q$ and $P R$ of $△ P Q R$ . In each case, is $E F ∥ Q R$ ? (Lengths in cm.)
1. $P E = 2.4$ , $E Q = 3.2$ , $P F = 1.8$ , $F R = 2.4$
  parallelnot parallelcannot tell
  Show answer
  Answer (A) parallel
  $\frac{2.4}{3.2} = 0.75 = \frac{1.8}{2.4},$ so $E F ∥ Q R$ by Theorem 6.2.
2. $P E = 5$ , $E Q = 4$ , $P F = 6$ , $F R = 5$
  parallelnot parallelcannot tell
  Show answer
  Answer (B) not parallel
  $\frac{5}{4} = 1.25$ but $\frac{6}{5} = 1.2.$ The ratios differ, so $E F$ is not parallel to $Q R$ .
3. $P Q = 2.25$ , $P E = 0.75$ , $P R = 3.6$ , $P F = 1.2$
  parallelnot parallelcannot tell
  Show answer
  Answer (A) parallel
  $\frac{P E}{P Q} = \frac{1}{3} = \frac{P F}{P R},$ so each side is cut in the ratio $1 : 2$ and $E F ∥ Q R$ .
3
ProofMediumNCERT Exercise 6.2, Q3 and Q5
$P Q R S$ is a quadrilateral and $X$ is a point on its diagonal $P R$ . The line through $X$ parallel to $R Q$ meets $P Q$ at $Y$ , and the line through $X$ parallel to $R S$ meets $P S$ at $Z$ .
1. Prove that $\frac{P Y}{Y Q} = \frac{P Z}{Z S} .$
  Show solution
  In $△ P Q R$ , $X Y ∥ R Q$ , so $\frac{P Y}{Y Q} = \frac{P X}{X R}$ (Theorem 6.1).
  In $△ P S R$ , $X Z ∥ R S$ , so $\frac{P Z}{Z S} = \frac{P X}{X R} .$ Both equal $\frac{P X}{X R} .$
2. Hence prove that $Y Z ∥ Q S$ .
  Show solution
  In $△ P Q S$ , $Y$ and $Z$ divide $P Q$ and $P S$ in the same ratio (part a), so $Y Z ∥ Q S$ by Theorem 6.2.
NumericalHardNCERT Exercise 6.2, Q4
In $△ X Y Z$ , $L$ is a point on $X Y$ . The line through $L$ parallel to $X Z$ meets $Y Z$ at $M$ , and the line through $L$ parallel to $X M$ meets $Y M$ at $N$ . If $Y N = 9$ cm and $N M = 3$ cm, find $M Z$ in cm.
cm
Show solution
In $△ Y X M$ , $L N ∥ X M$ , so $\frac{Y N}{N M} = \frac{Y L}{L X} .$
In $△ Y X Z$ , $L M ∥ X Z$ , so $\frac{Y L}{L X} = \frac{Y M}{M Z} .$
So $\frac{Y N}{N M} = \frac{Y M}{M Z}$ with $Y M = 9 + 3 = 12$ : $\frac{9}{3} = \frac{12}{M Z},$ and $M Z = 4$ cm.
ProofMediumNCERT Exercise 6.2, Q6
$O$ is a point inside $△ P Q R$ . Points $A$ , $B$ , $C$ lie on $O P$ , $O Q$ , $O R$ with $A B ∥ P Q$ and $B C ∥ Q R$ . Prove that $A C ∥ P R$ .
Show solution
In $△ O P Q$ , $A B ∥ P Q$ gives $\frac{O A}{A P} = \frac{O B}{B Q} .$
In $△ O Q R$ , $B C ∥ Q R$ gives $\frac{O B}{B Q} = \frac{O C}{C R} .$
So $\frac{O A}{A P} = \frac{O C}{C R},$ and in $△ O P R$ Theorem 6.2 gives $A C ∥ P R$ .
ProofEasyNCERT Exercise 6.2, Q7
In $△ P Q R$ , $M$ is the mid-point of $P Q$ , and the line through $M$ parallel to $Q R$ meets $P R$ at $N$ . Use Theorem 6.1 to show that $N$ is the mid-point of $P R$ .
Show solution
$M N ∥ Q R$ , so $\frac{P N}{N R} = \frac{P M}{M Q} = 1$ , because $P M = M Q$ . Hence $P N = N R$ and $N$ is the mid-point of $P R$ .
ProofEasyNCERT Exercise 6.2, Q8
In $△ P Q R$ , $M$ and $N$ are the mid-points of $P Q$ and $P R$ . Use Theorem 6.2 to show that $M N ∥ Q R$ .
Show solution
$\frac{P M}{M Q} = 1 = \frac{P N}{N R},$ so $M$ and $N$ divide the two sides in the same ratio, and $M N ∥ Q R$ by Theorem 6.2.
8
Long answerHardNCERT Exercise 6.2, Q9
In trapezium $P Q R S$ , $P Q ∥ S R$ , and the diagonals $P R$ and $Q S$ meet at $O$ .
1. Prove that $\frac{P O}{O R} = \frac{Q O}{O S} .$
  Show solution
  Draw $O T ∥ P Q$ with $T$ on $P S$ ; then $O T ∥ S R$ too.
  In $△ P S R$ , $T O ∥ S R$ : $\frac{P T}{T S} = \frac{P O}{O R} .$
  In $△ S P Q$ , $T O ∥ P Q$ : $\frac{S T}{T P} = \frac{S O}{O Q},$ so $\frac{P T}{T S} = \frac{Q O}{O S} .$
  Hence $\frac{P O}{O R} = \frac{Q O}{O S} .$
2. If $P O = 2 x - 3$ , $O R = 2 x - 2$ , $Q O = 2 x + 4$ and $O S = 3 x + 1$ , find $x$ .
  Show solution
  Cross-multiply: $(2 x - 3) (3 x + 1)$ $= (2 x + 4) (2 x - 2)$ .
  So $6 x^{2} - 7 x - 3 = 4 x^{2} + 4 x - 8$ , which gives $2 x^{2} - 11 x + 5 = 0$ , that is $(2 x - 1) (x - 5) = 0$ .
  $x = \frac{1}{2}$ makes $P O$ negative, so $x = 5$ . Check: $\frac{7}{8} = \frac{14}{16} .$
ProofHardNCERT Exercise 6.2, Q10
The diagonals of a quadrilateral $A B C D$ meet at $O$ , and $\frac{A O}{O C} = \frac{B O}{O D} .$ Prove that $A B C D$ is a trapezium.
Show solution
Draw $O E ∥ D C$ with $E$ on $A D$ . In $△ A D C$ , Theorem 6.1 gives $\frac{A E}{E D} = \frac{A O}{O C} = \frac{B O}{O D} .$
So in $△ D A B$ , $E$ and $O$ divide $D A$ and $D B$ in the same ratio, and $E O ∥ A B$ (Theorem 6.2).
Then $A B ∥ E O ∥ D C$ , so $A B C D$ has a pair of parallel sides: it is a trapezium.
NumericalMedium
In $△ A B C$ , $D E ∥ B C$ with $D$ on $A B$ and $E$ on $A C$ . $A D = x + 1$ , $D B = 3 x - 1$ , $A E = x + 3$ and $E C = 3 x + 4$ (in cm). Find $x$ .
Show solution
$\frac{x + 1}{3 x - 1} = \frac{x + 3}{3 x + 4} .$ Cross-multiplying: $3 x^{2} + 7 x + 4 = 3 x^{2} + 8 x - 3$ , so $x = 7$ .
Check: $\frac{8}{20} = \frac{10}{25} = 0.4.$

6.4 Criteria for Similarity of Triangles

By the definition, $△ A B C$ and $△ D E F$ are similar when all three pairs of angles match, $∠ A = ∠ D$ , $∠ B = ∠ E$ and $∠ C = ∠ F$ , and all three side ratios agree:

\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D}

(6.4)

For this we use the symbol $\sim$ , said as "is similar to". So $△ A B C \sim △ D E F$ is a short way of saying that the two triangles are similar, with $A, B, C$ matching $D, E, F$ . In Class 9 the symbol $≅$ played the same role for congruence.

Checking six facts every time is slow. In Class 9 you saw that three well-chosen facts settle congruence (SSS, SAS, ASA, RHS). Fewer facts are enough for similarity too; each such shortcut is a similarity criterion. The next three activities find them.

Activity 6.4NCERT Activity 4 · p. 86

Same angles, different bases

You’ll need

a ruler
a protractor
a sharp pencil

Predict first

Two triangles each have a

50^{\circ}

angle and a

70^{\circ}

angle at the ends of their bases, which are 4 cm and 6 cm long. What do you expect for the other sides?

only the bases are in the ratio

\frac{2}{3}

the other sides are equalevery side ratio is

\frac{2}{3}

there is no pattern

Draw $B C = 4$ cm. At $B$ draw a ray at $50^{\circ}$ to $B C$ , and at $C$ a ray at $70^{\circ}$ to $C B$ , on the same side. The rays meet at $A$ .
Draw $E F = 6$ cm and repeat with $50^{\circ}$ at $E$ and $70^{\circ}$ at $F$ . The rays meet at $D$ .
Measure $∠ A$ and $∠ D$ , then $A B$ , $C A$ , $D E$ and $F D$ , and fill in the table.

Table 6.5

Ratio	Value
$B C \div E F$	$4 \div 6 \approx 0.67$
$A B \div D E$
$C A \div F D$

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3/3 $∠ D P Q = ∠ E$ , so $P Q ∥ E F$ , and Theorem 6.1 gives $\frac{D P}{D E} = \frac{D Q}{D F} .$

Fig. 6.12Placing a copy of

△ A B C

in the corner

D

turns the problem into Theorem 6.1.NCERT Fig. 6.24 · p. 87

A small triangle ABC on the left and a larger triangle DEF on the right, with matching arc marks at B and E, C and F, and A and D. In the first step they sit side by side. In the second step the small triangle slides so that A lands on D, with its sides along DE and DF; its other corners land at P on DE and Q on DF. In the third step PQ is highlighted with parallel marks matching EF, and the note PQ ∥ EF appears.

Theorem 6.3AAA similarity criterion

If two triangles have their angles equal in pairs, their corresponding sides are proportional, so the triangles are similar.

Given: $△ A B C$ and $△ D E F$ with $∠ A = ∠ D$ , $∠ B = ∠ E$ and $∠ C = ∠ F$ . Suppose $D E > A B$ . (If $D E = A B$ , the triangles are congruent by ASA; if $D E < A B$ , swap their roles.)

Construction: mark $P$ on $D E$ with $D P = A B$ , and $Q$ on $D F$ with $D Q = A C$ . Join $P Q$ . This places a copy of $△ A B C$ in the corner $D$ (Fig. 6.12).

Proof: $△ A B C ≅ △ D P Q$ by SAS ( $A B = D P$ , $∠ A = ∠ D$ , $A C = D Q$ ). So $∠ D P Q = ∠ B = ∠ E$ .

$∠ D P Q$ and $∠ D E F$ are corresponding angles for the lines $P Q$ and $E F$ with transversal $D E$ , and they are equal. So $P Q ∥ E F$ .

Theorem 6.1 in the form of Example 6.1 gives $\frac{D P}{D E} = \frac{D Q}{D F},$ that is, $\frac{A B}{D E} = \frac{A C}{D F} .$

Placing the copy in the corner $E$ instead gives $\frac{A B}{D E} = \frac{B C}{E F}$ in the same way. So all three ratios are equal and $△ A B C \sim △ D E F$ . ∎

CorollaryAA similarity criterion

When two angles of one triangle equal two angles of another, the triangles are similar.

Activity 6.5NCERT Activity 5 · p. 88

Same ratios, then measure the angles

You’ll need

a ruler
a pair of compasses
a protractor

Predict first

Triangle 1 has sides 3.5, 4 and 5 cm; triangle 2 has sides 7, 8 and 10 cm. What do you expect when you measure the angles?

the bigger triangle has bigger anglesmatching angles are equalthe angles of triangle 2 are doubledno pattern

Draw $△ A B C$ with $C A = 5$ cm, $A B = 3.5$ cm and $B C = 4$ cm (draw $C A$ first, then use compasses to find $B$ ).
Draw $△ D E F$ with $F D = 10$ cm, $D E = 7$ cm and $E F = 8$ cm.
Check that $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D} .$ Then measure all six angles.

Table 6.6

Vertices	In $△ A B C$	In $△ D E F$
$A$ and $D$
$B$ and $E$
$C$ and $F$

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Theorem 6.4SSS similarity criterion

If the three sides of one triangle are proportional to the three sides of another, their corresponding angles are equal, so the triangles are similar.

Given: $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D},$ with this common ratio less than 1.

Construction: mark $P$ on $D E$ and $Q$ on $D F$ with $D P = A B$ and $D Q = A C$ . Join $P Q$ .

Proof: now $\frac{D P}{D E} = \frac{D Q}{D F},$ so $\frac{D P}{P E} = \frac{D Q}{Q F}$ and, by Theorem 6.2, $P Q ∥ E F$ . Corresponding angles give $∠ D P Q = ∠ E$ and $∠ D Q P = ∠ F$ , so $△ D P Q \sim △ D E F$ by AA.

Hence $\frac{P Q}{E F} = \frac{D P}{D E} = \frac{A B}{D E} = \frac{B C}{E F},$ which gives $P Q = B C$ .

So $△ A B C ≅ △ D P Q$ by SSS. Therefore $∠ A = ∠ D$ , $∠ B = ∠ D P Q = ∠ E$ and $∠ C = ∠ D Q P = ∠ F$ . ∎

Theorems 6.3 and 6.4 together say something special about triangles: equal angles force proportional sides, and proportional sides force equal angles. For quadrilaterals this failed (Fig. 6.5), but for triangles you only ever need to check one of the two conditions.

Activity 6.6NCERT Activity 6 · p. 89

One equal angle between proportional sides

You’ll need

a ruler
a protractor
a sharp pencil

Predict first

Take

∠ A = ∠ D = 70^{\circ}

with

A B = 2.5

cm,

A C = 4

cm,

D E = 5

cm and

D F = 8

cm. The other angles of the two triangles will also match.

TrueFalse

Draw $∠ A = 70^{\circ}$ . On its arms mark $B$ with $A B = 2.5$ cm and $C$ with $A C = 4$ cm. Join $B C$ .
Draw $∠ D = 70^{\circ}$ with $D E = 5$ cm and $D F = 8$ cm. Join $E F$ .
Measure $∠ B$ , $∠ C$ , $∠ E$ and $∠ F$ , and the third sides $B C$ and $E F$ .

Table 6.7

Pair	In $△ A B C$	In $△ D E F$
$∠ B$ and $∠ E$
$∠ C$ and $∠ F$
$B C$ and $E F$

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Theorem 6.5SAS similarity criterion

Two triangles are similar if an angle of one equals an angle of the other and the sides enclosing these angles are proportional.

3/3SAS: one pair of equal angles, and the sides that form them in one ratio.

Fig. 6.13Each criterion needs only three facts.NCERT p. 90

A small triangle ABC and a larger triangle DEF, one and a half times as big. In the AA step, matching arcs mark the equal angles at B and E and the equal angles at C and F. In the SSS step, the sides are labelled 4, 6 and 5 on the small triangle and 6, 9 and 7.5 on the large one, every side multiplied by 1.5. In the SAS step, only the equal angles at B and E are marked, with the sides around them labelled 4 and 6 and then 6 and 9.

Compare

Congruence and similarity tests

Congruent (Class 9)

SSS: three sides equal
SAS: two sides and the angle between them equal
ASA or AAS: two angles and a side equal
RHS: right angle, hypotenuse and a side equal

Similar (this chapter)

SSS: three sides proportional
SAS: an equal angle between proportional sides
AA: two angles equal (no side needed)
RHS: right angle, hypotenuse and a side proportional

Now let us use the criteria. In each example, first find the pairs of equal angles or proportional sides, then name the criterion, then write the similarity with the vertices in matching order.

2/3Alternate angles are equal (same marks), and so are the vertically opposite angles at $C$ .

Fig. 6.14Parallel lines and vertically opposite angles give two similar triangles.NCERT Fig. 6.29 · p. 91

Segments AE and BD cross at C. On the left, AB is a vertical side of triangle ABC; on the right, DE is a shorter vertical side of triangle EDC, parallel to AB. AB is 9 cm, AC is 6 cm, CE is 4 cm and DE is unknown. In the first step only the figure and lengths are shown. In the second step matching arcs mark the equal alternate angles at A and E, at B and D, and the vertically opposite angles at C. In the third step a dashed copy of triangle EDC turns half a turn about C and lands inside triangle ABC, with its side parallel to AB.

