NAME = SAGAR MANDAL

CLASS = X - A

SUBJECT = MATHS

SCHOOL = K.V A.F.S T.K.D

A Triangle is formed by three lines segments obtained by joining three pairs of points taken from a set of three non collinear points in the plane.

In fig 1.1 , three non collinear points A, B, C have been joined and the figure ABC, enclosed by three line segments , AB,BC, and CA is called a triangle.

The symbol ▲ (delta) is used to denote a triangle. The three given points are called the vertices of a triangle. The three line segments are called the sides of a triangle. The angle made by the line segment at the vertices are called the angles of a triangle.

If the sides of a ▲ABC are extended in order, then the angle between the extended and the adjoining side is called the exterior angles of the triangle.

In the fig. 1.2

1 ,2 and 3 are the exterior angles while 4 ,5,and 6 are the interior angles of the triangles.

1 4

52

6 3

A

BB C

A

CBFig. 1.1

Fig. 1.2

• ON THE BASIS OF SIDES

1. SCALENE TRIANGLE – If all the sides of triangle are unequal then it is a scalene tirangle.

2. ISOSCELES TRIANGLE –If any two sides of a triangle are equal then it is a isosceles triangle.

3. EQUILATERAL TRIANGLE- If all the sides of a triangle are equal then it is an equilateral triangle.

• ON THE BASIS OF ANGLES

1. ACUTE ANGLED TRIANGLE –If all the three angles of a triangle are less than 900 then it is an acute anglesd triangle.

2. RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal to 900 then it is a right angled triangle.

3. OBTUSE ANGLED TRIANGLE- If one angle of a triangle is greater then 900 then it is a obtuse angled triangle.

Two triangles are congruent if three sides and three angles of one triangle are equal to the corresponding sides and angles of other triangle.

The congruence of two triangles follows immediately from the congruence of three lines segments and three angles.

1. SIDE – ANGLE – SIDE RULE (SAS RULE)

Two triangles are congruent if any two sides and the includes angle of one triangle is equal to the two sides and the included angle of other triangle.

EXAMPLE :- (in fig 1.3)

GIVEN: AB=DE, BC=EF ,

B= E

SOLUTION: IF AB=DE, BC=EF , B= E then by SAS Rule

▲ABS = ▲DEF

4 cm4 cm

600 600

A

B C

D

FE

Fig. 1.3

2. ANGLE – SIDE – ANGLE RULE (ASA RULE )

Two triangles are congruent if any two angles and the included side of one triangle is equal to the two angles and the included side of the other triangle.

EXAMPLE : (in fig. 1.4)

GIVEN: ABC= DEF,

ACB= DFE,

BC = EF

TO PROVE : ▲ABC = ▲DEF

ABC = DEF, (GIVEN)

ACB = DFE, (GIVEN)

BS = EF (GIVEN)

▲ABC = ▲DEF (BY ASA RULE)

A

B C

D

E F

Fig. 1.4

3. ANGLE – ANGLE – SIDE RULE (AAS RULE)

Two triangles are congruent if two angles and a side of one triangle is equal to the two angles and one a side of the other.

EXAMPLE: (in fig. 1.5)

GIVEN: IN ▲ ABC & ▲DEF

B = E

A= D

BC = EF

TO PROVE :▲ABC = ▲DEF

B = E

A = D

BC = EF

▲ABC = ▲DEF (BY AAS RULE)

D

E F

A

B C

Fig. 1.5

4. SIDE – SIDE – SIDE RULE (SSS RULE)

Two triangles are congruent if all the three sides of one triangle are equal to the three sides of other triangle.

Example: (in fig. 1.6)

Given: IN ▲ ABC & ▲DEF

AB = DE , BC = EF , AC = DF

TO PROVE : ▲ABC = ▲DEF

AB = DE (GIVEN )

BC = EF (GIVEN )

AC = DF (GIVEN )

▲ABC = ▲DEF (BY SSS RULE)

D

E F

A

B C

Fig. 1.6

5. RIGHT – HYPOTENUSE – SIDE RULE (RHS RULE )

Two triangles are congruent if the hypotenuse and the side of one triangle are equal to the hypotenuse and the side of other triangle.

EXAMPLE : (in fig 1.7)

GIVEN: IN ▲ ABC & ▲DEF

B = E = 900 , AC = DF , AB = DE

TO PROVE : ▲ABC = ▲DEF

B = E = 900 (GIVEN)

AC = DF (GIVEN)

AB = DE (GIVEN)

▲ABC = ▲DEF (BY RHS RULE)

D

E F

A

B C

900

900

Fig. 1.7

1. The angles opposite to equal sides are always equal.

Example: (in fig 1.8)

Given: ▲ABC is an isosceles triangle in which AB = AC

TO PROVE: B = C

CONSTRUCTION : Draw AD bisector of BAC which meets BC at D

PROOF: IN ▲ABC & ▲ACD

AB = AD (GIVEN)

BAD = CAD (GIVEN)

AD = AD (COMMON)

▲ABD = ▲ ACD (BY SAS RULE)

B = C (BY CPCT)

A

B D C

Fig. 1.8

2. The sides opposite to equal angles of a triangle are always equal.

Example : (in fig. 1.9)

Given : ▲ ABC is an isosceles triangle in which B = C

TO PROVE: AB = AC

CONSTRUCTION : Draw AD the bisector of BAC which meets BC at D

Proof : IN ▲ ABD & ▲ ACD

B = C (GIVEN)

AD = AD (GIVEN)

BAD = CAD (GIVEN)

▲ ABD = ▲ ACD (BY ASA RULE)

AB = AC (BY CPCT)

A

B D C

Fig. 1.9

When two quantities are unequal then on comparing these quantities we obtain a relation between their measures called “ inequality “ relation.

Theorem 1 . If two sides of a triangle are unequal the larger side has the greater angle opposite to it. Example: (in fig. 2.1)

Given : IN ▲ABC , AB>AC

TO PROVE : C = B

Draw a line segment CD from vertex such that AC = AD

Proof : IN ▲ACD , AC = AD

ACD = ADC --- (1)

But ADC is an exterior angle of ▲BDC

ADC > B --- (2)

From (1) &(2)

ACD > B --- (3)

ACB > ACD ---4

From (3) & (4)

ACB > ACD > B , ACB > B ,

C > B

A

B

D

C

Fig. 2.1

THEOREM 2. In a triangle the greater angle has a large side opposite to it

Example: (in fig. 2.2)

Given: IN ▲ ABC B > C

TO PROVE : AC > AB

PROOF : We have the three possibility for sides AB and AC of ▲ABC

(i) AC = AB

If AC = AB then opposite angles of the equal sides are equal than

B = C

AC ≠ AB

(ii) If AC < AB

We know that larger side has greater angles opposite to it.

AC < AB , C > B

AC is not greater then AB

(iii) If AC > AB

We have left only this possibility AC > AB

A

CB

Fig. 2.2

THEOREM 3. The sum of any two angles is greater than its third side

Example (in fig. 2.3) TO PROVE : AB + BC > AC

BC + AC > AB

AC + AB > BC

CONSTRUCTION: Produce BA to D such that AD + AC .

Proof: AD = AC (GIVEN)

ACD = ADC (Angles opposite to equal sides are equal )

ACD = ADC --- (1)

BCD > ACD ----(2)

From (1) & (2) BCD > ADC = BDC

BD > AC (Greater angles have larger opposite sides )

BA + AD > BC ( BD = BA + AD)

BA + AC > BC (By construction)

AB + BC > AC

BC + AC >AB

A

CB

D

Fig. 2.3

THEOREM 4. Of all the line segments that can be drawn to a given line from an external point , the perpendicular line segment is the shortest.

Example: (in fig 2.4)

Given : A line AB and an external point. Join CD and draw CE AB

TO PROVE CE < CD

PROOF : IN ▲CED, CED = 900

THEN CDE < CED

CD < CE ( Greater angles have larger side opposite to them. )

BA

C

ED Fig. 2.4

1. If the altitude from one vertex of a triangle bisects the opposite side, then the triangle is isosceles triangle.Example : (in fig.2.5)

Given : A ▲ABC such that the altitude AD from A on the opposite side BC bisects BC i. e. BD = DC

To prove : AB = AC

SOLUTION : IN ▲ ADB & ▲ADC

BD = DC

ADB = ADC = 900

AD = AD (COMMON )

▲ADB = ▲ ADC (BY SAS RULE )

AB = AC (BY CPCT)

A

CDBFig. 2.5

THEOREM 2. In a isosceles triangle altitude from the vertex bisects the base .

