The document describes how to determine if three lines will form a triangle based on their lengths. It states that if the sum of the lengths of the two shortest lines is less than the longest line, the lines will not form a triangle. If the sum is equal to the longest line, the two shortest lines will overlap the longest line. Only if the sum is greater than the longest line will the three lines form a triangle.

1. Triangle is one of the basic shapes in geometry. Can you form a triangle
using any three lines? Here is one of the simple ways to know whether the
given lines will form a triangle or not.
Take 3 lines. Consider their lengths as x, y, and z; z be the longest side. If,
x + y < z, the triangle will not be formed. If x + y = z, the two lines will
join and will overlap on the line z. Finally when x + y > z, only then these
three lines will form a triangle. You can also give it a try and let us know
about your little experiment.

2. INTRODUCTION
Triangle: A shape having three sides and forming three
angles is called as Triangle.
In given diagram point A , B and C are the vertices of
∆ ABC and AB, BC, AC are the sides of ∆ ABC. Sum of
Measures of angles of any triangle is °.
i.e. ∠ A + ∠B + ∠C = °
TYPES OF TRIANGLES
1) Equilateral Triangle: Triangles having three equal
sides are called Equilateral Triangles. Each angle of an
equilateral triangle is 60°.
2) Isosceles Triangle: Triangles having two equal sides
are called as Isosceles Triangle.
3) Scalene Triangle: Triangles in which all sides have
different size of length are called a Scalene Triangle.
4) Right Angled Triangle: Triangle in which one angle is
90° are called as Right Angled Triangle.
SIMILAR TRIANGLES
Two triangles are similar, if their
corresponding angles are equal and
their corresponding sides are in the
same ratio.
Side AB, Side BC, Side AC
∠ + ∠ + ∠ = °
° + ° + ° = °
A
B C
°
°
°
A
B C
Equilateral Isosceles Scalene Right
angleTriangle Triangle Triangle Triangle
A
B C
°
° °
4 7
6
P
Q R
° °
3
2 3°
6.1 Similarity

3. In ∆ ABC and ∆ PQR,
∠ A = ∠P, ∠B = ∠Q, ∠ C = ∠R ………… …… ( I )
= = , = = , =
.
= …………..( II )
From statement (I) and (II)
∆ ABC and ∆ PQR are similar triangles as the ratios of their
corresponding sides are same and corresponding angles are
congruent.
We write it as ΔABC ~ ΔPQR
If both triangles are similar then find the
height of tree by using the property of
similar triangle.
=
𝐚 𝐰
𝐚 𝐰
𝒉
=
𝒉 =
×
𝒉 =
CRITERIA FOR SIMILARITY OF TRIANGLES
1) AAA or AA criteria: (Angle - Angle - Angle)
If two triangles are equiangular
(corresponding angles are equal to each other),
then they are similar.
Example:
In ΔABC and ΔPQR,
∠A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R
then Δ ABC ~ Δ PQR (by AAA criteria).
Remark: If two angles of one triangle are equal to two angles of another triangle respectively
then the third angle has to be same. Hence an AA criterion is the same as AAA criteria.
A
B C
P
Q R
° °
°
° °
°
=
𝐑
=
𝐑
=
( is the scale factor) A
scale factor is a number
which scales, or
multiplies, some quantity
to give similar shape of
figures or we can say the
ratio of corresponding
sides of two similar
figures

4. 2) SSS criteria: (Side - Side - Side) If in two triangles,
the sides of one triangle are proportional to the
corresponding sides of other triangle then they are
similar.
Example: In ΔABC and ΔPQR,
= =
Then ΔABC ~ ΔPQR (by SSS criteria)
3) SAS criteria: (Side - Angle - Side)
If in two triangles, one pair of corresponding sides is
proportional and the included angles are equal then the
two triangles are similar.
Example: In ΔABC and ΔPQR,
if ∠ A = ∠ P
And =
then Δ ABC ~ Δ PQR ( by SAS criteria)
A
B
C
Q
P
R
2
3
3.5
4
6
7
= = , = = , =
.
=
A
P
6 4
12
8
C
R
B
Q
Charts

5. Q.1. AAA test and AA test can be considered as same criteria for similarity of triangles.
A) True B) False
Q.2. Similar triangles always have equal lengths of corresponding sides.
A) True B) False
Q.3. A figure with three sides and three angles is called a …….
A) square B) triangle C) rectangle D) circle
Q.4. Sum of measures of angles of triangle is …….
A) ° B) ° C) ° D) °
Q.5. Angles of equilateral triangle are ……. each
A) ° B) ° C) ° D) None of these
Q.6. All …….. triangles are similar.
A) right angled B) isosceles C) equilateral D) None of these
Q.7. If ΔABC ~ ΔPQR then =
…….
A) PQ B) QR C) AC D) PR
Q.8. What is similarity criteria for Δ ABC ~ Δ PQR ?
A) SAS B) AAA
C) SSS D) ASA
A
B
C
Q
P
R
2
3
3.5
4
6
7
MCQs (Multiple Choice Questions)

6. Q.9. What are the similarity criteria
for Δ DEF and Δ XYZ ?
A) SAS C) SSS
B) AAA D) Side-Hypo
Q.10. ΔABC is scaled to form new ΔDEF. = , If scaling factor is 2.5, then find
A) 8 B) 7.5 C) 10.5 D) 12.5
Q.1. Observe the pairs of triangle given below
and write the similarity criterion.
Solution:
(i) In ΔABC, ∠ = ° and ∠ = ° ,
∴ ∠ = °
In ΔLMN, ∠ = ° and ∠ = ° ,
∴ ∠ = °
In ΔABC and ΔLMN
∠ = ∠ and ∠ = ∠
∴ ΔABC ~ ΔLMN…………(By AA criteria)
(ii) In Δ DEF and Δ PQR,
= =
= = ∴ ∠E = ∠Q = 50o
∴ = ∴ Δ DEF ~ Δ PQR ………. (By SAS criteria)
Answers:
1.(A), 2.(B), 3.(B), 4.(C), 5.(A), 6.(C), 7.(D), 8.(C), 9.(A), 10.(D)
Solved Questions
D
E F
X
Z
6
4
3
2
3

7. (iii) In Δ RST, ∠R=90° and ∠S=50ᵒ , ∴ ∠T=40ᵒ
In Δ XYZ, ∠X=90° and ∠Y=70° , ∴ ∠N=20°
In Δ RST and Δ XYZ,
∠R=∠X, But ∠S≠∠Y and ∠T≠∠Z
Δ RST and Δ XYZ are not similar.
Q.2. Using the information in the given figure, find ∠𝐑.
Solution
In Δ ABC and Δ PQR,
=
.
.
=
= =
= =
∴ AB/PQ= BC/QR= AC/PR
Δ ABC ~ Δ PQR ……… (By SSS criteria)
∠C=∠R ……………. (c.a.s.t)
In Δ ABC, ∠A =70ᵒ and ∠B = 50° , ∴ ∠C= 60ᵒ
Therefore, ∠R = 60ᵒ
Q.3. A building of length 50 m casts a shadow of 20 m on
the ground. At a same time a lamp post casts a shadow 10
m long. What is the height of the lamp post?
Solution:
Let AB be the height of building and PQ be the height of lamp post.
Therefore BC represents the shadow of building and QR represents shadow of lamp post.
∴ AB = 50 m , BC = 20 m , QR = 10 m , PQ=?
In ΔABC and ΔPQR
An Approach: Here the triangles
may not look similar because of
rotation but when we see the ratios of
their corresponding sides, we get the
same ratio

An Approach: Here the key point is the diagram. When
the sun rays fall, they make the same angle all the time. By
looking this only, we can say that the triangles are similar
by AA test of similarity.

8. ∠ABC = ∠PQR ……….(Each 90ᵒ)
∠BAC = ∠QPR ……… .(At the same time, the light rays from the sun will fall
on the building and the lamp post at the same angle)
ΔABC ~ ΔPQR ………..(By AA criteria)
∴ = ….. (Ratio of sides of similar triangles)
=
PQ =
PQ = 25 m
Height of lamp post is 25 m.
Q.4. ΔABC is a right angled triangle at B. BD⏊ AC.
Prove that i) Δ ADB ~ Δ BDC ~ Δ ABC
ii) BD2
= AD × DC
Solution:
(i) In Δ ADB and Δ ABC
∠ADB = ∠ABC ………. (Each °)
∠DAB = ∠BAC ………..(Common angle)
ΔADB ~ ΔABC ……..(By AA criteria) (1)
In ΔBDC and ΔABC
∠BDC = ∠ABC ………..(Each °)
∠BCD = ∠ACB ………..(Common angle)
Δ BDC ~ Δ ABC……... (By AA criteria) (2)
From (1) and (2) we can say,
Δ ADB ~ Δ BDC ~ Δ ABC
Remember: See carefully the
correspondence between the vertices of
triangles.
An Approach: When we’ve have right angled triangles and
information about the sides is given, directly look for AA test of
similarity.
Here in the 1st
question if we try to prove ∆ABD similar to ∆BDC
it would be difficult as the information about the sides and angles
both are not provided. But if we take ∆ABC and prove it similar to
∆ABD and ∆ABD

9. (ii) Δ ADB ~ Δ BDC
∴ = …………… (Corresponding sides of similar triangles)
= ………….... (We can write DB as BD)
= × …….. (By cross multiplying)
Q.5. In Fig.(a), prove that
(i) ΔACB ~ ΔDEB (ii) ΔAME ~ ΔDMC
Solution: (i) In Δ ACB and Δ DEB
∠ACB = ∠DEB ………….(Each 90ᵒ)
∠ABC = ∠DBE ………….(Common angle)
Δ ACB ~ Δ DEB………..(By AA criteria)
(ii) In Δ AME and Δ DMC
∠AEM = ∠DCM ……….. (each 90ᵒ)
∠AME = ∠DMC ……….(Vertically opposite angles)
Δ AME ~ Δ DMC …………..(By AA criteria)
At a Glance
 Similar triangles can have same area because even congruent triangles are called similar.
 Order is Important: When you are saying Triangle ABC is similar to PQR it means the angle A is equal
to angle P, angle B is equal to angle Q and angle C is equal to angle R.
And one will have to maintain the same order while writing the ratios of the sides.
 While drawing diagram always make sure either you are naming it from left to right or right to left (This
is the standard way of naming any geometrical shape)
 A fundamental property of Triangle: A Triangle can be formed when the addition of the lengths two
smaller line segments is greater than the third larger one.
A fundamental property of Right triangle: This triangle can only be formed if the addition of the
squares of the lengths of any two line segments is equal to the third one.

10. Q.1. Find value of x, if given pair of triangles are similar.
Ans: x = 12 units
Q.2. Is the given pair of triangles similar?
Justify your answer.
Ans: Not similar
Q.3. In the given figure, find the value of x if JK ‖ MN.
Ans: 3.33 cm
(Hint: Try to analyze whatever is given try to remember all the
properties related to it. Here it is given JK || MN this reminds us that
alternate angles are equal between parallel lines)
Q.4. A person is standing 35 feet away from a street light that is 24 feet tall. How tall is he, if his
shadow is 7 feet long?
Ans: 4 feet
Q.5. In figure, ∠ADC =∠BAC. Prove that CA2
= BC.DC
(Hint: First prove Δ ADC ~ Δ BAC )
Q.6. Show Δ PQR~ Δ PDE, If DE || QR.
( Hint: Use corresponding angles property)
8
Practice Yourself

11. PROPERTIES
1st
Property:
The ratio of areas of two triangles is equal to
the square of the ratio of their corresponding
sides.
If ΔABC and ΔPQR are similar triangles then
A ΔABC
A ΔPQR
=
AB
PQ
=
BC
QR
=
AC
PR
2nd Property:
Triangles drawn between two parallel lines with same
base have equal areas.
A(∆ABC) = A(∆PBC)…..
(Triangles having equal height and base)
3rd
Property:
Ratio of areas of triangles with equal height is equal to the ratio of their bases. We can also say,
ratio of areas of triangles with equal base is equal to the ratio of their heights.
∆
∆
= …(Triangles having equal height)
∆
∆
= …(Triangles having equal base)
6.2 Properties and Theorem
Remember: Height of triangles drawn
between two parallel lines is always
same
Scan to watch BPT Theorem

12. 4th
Property:
Basic Proportionality theorem: If a line is parallel to
a side of a triangle to intersect the other sides in two
distinct points, then the other two sides are divided in the
same ratio. If in ΔPQR, MN ‖ QR then =
𝐑
5th
Property:
Pythagoras Theorem: In a right angle triangle, the square of
the hypotenuse is equal to the sum of the squares of the other
two sides. Let ∆ ABC be right angled at B, then according to
Pythagoras theorem AC = AB + BC
Q.1 The ratio of areas of any two triangles is equal to the ratio of their corresponding
sides.
A) True B) False
Q.2 If a line is parallel to a side of a triangle to intersect the other sides in two distinct
points, then the other two sides are divided in the same ratio.
A) True B) False
Q.3 The areas of similar triangles are always same.
A) True B) False
Q.4 In given figure PS || QR, then A (∆ PQR)
and A (∆SQR) will be equal.
A) True B) False
Q.5 In the given figure, PQ || BC, then what would be
the values of AQ & QC respectively?
A) 2 cm & 6 cm B) 6 cm & 2 cm
C) 3 cm & 5 cm D) None of these
B
C
A
P
Q
1 cm
3 cm
MCQs (Multiple Choice Questions)
P
Q
S
R
Scan to watch Pythagoras
Theorem activity video

13. Q.6 On the basis of given information in the figure, we can say
A) PQ || BC B) PQ ⏊ BC C) PQ = BC
Q.7 In given figure, PQ || XZ, then what is the
value of x?
A) 10 cm B) 9 cm
C) 10.5 cm D) 12
Q.8 ∆ ABC ~ ∆ PQR, if A (∆ ABC) = 24 cm2
, AB = 2 cm and PQ = 3cm what is
A(∆ PQR)?
A) 36 cm2
B) 50 cm2
C) 24 cm2
D) 54 cm2
Q.9 ∆ ABC ~ ∆ DEF, if A (∆ ABC) =16 cm2
and A (∆ DEF) =36 cm2
then AB:DE is ___?
A) B) C)
9
D)
9
Q.10 In given figure A (∆ ABD) = 9 cm2
and A
(∆ADC) = 15 cm2
,then length of BD=?
A) 2 cm B) 5 cm
C) 3 cm D) 7 cm
Q.11 Pythagoras theorem holds true for ……….
A) Obtuse triangle B) Right angled triangle C) Acute triangle D) None of these
Q.12 ∆ ABC is right angled at B. If AB = 3 cm and BC = 4 cm then AC =?
A) 6 cm B) 3 cm C) 4 cm D) 5 cm
Q.13 If AB = 5 and BC= 12 then what should be AC to make ∆ ABC right triangle
at B.
A) 6 cm B) 13 cm C) 5 cm D) 15
X
Y
Z
P
Q
2 cm
3 cm
7 cm
x
D
A
B C
5 cm
Answers:
1.(B), 2.(A), 3.(B), 4.(B), 5.(A), 6.(A), 7.(C), 8.(D), 9.(B), 10.(C), 11.(B),
12.(C), 13.(B)

14. Important Theorems

15. Q.1. In given triangles DE || BC, find the values of unknown line segments (x).
Solution (a):
In the figure (a), DE || BC
= …….. (By basic proportionality theorem)
=
.
𝑥
x =
. ×
x = 7.5 cm
Solution (b):
In the figure (b), DE || BC
= ……… (By basic proportionality theorem)
=
𝑥
𝑥 =
x = 2.5 cm
Solution (c):
In the figure (c), DE || BC
= ……….. (By basic proportionality theorem)
𝑥
=
𝑥 =
𝑥 = 2.4 cm
An Approach: This is a
straight forward question.
One parallel line is given
which is enough to use
BPT.
Solved Questions