Example 6.4NCERT Example 4 · p. 91

Bow-tie triangles

In Fig. 6.14, $A B ∥ E D$ , and the segments $A E$ and $B D$ cross at $C$ . $A B = 9$ cm, $A C = 6$ cm and $C E = 4$ cm. Show that $△ A B C \sim △ E D C$ and find $D E$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6.5NCERT Example 5 · p. 91

Find the matching angle

In $△ A B C$ , $A B = 3$ cm, $B C = 6$ cm, $C A = 3 \sqrt{3}$ cm, $∠ A = 90^{\circ}$ and $∠ B = 60^{\circ}$ . In $△ P Q R$ , $P Q = 12$ cm, $Q R = 6 \sqrt{3}$ cm and $R P = 6$ cm. Find $∠ Q$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6.6NCERT Example 6 · p. 92

From a product to similar triangles

Segments $A C$ and $B D$ cross at $O$ , with $O A = 6$ cm, $O C = 4$ cm, $O B = 3$ cm and $O D = 8$ cm. (a) Check that $O A \cdot O C = O B \cdot O D$ . (b) Prove that $∠ O A B = ∠ O D C$ . (c) If $D C = 10$ cm, find $A B$ .

Try it yourself first. Then reveal the solution one step at a time.

When a question gives a product of lengths, as in Example 6.6, divide to turn it into a ratio. Then look for an equal angle between those sides.

3/3Right triangles $A B E$ and $C D E$ share $∠ E$ , so they are similar (AA).

Fig. 6.15The farther the child walks, the longer the shadow.NCERT Fig. 6.32 · p. 92

A street lamp A on top of a post AB on the left. In the first step a child stands close to the post and casts a short shadow; a dashed ray from the lamp grazes the child’s head and meets the ground at the shadow’s tip. In the second step the child has walked 6 m away from the post in 5 seconds, standing at D with head C, and casts a longer shadow DE. In the third step the two right triangles ABE and CDE are shaded, with AB = 4.5 m, CD = 1.5 m, BD = 6 m and the shadow DE = x.

Example 6.7NCERT Example 7 · p. 92

A shadow under a street lamp

A child 1.5 m tall walks away from the foot of a 4.5 m high street lamp at 1.2 m/s. How long is the child’s shadow after 5 seconds?

Try it yourself first. Then reveal the solution one step at a time.

Notice that here the shadow is always half the child’s distance from the post, since $\frac{B D + x}{x} = 3$ gives $x = \frac{B D}{2} .$ The shadow grows as the child walks away.

Example 6.8NCERT Example 8 · p. 93

Altitudes of similar triangles

$△ A B C \sim △ P Q R$ , and $A D$ and $P S$ are altitudes ( $A D ⊥ B C$ , $P S ⊥ Q R$ ). (a) Prove that $△ A B D \sim △ P Q S$ . (b) Prove that $\frac{A D}{P S} = \frac{A B}{P Q} .$ (c) If $A B = 10$ cm, $P Q = 6$ cm and $A D = 8$ cm, find $P S$ .

Try it yourself first. Then reveal the solution one step at a time.

The same holds for medians and for angle bisectors, as you will prove in Practice 6.3. Matching segments of similar triangles are in the ratio of the sides.

Quick check

1 of 3

At the same moment, a boy 1.2 m tall casts a 0.9 m shadow and a neem tree casts a 6 m shadow. How tall is the tree, in metres?

Show answer

Answer 8 m

Height ÷ shadow is the same for both:

\frac{h}{6} = \frac{1.2}{0.9},

h = 8

6.5 Summary

2/3Corresponding angles are equal, so $△ A D E \sim △ A B C$ (AA).

Fig. 6.17The whole chapter in one triangle.

Triangle ABC with D on AB and E on AC, and DE parallel to BC. In the first step DE is highlighted and the note says AD : DB = AE : EC, the Basic Proportionality Theorem. In the second step the small triangle ADE is shaded and matching angle marks at D and B, and at E and C, show that triangle ADE is similar to triangle ABC by AA. In the third step the note says DE ÷ BC = AD ÷ AB = AE ÷ AC, so the parallel segment is the same fraction of BC.

Recap

Triangles: key points and summary

Similar figures have the same shape; congruent figures also have the same size (Section 6.2).
Polygons with the same number of sides are similar when matching angles are equal and matching sides proportional; in general both are needed (Section 6.2).
The common ratio of matching sides is the scale factor (Section 6.2).
BPT: a line parallel to one side of a triangle cuts the other two sides proportionally (Theorem 6.1).
Converse: a line cutting two sides proportionally is parallel to the third side (Theorem 6.2).
AA: two pairs of equal angles give similar triangles (Theorem 6.3).
SSS: three pairs of proportional sides give similar triangles (Theorem 6.4).
SAS: an equal angle between proportional sides gives similar triangles (Theorem 6.5).
For triangles, equal angles and proportional sides come together, so one condition is enough (Section 6.4).
Write similarity statements with matching vertices in order (Order matters).
Similar triangles measure what you cannot reach, such as heights from shadows (Example 6.7).

Flash cards · 6

Test yourself

1 / 6

Board exam practice

These questions mix the chapter’s ideas in the styles used in board papers: multiple choice, assertion and reason, a case study, and longer proofs.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 6.1, Q1	Practice 6.1 · Q1
NCERT Exercise 6.1, Q2	Practice 6.1 · Q2
NCERT Exercise 6.1, Q3	Practice 6.1 · Q3
NCERT Exercise 6.2, Q1	Practice 6.2 · Q1
NCERT Exercise 6.2, Q2	Practice 6.2 · Q2
NCERT Exercise 6.2, Q3 and Q5	Practice 6.2 · Q3
NCERT Exercise 6.2, Q4	Practice 6.2 · Q4
NCERT Exercise 6.2, Q6	Practice 6.2 · Q5
NCERT Exercise 6.2, Q7	Practice 6.2 · Q6
NCERT Exercise 6.2, Q8	Practice 6.2 · Q7
NCERT Exercise 6.2, Q9	Practice 6.2 · Q8
NCERT Exercise 6.2, Q10	Practice 6.2 · Q9
NCERT Exercise 6.3, Q1	Practice 6.3 · Q1
NCERT Exercise 6.3, Q2	Practice 6.3 · Q2
NCERT Exercise 6.3, Q3	Practice 6.3 · Q3
NCERT Exercise 6.3, Q4	Practice 6.3 · Q4
NCERT Exercise 6.3, Q5	Practice 6.3 · Q5
NCERT Exercise 6.3, Q6	Practice 6.3 · Q6
NCERT Exercise 6.3, Q7	Practice 6.3 · Q7
NCERT Exercise 6.3, Q8	Practice 6.3 · Q8
NCERT Exercise 6.3, Q9	Practice 6.3 · Q9
NCERT Exercise 6.3, Q10	Practice 6.3 · Q10
NCERT Exercise 6.3, Q11	Practice 6.3 · Q11
NCERT Exercise 6.3, Q12	Practice 6.3 · Q12
NCERT Exercise 6.3, Q13	Practice 6.3 · Q13
NCERT Exercise 6.3, Q14	Practice 6.3 · Q14
NCERT Exercise 6.3, Q15	Practice 6.3 · Q15
NCERT Exercise 6.3, Q16	Practice 6.3 · Q16

Quick check: MCQs on Triangles

6.2 Similar Figures3 questions
6.3 Similarity of Triangles4 questions
6.4 Criteria for Similarity of Triangles4 questions
6.4 Criteria for Similarity of Triangles3 questions

Glossary · 8

Key terms (English–Hindi)

Basic Proportionality Theoremआधारभूत समानुपातिकता प्रमेयaadhaarbhoot samaanupaatikta pramey: A line parallel to one side of a triangle divides the other two sides in the same ratio.First seen in the chapter →

congruent figuresसर्वांगसम आकृतियाँsarvaangsam aakritiyaan: Figures with the same shape and the same size.First seen in the chapter →
converseविलोमvilom: The statement made by swapping what is given and what is concluded in a theorem.First seen in the chapter →
corresponding partsसंगत भागsangat bhaag: Matching vertices, sides and angles of two figures under a stated pairing of their vertices.First seen in the chapter →

equiangular trianglesसमानकोणिक त्रिभुजsamaankonik tribhuj: Two triangles whose angles are equal in pairs.First seen in the chapter →

scale factorस्केल गुणकscale gunak: The common ratio of matching sides of two similar figures.First seen in the chapter →
similar figuresसमरूप आकृतियाँsamroop aakritiyaan: Figures that have one shape, whatever their sizes: one is an exact enlargement or reduction of the other.First seen in the chapter →
similarity criterionसमरूपता की कसौटीsamroopta ki kasauti: A short set of facts (AA, SSS or SAS) that is enough to prove two triangles similar.First seen in the chapter →

Triangles: Class 10 Maths · Chapter 6

What you will learn in Triangles

6.1 Introduction

6.2 Similar Figures

Practice 6.1: questions with answers

6.3 Similarity of Triangles

Practice 6.2: questions with answers

6.4 Criteria for Similarity of Triangles

Practice 6.3: questions with answers

6.5 Summary

Triangles: key points and summary

Board exam practice

Practice questions with answers (NCERT “Board exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Triangles

Key terms (English–Hindi)

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