EXAMPLE: (in fig. 2.6)

GIVEN: An isosceles triangle

AB = AC

To prove : D bisects BC i.e. BD = DC

Proof:

IN ▲ ADB & ▲ADC

ADB = ADC

AD = AD

B = C ( AB = AC ; B = C)

▲ADB = ▲ ADC

BD = DC (BY CPCT)

A

CDB

Fig. 2.6

THEOREM 3. If the bisector of the vertical angle of a triangle bisects the base of the triangle, then the triangle is isosceles. EXAMPLE: (in fig. 2.7)

GIVEN: A ▲ABC in which AD bisects A meeting BC in D such that BD = DC, AD = DE

To prove : ▲ABC is isosceles triangle .

Proof: In ▲ ADB & ▲ EDC

BD = DC

AD = DE

ADB = EDC

▲ADB = ▲EDC

AB = EC

BAD = CED (BY CPCT)

BAD = CAD (GIVEN)

CAD = CED

AC = EC (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL)

AC = AB , HENCE ▲ABC IS AN ISOSCELES TRIANGLE.

E

D CB

A

Fig. 2.7

QUES. IN FIG. AB = AC, D is the point in interior of ▲ABC such that

DBC = DCB. Prove that AD bisects BAC of ▲ ABC

SOLUTION: IN ▲BDC,

DBC = DCB (GIVEN)

DC = DB (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL.)

IN ▲ ABD & ▲ ACD

AB = AC (GIVEN)

BD = CD (GIVEN)

AD = AD (COMMON)

▲ABD = ▲ACD (BY SSS RULE)

BAD = CAD (BY CPCT)

Hence, AD is the bisector of ▲BAC

A

D

B C

QUES. In fig. PA AB , QB AB & PA = QB.If PQ intersects AB at O, show that o is the mid point of AB as well as that of PQ

SOLUTION: In ▲OAP & ▲OBQ

PA = QB (GIVEN)

PAO = OBQ = 900 & AOP = BOQ

▲ OAP = ▲OBQ (BY AAS RULE)

OA = OB ( BY CPCT)

OP = OQ ( BY CPCT)

O BA

P

Q

QUES. In fig. PS = PR, TPS = QPR.Prove that PT = PQ

SOLUTION: IN ▲PRS

PS = PR

PRS = PSR (ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL)

1800 - PRS = 1800 - PSR

PRQ = PST

Thus , in ▲ PST & ▲PRQ

TPS = QTR

PS = PR

PST = PRQ

▲PRT = ▲PSQ ( BY ASA RULE)

PT = PQ (BY CPCT)

TS R

Q

P

QUES. In fig. BM and DN are both perpendicular to the line segment AC and BM = DN. Prove that AC bisects BD.

SOLUTION IN ▲BMR and ▲DNR

BMR = DNR (GIVEN)

BRM = DRN (VERTICALLY OPPOSITE ANGLES)

BM = DN (GIVEN)

▲ BMR = ▲DNR (BY AAS RULE)

BR = DR

R is the mid point of BD

Hence AC bisects BD

B A

CD

M

N

R

Deduction 1: All angles at the circumference on the same arc are equal in measure.

To prove: bac = bdc

Proof: 3 = 2 1

Angle at the centre is twice the angle on the circumference (both

on the arc bc)3 = 2 2 Angle at the centre is twice the angle on the circumference (both

on arc bc) 2 1 = 2 2

1 = 2

i.e. bac = bdc

Q.E.D.

a

bc

d

3

1 2

.o

QUES. In fig . It is given that BC = CE and 1 = 2. Prove that

▲ GCB = ▲ DCE

SOLUTION:

IN ▲ GCB & ▲ DCE

1 + GBC = 1800 (LINEAR PAIR )

2 + DEC = 1800 ( LINEAR PAIR)

1 + GBC = 2 + DEC -- (1) GBC = DEC (From (1) )

GBC = DEC (GIVEN)

BC = CE (GIVEN)

GCB = DCE (VERTICALLY OPPOSITE ANGLES )

▲GCB = ▲ DCE (BY ASA RULE)

*----- *---- *------ * -----* ----- * ----* ---- *

A

B

C

F

G

E

D

1 2

a

b

c Ld∟∟.

Theorem: A line through the centre of a circle perpendicular to a chord bisects the chord.

Given:

Circle, centre c, a line L containing c, chord [ab], such that L ab and L ab = d.

To prove:

ad = bdConstructi

on:Label right angles 1 and

2.Proof: 1 = 2 =

90 Given

ca = cb

Both radii cd =

cd

commonR H S Corresponding sides

Consider cda and cdb:

cda cdb ad =

bd

1

2

Q.E.D.

b

a

c

d

e f

2

1

2

3

1 3

Theorem: If two triangles are equiangular, the lengths of the corresponding sides are in proportion.

Given :

Two triangles with equal angles.

To prove: |df|

|ac|=

|de|

|ab|

|ef|

|bc|=

Construction:

On ab mark off ax equal in length to de. On ac mark off ay equal to df and label the angles 4 and 5.

Proof: 1 = 4

[xy] is parallel to [bc]

|ay|

|ac|=

|ax|

|ab|As xy is parallel to bc.

|df|

|ac|=

|de|

|ab|Similarly

|ef|

|bc|=

x y4 5

Q.E.D.

Theorem: In a right-angled triangle, the square of the length of the side opposite to the right angle is equal to the sum of the squares of the other two sides.

Q.E.D.

cb

a

c

b

a

b

a c

b

a

c

1

2

3

45

To prove that angle 1 is 90º

Proof:3+ 4+ 5 = 180º ……Angles in a triangle

But 5 = 90º => 3+ 4 = 90º

=> 3+ 2 = 90º ……Since 2 = 4

Now 1+ 2+ 3 = 180º ……Straight line

=> 1 = 180º - ( 3+ 2 )

=> 1 = 180º - ( 90º ) ……Since 3+ 2 already proved to be 90º

=> 1 = 90º

The HypotenuseThe Hypotenuse

The hypotenuse The hypotenuse of a right triangle of a right triangle is the triangle's is the triangle's longest side, i.e., longest side, i.e., the side opposite the side opposite the right angle.the right angle.

To day we are teaching aboutTo day we are teaching about

Pythagorean Theorem

Pythagoras (~580-500 B.C.)

He was a Greek philosopher responsible for important developments in mathematics,

astronomy and the theory of music.

1. cut a triangle with base 4 cm and height 3 cm

0 1 2 3 4 5

4 cm

01

23

45

3

cm

2. measure the length of the hypotenuse

0

1

2

3

4

5

Now take out a square paper and a ruler.

5 cm

Consider a square PQRS with sides a + ba

a

a

a

b

b

b

b cc

cc

Now, the square is cut into

- 4 congruent right-angled triangles and

- 1 smaller square with sides c

Proof of Pythagoras’ Theorem

P Q

R S

a + b

a + b

A B

CD

Area of square ABCD

= (a + b) 2

b

b

a b

b

a

a

a

cc

cc

P Q

RS

Area of square PQRS

= 4 + c 2ab

2

a 2 + 2ab + b 2

= 2ab + c 2

a2 + b

2 = c

2

Theorem states that:"The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas

of the squares upon the remaining sides."

The Pythagorean Theorem asserts that for a right triangle, the square of the hypotenuse is equal to the sum of the

squares of the other two sides: a2 + b2 = c2

The figure above at the right is a visual display of the theorem's conclusion. The figure at the left contains a

proof of the theorem, because the area of the big, outer, green square is equal to the sum of the areas of the four

red triangles and the little, inner white square: c2 = 4(ab/2) + (a - b)2 = 2ab + (a2 - 2ab + b2) = a2 +

b2

Animated Proof of the Pythagorean TheoremBelow is an animated proof of the Pythagorean Theorem. Starting with a right triangle and squares on each side, the middle size square is cut into congruent quadrilaterals (the cuts through the center and parallel to the sides of the biggest square). Then the

quadrilaterals are hinged and rotated and shifted to the big square. Finally the smallest square is translated to cover the remaining middle part of the biggest square. A perfect

fit! Thus the sum of the squares on the smaller two sides equals the square on the biggest side.

Afterward, the small square is translated back and the four quadrilaterals are directly translated back to their

original position. The process is repeated forever.

Pythagorean TheoremOver 2,500 years ago, a Greek mathematician named Pythagoras developed a

proof that the relationship between the hypotenuse and the legs is true for all right triangles.

In any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs."

This relationship can be stated as:

and is known as the

 Pythagorean Theorem

  a, b are legs.c is the hypotenuse (across

from the right angle).

There are certain sets of numbers that have a very special property.  Not only do these numbers satisfy the Pythagorean Theorem, but any multiples of these

numbers also satisfy the Pythagorean TheoremFor example:  the numbers 3, 4, and 5 satisfy the Pythagorean Theorem.  If you multiply all three numbers by 2  (6, 8, and 10), these new numbers ALSO

satisfy the Pythagorean theorem.    

If we think about a right triangle we know of course that one of the angles is a right angle. We also know that the other two angles are acute angles (why?). In fact we know that the other two angles are complementary angles. Therefore there is a relationship between the sizes of the angles that

the two acute angles have measures that add up to ninety degrees. What about sides? Is there a relationship between the sides of a right triangle? We know from

previous lessons that if we have the lengths of just two of the sides we can construct the triangle so it is enough to know the lengths of two sides to determine the length of the third side. We shall now try to figure out the relationship. We shall, to make it easy to communicate assume that the

length of the hypotenuse is c units and that the two legs are of length a and b units. So according to the Pythagorean Theorem, the area of square A, plus the area of square B should equal the area of square C.

The special sets of numbers that possess this property are called 

Pythagorean Triples. The most common Pythagorean Triples are:

3, 4, 5

5, 12, 13

8, 15, 17

The Pythagorean Theorem                                                                                                          

The Pythagorean Theorem is one of Euclidean Geometry's most beautiful theorems. It is simple, yet obscure, and is used continuously in mathematics and physics. In short, it is really cool.

This first method is one of the ways the Pythagoreans would have proved the theorem. Unfortunately, it lacks glamour. In the following picture let ABC be a right triangle and BD be a

segment drawn perpendicular to AC.

Since the triangles are similar, the sides must be of proportional lengths. •AB/AD=AC/AB, or AB x AB = AD x AC •BC/CD=AC/BC, or BC x BC= AC x CD

•Then, adding the two together, BC^2 + AB^2 = (AD + DC) x AC= AC^2•

Pythagorean Theorem in text Pythagorean Theorem in text bookbook

of 10of 10thth class classGiven:- ABC is a right angle TriangleGiven:- ABC is a right angle Triangle

. .

angle B =90angle B =9000

R.T.P:- ACR.T.P:- AC22 = AB = AB2 2 +BC+BC22

Construction:- To draw BD Construction:- To draw BD AC . AC .

A

B C

D

Proof:- In Triangles ADB and ABC AngleA=Angle A (common)

Angle ADB=Angle ABC (each 900 ) ADB ~ ABC ( A.A corollary )

So that AD/AB=AB/AC (In similar triangles corresponding sides are proportional )

AB2 = AD X AC _________(1) Similarly BC2 = DCXAC _________(2)

Adding (1)&(2) we get AB2 +BC2 = AD X AC + DCXAC

= AC (AD +DC) = AC . AC

=AC2

There fore AB2 +BC2 =AC2

Typical Examples

Example 1. Find the length of AC.

Hypotenuse

AC2 = 122 + 162 (Pythagoras’ Theorem)AC2 = 144 + 256

AC2 = 400AC = 20

A

CB

16

12Solution :

Example 2. Find the length of diagonal d .

10

24 d

Solution:

d2 = 102 + 242 (Pythagoras’ Theorem)

d

10 24

26

2 2

676

16km

12km

1.A car travels 16 km from east to west. Then it turns left and travels a further 12 km. Find the displacement between the starting point and

the destination point of the car.

N

?

Application of Pythagoras’ Theorem

16 km

12 km

AB

C

Solution :

In the figure,AB = 16BC = 12

AC2 = AB2 + BC2 (Pythagoras’ Theorem)AC2 = 162 + 122

AC2 = 400AC = 20

The displacement between the starting point and the destination point of the car is 20 km

2. The height of a tree is 5 m. The distance between the top of it and the tip of its shadow is 13 m.

Solution:132 = 52 + L2 (Pythagoras’ Theorem)

L2 = 132 - 52

L2 = 144L = 12

Find the length of the shadow L.

5 m13 m

L