16. Q. 2. Check whether the line m is parallel to one of the side of triangles
Solution (a):
In the figure (a)
= =
= =
∴ ≠
∴ DE is not parallel to BA
Solution (b):
In the figure (b)
= =
=
∴ =
∴ DE || BA (By converse of BPT)
Q.3. In a trapezium ABCD, where AB ‖ CD if AB = 3 cm and CD = 5
cm. Show that ∆ AOB ~ ∆ DOC and also find ratio of their areas.
Solution: In ∆ AOB and ∆ DOC
∠ AOB = ∠ DOC ……… (Vertically opposite angles)
∠ BAO = ∠ ODC ……………... (Alternate angles)
∆AOB ~ ∆DOC …………….... (By AA criteria)
∆
∆
= ……… (Ratio of areas of similar triangles is equal to the square of the
ratio of their corresponding sides.)
∆
∆
=
∆
∆
=
An Approach: Just
use the converse of
BPT
An Approach: Whenever we have parallel lines, try to use the
alternate angles or corresponding angles properties

17. Q.4. In given figure ABCD is quadrilateral. BC intersects AD at
O. Show that
∆
∆
=
Construction: Draw AM and DN perpendicular to BC
Solution:
∆
∆
=
∗ ∗
∗ ∗
= (1)
If we prove that = , then we will be done.
So, let’s take the triangles associated with these sides and prove them
similar, so that we can use the property of corresponding sides of similar triangle.
Now in ∆ AOM and ∆ DON
∠ AOM = ∠ DON…….. (Vertically opposite angles)
∠ AMO = ∠ DNO ……. (Each 90ᵒ)
∆ AOM ~ ∆ DON …….. (By AA criteria)
= (2 )…….. (Ratio of sides of similar triangles)
Therefore from equation (1) and equation (2), we can get
∆
∆
=
Q.5. ∆ DEF ~ ∆ XYZ. If A (∆DEF) = 64 cm2
. DE = 4 cm and XY = 3 cm then find the value
of A (∆XYZ).
Solution: ∆DEF ~ ∆XYZ ………….. (Given)
∴
∆
∆
= …………. (Ratio of areas of similar triangles)
∆
=
∆
=
9
∴ A ∆XYZ = 𝒄
An Approach: This is a little tricky question, now here we need to find area of a triangle and formula for
the same is ½ * b * h. when we see ∆ABC & ∆BDC, we have BC as base but we do not have the height for
either of the triangles. This is reason we have done the construction of AM & DN.
An Approach: Just
use the Ratio of
areas of similar
triangles property.
This a simple
question.

18. Q.6. In given figure A, B and C are points on OP, OQ and OR
respectively such that AB || PQ and AC || PR.
Show that BC || QR.
Solution: In ∆ POQ
AB || PQ (Given)
= (1) ………… (By Basic proportionality theorem)
In ∆ POR
AC || PR (Given)
= (2)…………. (By Basic proportionality theorem)
From equation (1) and equation (2) we can say
= (3)
In ∆ OQR
= ……………. (From equation 3)
Therefore by converse of BPT, we get BC || QR
Q.7. Sides of triangles are given below. Check whether they form right angled triangle or
not.
(i) 10 cm, 6 cm and 8 cm (ii) 4 cm, 5 cm and 6 cm
Solution (i) We know that hypotenuse is the longest side of a right triangle,
We will take hypotenuse as 10 cm
(10)2
= 100
(6)2
= 36 and (8)2
= 64
Here, 100 = 36 + 64
(10)2
= (6)2
+ (8)2
Means Pythagoras theorem is satisfied
Hence triangle of sides 10 cm, 6 cm and 8 cm forms a right angled triangle.
An Approach: The diagram may scare you, but it’s simple. If
we can = 𝐑
, then we can say that BC || QR by converse
of BPT. The above ratios can be obtained by applying BPT
in ∆ and ∆

19. Solution (ii) Here the longest side is 6 cm
62
= 36
52
= 25 and = 16
But 36 ≠ 25 + 16
Means 62
≠ 52
+ 42
Here Pythagoras theorem is not satisfied
These sides do not form a right triangle.
Q.8. Prove that the sum of the squares of the sides of a rhombus is
equal to the sum of the squares of its diagonals.
Solution: Let ABCD is a rhombus.
∴ AB = BC = CD = AD
We have to prove,
AB + BC + CD + AD = AC + BD
The diagonals of a rhombus are perpendicular and bisect each other.
∴ AO = OC = 𝑎𝑛𝑑 BO = OD =
In ∆ AOB,
AB = AO + BO …………. (By Pythagoras theorem)
AB = +
AB = +
AB =
+
AB = AC + BD
AB + AB + AB + AB = AC + BD ……. ( AB can be written as AB +AB +AB +AB )
+ + + = + …….. ( = = =
An Approach: First of all convert the question in mathematical
form as AB2
+ BC2
+ CD2
+ AD2
= AC2
+BD2
. By looking at the
terms, the problem may look complicated but this is a simple one.
Just have a look at the first term, AB2
which is a hypotenuse of
∆ AOB and hence AB2
= AO2
+ BO2
Now, AO2
& BO2
is nothing but AC 2
& BD 2
.

20. Q.9. D and E are points on the sides CA and CB respectively of a ∆ ABC right angled at C.
Prove that: AE2
+ BD2
= AB2
+ DE2
Solution: ∆ABC is right triangle at C.
Points D and E are on sides CA and CB.
Join AE, DE and BD
∆ ACE and ∆ DCB are right triangles
By Pythagoras theorem we have
AE2
= AC2
+ CE2
……..equation I
BD2
= CD2
+ BC2
……..equation II
Adding equation I and equation II
AE2
+ BD2
= AC2
+ CE2
+ CD2
+ BC2
AE2
+ BD2
= (AC2
+ BC2
) + (CE2
+ CD2
)
…………………… {AC2
+ BC2
= AB2
and CE2
+ CD2
= DE2
}
AE2
+ BD2
= AB2
+ DE2
Q.1 If in the given figure line l is parallel to side BC of ∆ ABC, then find the value of x.
Ans: (a) 3.1 (b) 2.5
Q.2. Using Basic proportionality theorem check whether the line m is parallel to one of the
sides of the triangle.
Ans : (a) Line m is not parallel to BC (b) Line m is not parallel to BC
B
C
A
l
2.5
7.5
9.3
(a)
x
B
C
A
l
2.3
4.6
5
(b)
x
B
C
A
m
2.1 8.4
8
(a)3
B
C
A
m
2.3
4.6
3.5
(b)
7
Tips: See L.H.S, take the required
terms and add them.
Practice Yourself

21. Q.3. In triangle ABC, Points D, E and F are on the sides AB,
AC and BC respectively such that DF || AC and EF || AB.
Show that DE || BC.
Q.4. ∆ABC ~ ∆PQR. If A(∆ ABC) = 36 cm2
. AB = 3 cm and PQ = 8 cm then find the
value of A (∆PQR).
Ans: 256 cm2
Q.5. In the figure seg PB and seg QA are perpendicular to
seg AB. If PO = 5 cm and QO = 6 cm, A (∆POB) = 200
cm2
, find the area of ∆ QOA.
(Hint: First prove that both the triangles are similar.)
Ans: 288 cm2
Q.6. A ladder 10 m long is placed on the ground in such a way that it touches the top of
a vertical wall 6 m high. Find the distance of the foot of the ladder from the bottom of the
wall. Ans: 4 m
Q.7. The height of two building is 40 m and 35m respectively. If the distance between
the two buildings is 12 m, find the distance between their tops. Ans: 13 m
Q.8. In an equilateral triangle, prove that three times the square of one side is equal to
four times the square of one of its altitudes.
(Hint: Altitudes of equilateral triangle bisects the opposite side of triangle)
A
B C
ED
F
0
B
P
A
Q
At a Glance
 When sides of a triangle are changed, the area of newly formed triangle gets changed by the
square of the scale of change.
 When sides of a triangle are changed, the perimeter of newly formed triangle gets changed by
the same scale of change.
 BPT theorem says if a line is drawn parallel to one side of a triangle then it divides the other
two sides in the same ratio.
And if a line divides other two sides of a triangle into two halves then that line segment will be
parallel to and half of the third side. This is known as Midpoint theorem.
 Did you know the Rule of Pythagoras theorem is also used in coordinate geometry to
establish the distance formula which is used for finding the distances between two coordinate
points?

22. Make your own notes: