SUBMITTED BY-: HANIESH

CLASS-: 9th B

SUBJECT-: MATHS

A triangle is one of the basic shapes of

geometry: a polygon with three corners or

vertices and three sides or edges which are

line segments.

In Euclidean geometry any three non-

collinear points joined together with three

line segments is called a triangle…

What is a Triangle?

A

B

C

Area of triangle is always calculated by

half its base multiplied to its corresponding

height…

TriangleArea = ½b × h

b = baseh = vertical height

TRIANGLES

1.Triangles on the same base and between the

same parallels are equal in area.

2.Triangles having equal areas and having one side

of one of the triangles equal to one side of the other

have their corresponding altitudes equal.

3.Two triangles having the same base and equal

areas lie between the same parallels.

The area of the parallelogram is got by

multiplying its base to its

corresponding altitude (height)…

ParallelogramArea = b × h

b = baseh = vertical height

PARALLELOGRAMS 1.Parallelograms on the same base and the between the

same parallels are equal in area.

2.A diagonal of a parallelogram divides it into two triangles of equal area.

3.Parallelograms and a rectangle on the same base and between the same parallels are equal in area.

PARALLELOGRAMS

4.The area of a parallelogram is the product of its base and the corresponding altitude.

5.The area of a triangle is half he product of any of its side and the corresponding altitude.

6.If a triangle and a parallelogram are on the same base and between he same parallels,the area of the triangle is equal to half of he parallelogram.

If two figures have the same shape and the

same size, then they are said to be

congruent figures…

Congruent figures are exact duplicates of

each other. One could be fitted over the

other so that their corresponding parts

coincide.

The concept of congruent figures applies to

figures of any type.

If two figures have the same shape and

the same size, then they are said to be

congruent figures.

For example, rectangle ABCD and

rectangle PQRS are congruent

rectangles as they have the same shape

and the same size.

Side AB and side PQ are in the same

relative position in each of the figures.

We say that the side AB and side PQ

are corresponding sides.

DC

BA

R

QP

S

Congruent triangles have the same size

and the same shape. The corresponding

sides and the corresponding angles of

congruent triangles are equal.

A B

C

D E

F

There are five

types of

congruence rules.

They are SSS, SAS,

AAS=SAA, ASA and

RHS congruence

Rules.

THEOREM 1

Given: Two parallelograms ABCD and EFCD On the same base DC and between the same

parallels AF and DC.

TO PROVE: ar(ABCD)=ar(EFCD)

PROOF: In ADE and BCF,

Angle DAE= angle CBF (corresponding angles)

Angle AED= Angle BFC (corresponding angles)

Angle ADE= Angle BCF (angle sum property)

Also, AD =BC (opposite sides of the parallelogram ABCD)

Triangle ADE is congruent to triangle BCF (By ASA rule)

Therefore, ar(ADE)=ar(BCF) (congruent figures have equal areas)

Now, ar (ABCD) = ar(ADE) + ar(EDCB)

ar (ABCD) = ar (BCF)=ar(EDCB)

ar (ABCD) = ar (EFCD)

Hence they are equal in area

D C

A E B F

CONCLUSION OF THE THEOREM 1

Parallelograms on the same base or equal bases and

lie between the same parallels are equal in area.

Parallelograms on the same base or equal bases

that have equal areas lie between the same

parallels.

Parallelograms which lie between same parallels

and have equal area lie on the same base.

THEOREM 2

GIVEN : ABCD is a parallelogram, AC is diagonal, Two

Triangles ADC and ABC on the same base AC and between the

Same parallels BC and AD

TO PROVE : ar (Δ ABC ) = ar (Δ CDA )

Proof: In triangles ABC , CDA

AB = CD (opposite sides of a parallelogram )

BC = DA (opposite sides of a parallelogram )

AC = AC (common )

Therefore Δ ABC Δ CDA (S-S-S congruency)

There fore Δ ABC = Δ CDA (if two triangles are congruent then their areas are equal ).

A B

CD

CONCLUSION OF THEOREM 2

Triangles on the same base and between the same

parallels are equal in area.

Triangles on the same base and having equal areas

lie between the same parallels.

Triangles between the same parallels and have

equal area lie on the same base.

THEOREM 3

IF A TRIANGLE AND A PARALLELOGRAM LIE ON THE SAME BASE AND BETWEEN THE SAME PARALLELS, THEN THE AREA OF THE TRIANGLE IS EQUAL TO HALF THE AREA OF THE PARALLELOGRAM.

Given: triangle ABP and parallelogram ABCD on the same base AB and between the same parallels AB and PC.

To prove: ar(PAB)=1/2 * ar (ABCD)

Solution: Draw BQ//AP to get another parallelogram ABQP. Now parallelograms ABQP and ABCD are on the same base AB and between the same parallels AB and PC

Therefore ar (ABQP) = ar (ABCD) [By theorem 1]……(1)

But triangle PAB is congruent to triangle BQP (since diagnol PB divides parallelogram ABQP into two congruent triangles)

So, ar (PAB) = ar (BQP)…………(2)

Therefore ar (PAB) = ½ * ar (ABQP) ………….(3)

This gives ar (PAB) = ½ * ar (ABCD)…………..(FROM 1 AND 3)

P D Q C

A B

THANK

YOU