# Triangles For Class 10 CBSE NCERT

### Transcript of Triangles For Class 10 CBSE NCERT

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.

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

angle Triangle Triangle Triangle Triangle

A

B C

°

° °

4 7

6

P

Q R ° °

3

2 3°

6.1 Similarity

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

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

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)

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

(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.

∠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

(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.

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

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 ΔABCA ΔPQR = ABPQ = BCQR = ACPR

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

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

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)

Important Theorems

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

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

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.

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 ∆

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.

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

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

E D

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?

Make your own notes: