area of triangles and parallelograms

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Area of tri angles and parallelograms

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Page 1: Area of triangles and parallelograms

Area of triangles

and parallelograms

Page 2: Area of triangles and parallelograms

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

Page 3: Area of triangles and parallelograms

Nature teaches us….

It is from nature that man has learnt everything and even maths…. Here are some of the triangles that occur in nature….

Page 4: Area of triangles and parallelograms

But man has used his intelligence to apply the shapes of nature ….

Page 5: Area of triangles and parallelograms

In Euclidean geometry, a parallelogram is a convex quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

What is a parallelogram?

Page 6: Area of triangles and parallelograms

Here are some parallelograms that occur in real life…

Page 7: Area of triangles and parallelograms

Understanding area of figures

Area is a count of how many unit squares fit inside a figure. To fully understand this classic definition of area, we need to picture the unit square. A unit square is a square that is one unit long by one unit wide. It can be 1'x1', 1 m x 1 m, 1 yd x 1 yd, 1" x 1", ...

1

1

Page 8: Area of triangles and parallelograms

Area of triangles

Area of triangle is always calculated by half its base multiplied to its corresponding height…

TriangleArea = ½b × h

b = baseh = vertical height

Page 9: Area of triangles and parallelograms

Area of parallelograms

The area of the parallelogram is got by multiplying its base to its corresponding altitude (height)…

ParallelogramArea = b × h

b = baseh = vertical

height

Page 10: Area of triangles and parallelograms

UNDERSTANDING CONGRUENCE…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.

Page 11: Area of triangles and parallelograms

NATURE TEACHES US…As always , nature is the best teacher.. Nature teaches us congruence by means of reflections and through BILATERAL SYMMETRY etc…Mostly, everything in this world is bilaterally symmetrical especially we, the humans.

Page 12: Area of triangles and parallelograms

CONGRUENCE IN MATHEMATICSIf 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

Page 13: Area of triangles and parallelograms

CONGRUENCE IN TRIANGLESCongruent 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.

Page 14: Area of triangles and parallelograms

SOME THEOREMS RELATED TO CONGRUENCE

If two figures are congruent , they have equal areas.

But its converse IS NOT TRUE.

Two figures having equal areas need not be congruent.

A B

Figures A and B are congruent and hence they have equal areas.

D

CFigures C and D have equal areas, but they are not congruent.

Page 15: Area of triangles and parallelograms

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

Page 16: Area of triangles and parallelograms

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.

Page 17: Area of triangles and parallelograms

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

Page 18: Area of triangles and parallelograms

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.

Page 19: Area of triangles and parallelograms

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

Page 20: Area of triangles and parallelograms

PRESENTED BY

K.M.Bharthe 9 B

THANK YOU