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Area of parallelogram and triangles
INTRODUCTION
In earliar classes ,we have learnt some formulae for
finding the areas of different plane figures such as
triangles, rectangle square etc……In this chapter ,we
will consolidate the knowledge about these formulae by
studying some relationship between the areas of these
figure under the condition when the lie on the same base
and between the same parallels. So, let us first
understand the figures on the same base and between
the same parallels.
PARALLELOGRAMS ON THE SAME BASE AND
BETWEEN THE SAME PARALLELS----
Let us look at the figures-----
A
B
C
D
U
A
B
C
D
P
U
In triangle PAB and parallelogram ABCD are on the same base
AB.
Similarly in parallelogram ABCD and ABPQ are on the same base AB
Figures on the same base and between the same parallels.
P
S
M Q N
R
In parallelogram PQRS and MNRS are on the same base SR and
between the same parallels PN and SR.
THEOREM 1----
PARALLELOGRAMES ARE ON THE SAME BASE
AND BETWEEN THE SAME PARALLELS ARE EQAL
IN AREA .
PROOF = Two parallelograms ABCD and EFCD, on the
same base DC and between the same parallels AF AND AD
are given---
We need to prove that ar(ABCD)=ar (EFCD)
In traingle ADE and traingle BSF
DAE=CBF
AED=BFC
Therefore, ADE=BCF (angle some property of a triangle)
Also AD= BC
So, tr. ADE congruent to tr. BCF (by ASA)
Therefore, ar (ADE=ar(BCF) (congruent figures have equal
areas.
Now ar (ABCD)=ar(ADE)+ar(EDCB)
=ar(BCF)+ar(EDCB)
=ar(EFCD)
So, parallelograms ABCD and EFCD are equal in area.
Triangles on the same base and between the same parallels
AA PP
BB CC
We have 2 triangles ABC and
PBC on the same base BC and
between the same parallels BC
and AP
What can you say about the areas of such triangles? To answer
this question , you may perform the activity of drawing several
pairs of triangles on the same base and between the same
parallels on the graph sheet and find there
Areas by the method of counting the squares
To obtain a logical answer we may proceed as follows =
AA
BB
CC
PP DD
RR
In fig. draw CD BA
CR BP such that D
and R lie on line AP .
From this , we obtain 2 parallelograms PBCR & ABCD
on the same base & between the same parallels BC&AR.
Therefore , ar(ABCD)=ar (PBCR)
Now, tr. ABC = tr. CDA and tr. PBC = tr. CRP
Therefore, AR(ABC) =AR (PBC)
In this way, we have arrived at the following theorem
Theorem 2---
The triangles on the same base (or equal base) & between the same
parallels are equal in area.
Theorem 3---
Two triangles having the same base (or equal bases) & equal
areas lie on the same parallels.
Let us now take some examples---
EX 1= In fig. ,ABCD is a parallelogram
& EFCD is a rectangle. Also AL
perpendicular to DC . Prove that—
1. ar (ABCD)=ar (EFCD)
EE
DD LL CC
FF BBAA
2. ar(ABCD)=DC AL
Solution--- 1.As a rectangle is also a parallelogram
therefore, ar(ABCD)=ar(EFCD) {by theorem 1}
2. From above result,
Ar (ABCD) =DC FCAr (ABCD) =DC FC
As AL perpendicular to DC therefore ,AFCL is also a rectangle.
So, AL=FC
Therefore, ar (ABCD)= DC AL
Practice questionPractice question
1.  What is the area of a triangle with base 10 cm and
height 25 cm?
A. 35 sq cm
B. 62.5 sq cm
C. 125 sq cm
D. 250 sq cm
 2.  What is the area of a parallelogram with base
30 ft and height 25 ft?
  A. 55 sq ft  
 B. 187.5 sq ft  
 C. 375 sq ft 
 D. 750 sq ft
 3.  Find the perimeter of a regular pentagon with
side length of 14 mm.
 A. 42 mm
 B. 42 sq mm
 C. 70 mm
 D. 70 sq mm
 4.  Find the area of a triangle with base of 4 in
and height of 6 in.
 A. 12 sq in
 B. 24 sq in
 C. 12 in
 D. 24 in
5. The area of a parallelogram with base of 40 cm
and height of 0.5 cm is _________.
 A. 200 sq cm
 B. 20 sq cm
 C. 10 sq cm
 D. 20 cm
                
answersanswers
 1-c1-c
 2-d2-d
 3-d3-d
 4-a4-a
 5-b5-b
Area of a ParallelogramArea of a Parallelogram

The area of a polygon is the number of square units inside the polygon.The area of a polygon is the number of square units inside the polygon.
Area is 2-dimensional like a carpet or an area rug. A parallelogram is aArea is 2-dimensional like a carpet or an area rug. A parallelogram is a
4-sided shape formed by two pairs of parallel lines. Opposite sides are4-sided shape formed by two pairs of parallel lines. Opposite sides are
equal in length and opposite angles are equal in measure. To find theequal in length and opposite angles are equal in measure. To find the
area of a parallelogram, multiply the base by the height. The formula is:area of a parallelogram, multiply the base by the height. The formula is:
 where is the base, is the height, and where is the base, is the height, and ·· means multiply. The base andmeans multiply. The base and
height of a parallelogram must be perpendicular. However, the lateralheight of a parallelogram must be perpendicular. However, the lateral
sides of a parallelogram are not perpendicular to the base. Thus, asides of a parallelogram are not perpendicular to the base. Thus, a
dotted line is drawn to represent the height. Let's look at some examplesdotted line is drawn to represent the height. Let's look at some examples
involving the area of a parallelogram.involving the area of a parallelogram.
Area of triangleArea of triangle
 The area of a polygon is the number of square units inside that polygon. Area is 2-dimensionalThe area of a polygon is the number of square units inside that polygon. Area is 2-dimensional
like a carpet or an area rug. Alike a carpet or an area rug. A triangletriangle is a three-sided polygon. We will look at several typesis a three-sided polygon. We will look at several types
of triangles in this lesson.of triangles in this lesson.
To find the area of a triangle, multiply the base by the height, and then divide by 2. The divisionTo find the area of a triangle, multiply the base by the height, and then divide by 2. The division
by 2 comes from the fact that a parallelogram can be divided into 2 triangles. For example, inby 2 comes from the fact that a parallelogram can be divided into 2 triangles. For example, in
the diagram to the left, the area of each triangle is equal to one-half the area of thethe diagram to the left, the area of each triangle is equal to one-half the area of the
parallelogram.parallelogram.
Since the area of a parallelogram is , the area of a triangle must be one-half the area of aSince the area of a parallelogram is , the area of a triangle must be one-half the area of a
parallelogram. Thus, the formula for the area of a triangle is:   or   where is the base, is theparallelogram. Thus, the formula for the area of a triangle is:   or   where is the base, is the
height andheight and ·· means multiply. The base and height of a triangle must be perpendicular to eachmeans multiply. The base and height of a triangle must be perpendicular to each
other. In each of the examples below, the base is a side of the triangle. However, depending onother. In each of the examples below, the base is a side of the triangle. However, depending on
the triangle, the height may or may not be a side of the triangle. For example, in the rightthe triangle, the height may or may not be a side of the triangle. For example, in the right
triangle in Example 2, the height is a side of the triangle since it is perpendicular to the base. Intriangle in Example 2, the height is a side of the triangle since it is perpendicular to the base. In
the triangles in Examples 1 and 3, the lateral sides are not perpendicular to the base, so a dottedthe triangles in Examples 1 and 3, the lateral sides are not perpendicular to the base, so a dotted
line is drawn to represent the height.line is drawn to represent the height.
 Definition of a Parallelogram
 A parallelogram is a quadrilateral that has two pairs of parallel sides. Properties of Parallelograms
 1. The opposite sides are equal in length.
 2. The opposite angles are congruent.
 3. The diagonals bisect each other.
 Conditions to be a Parallelogram
 If a quadrilateral has one of the following conditions, then it is a parallelogram.
 1. If a quadrilateral has two pairs of parallel sides, then it is a parallelogram.
 2. If a quadrilateral has two pairs of opposite sides of the same lengths, then it is a parallelogram. Drag point A.
 3. If two pairs of opposite angles of a quadrilateral are the same, then the figure is a parallelogram.
 4. If the diagonal of a quadrilateral bisect each other, the figure is a parallelogram. Drag point A and point B.
Point C is point symmetry with point A about point O.
Point D is point symmetry with point B about point O.
 5. If a quadrilateral has one pair of opposite sides which are parallel and equal in length, then the figure is a parallelogram. AD
and BC are parallel and equal in length.
Drag point A and point B.
 We know that Area of a parallelogram is equal to the product of its base and
height.
The diagonals of a parallelogram bisect each other.
A diagonal of the parallelogram divides it into two congruent triangles.
Opposite sides and angles of a parallelogram are equal.
Parallelograms on the same base and between the same parallels are equal in area.
Parallelograms on the same base that have equal areas are between the same
parallels.
All figures that are on the same base and between same parallels need not have equal
areas.
Two congruent figures have equal areas but the converse need not be true.
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.
If a parallelogram and a triangle are on the same base and they are between the
same parallels, then area of the triangle is half the area of the parallelogram.
 Two figures are said to be on the same base and between the same
parallel lines if they have a common base and the vertices
opposite to this common base of each figure lie on a line parallel to
the base. Out of these two parallel lines, one should be the line
containing the common base and the other is the line passing
through the vertices of both the figures opposite to the base. We
know that any two congruent figures have same area but the
converse is not true. Median of a triangle divides it into two
triangles of equal area. Two triangles with same base (or equal
bases) and equal areas will have Area of triangle is half the
product of the base and the corresponding altitude. Also,
Triangles on the same base and between the same parallels
are equal in area. equal corresponding altitudes. Triangles
on the same base and having equal areas lie between the same
parallels.

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CLASS IX MATHS 6 areas of parallelogram and triangles

  • 1. Area of parallelogram and triangles
  • 2. INTRODUCTION In earliar classes ,we have learnt some formulae for finding the areas of different plane figures such as triangles, rectangle square etc……In this chapter ,we will consolidate the knowledge about these formulae by studying some relationship between the areas of these figure under the condition when the lie on the same base and between the same parallels. So, let us first understand the figures on the same base and between the same parallels.
  • 3. PARALLELOGRAMS ON THE SAME BASE AND BETWEEN THE SAME PARALLELS---- Let us look at the figures----- A B C D U A B C D P U In triangle PAB and parallelogram ABCD are on the same base AB.
  • 4. Similarly in parallelogram ABCD and ABPQ are on the same base AB Figures on the same base and between the same parallels. P S M Q N R In parallelogram PQRS and MNRS are on the same base SR and between the same parallels PN and SR.
  • 5. THEOREM 1---- PARALLELOGRAMES ARE ON THE SAME BASE AND BETWEEN THE SAME PARALLELS ARE EQAL IN AREA . PROOF = Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF AND AD are given--- We need to prove that ar(ABCD)=ar (EFCD) In traingle ADE and traingle BSF DAE=CBF AED=BFC Therefore, ADE=BCF (angle some property of a triangle)
  • 6. Also AD= BC So, tr. ADE congruent to tr. BCF (by ASA) Therefore, ar (ADE=ar(BCF) (congruent figures have equal areas. Now ar (ABCD)=ar(ADE)+ar(EDCB) =ar(BCF)+ar(EDCB) =ar(EFCD) So, parallelograms ABCD and EFCD are equal in area.
  • 7. Triangles on the same base and between the same parallels AA PP BB CC We have 2 triangles ABC and PBC on the same base BC and between the same parallels BC and AP What can you say about the areas of such triangles? To answer this question , you may perform the activity of drawing several pairs of triangles on the same base and between the same parallels on the graph sheet and find there
  • 8. Areas by the method of counting the squares To obtain a logical answer we may proceed as follows = AA BB CC PP DD RR In fig. draw CD BA CR BP such that D and R lie on line AP . From this , we obtain 2 parallelograms PBCR & ABCD on the same base & between the same parallels BC&AR.
  • 9. Therefore , ar(ABCD)=ar (PBCR) Now, tr. ABC = tr. CDA and tr. PBC = tr. CRP Therefore, AR(ABC) =AR (PBC) In this way, we have arrived at the following theorem Theorem 2--- The triangles on the same base (or equal base) & between the same parallels are equal in area.
  • 10. Theorem 3--- Two triangles having the same base (or equal bases) & equal areas lie on the same parallels. Let us now take some examples--- EX 1= In fig. ,ABCD is a parallelogram & EFCD is a rectangle. Also AL perpendicular to DC . Prove that— 1. ar (ABCD)=ar (EFCD) EE DD LL CC FF BBAA 2. ar(ABCD)=DC AL
  • 11. Solution--- 1.As a rectangle is also a parallelogram therefore, ar(ABCD)=ar(EFCD) {by theorem 1} 2. From above result, Ar (ABCD) =DC FCAr (ABCD) =DC FC As AL perpendicular to DC therefore ,AFCL is also a rectangle. So, AL=FC Therefore, ar (ABCD)= DC AL
  • 12. Practice questionPractice question 1.  What is the area of a triangle with base 10 cm and height 25 cm? A. 35 sq cm B. 62.5 sq cm C. 125 sq cm D. 250 sq cm
  • 13.  2.  What is the area of a parallelogram with base 30 ft and height 25 ft?   A. 55 sq ft    B. 187.5 sq ft    C. 375 sq ft   D. 750 sq ft
  • 14.  3.  Find the perimeter of a regular pentagon with side length of 14 mm.  A. 42 mm  B. 42 sq mm  C. 70 mm  D. 70 sq mm
  • 15.  4.  Find the area of a triangle with base of 4 in and height of 6 in.  A. 12 sq in  B. 24 sq in  C. 12 in  D. 24 in
  • 16. 5. The area of a parallelogram with base of 40 cm and height of 0.5 cm is _________.  A. 200 sq cm  B. 20 sq cm  C. 10 sq cm  D. 20 cm                 
  • 17. answersanswers  1-c1-c  2-d2-d  3-d3-d  4-a4-a  5-b5-b
  • 18. Area of a ParallelogramArea of a Parallelogram  The area of a polygon is the number of square units inside the polygon.The area of a polygon is the number of square units inside the polygon. Area is 2-dimensional like a carpet or an area rug. A parallelogram is aArea is 2-dimensional like a carpet or an area rug. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Opposite sides are4-sided shape formed by two pairs of parallel lines. Opposite sides are equal in length and opposite angles are equal in measure. To find theequal in length and opposite angles are equal in measure. To find the area of a parallelogram, multiply the base by the height. The formula is:area of a parallelogram, multiply the base by the height. The formula is:  where is the base, is the height, and where is the base, is the height, and ·· means multiply. The base andmeans multiply. The base and height of a parallelogram must be perpendicular. However, the lateralheight of a parallelogram must be perpendicular. However, the lateral sides of a parallelogram are not perpendicular to the base. Thus, asides of a parallelogram are not perpendicular to the base. Thus, a dotted line is drawn to represent the height. Let's look at some examplesdotted line is drawn to represent the height. Let's look at some examples involving the area of a parallelogram.involving the area of a parallelogram.
  • 19. Area of triangleArea of triangle  The area of a polygon is the number of square units inside that polygon. Area is 2-dimensionalThe area of a polygon is the number of square units inside that polygon. Area is 2-dimensional like a carpet or an area rug. Alike a carpet or an area rug. A triangletriangle is a three-sided polygon. We will look at several typesis a three-sided polygon. We will look at several types of triangles in this lesson.of triangles in this lesson. To find the area of a triangle, multiply the base by the height, and then divide by 2. The divisionTo find the area of a triangle, multiply the base by the height, and then divide by 2. The division by 2 comes from the fact that a parallelogram can be divided into 2 triangles. For example, inby 2 comes from the fact that a parallelogram can be divided into 2 triangles. For example, in the diagram to the left, the area of each triangle is equal to one-half the area of thethe diagram to the left, the area of each triangle is equal to one-half the area of the parallelogram.parallelogram. Since the area of a parallelogram is , the area of a triangle must be one-half the area of aSince the area of a parallelogram is , the area of a triangle must be one-half the area of a parallelogram. Thus, the formula for the area of a triangle is:   or   where is the base, is theparallelogram. Thus, the formula for the area of a triangle is:   or   where is the base, is the height andheight and ·· means multiply. The base and height of a triangle must be perpendicular to eachmeans multiply. The base and height of a triangle must be perpendicular to each other. In each of the examples below, the base is a side of the triangle. However, depending onother. In each of the examples below, the base is a side of the triangle. However, depending on the triangle, the height may or may not be a side of the triangle. For example, in the rightthe triangle, the height may or may not be a side of the triangle. For example, in the right triangle in Example 2, the height is a side of the triangle since it is perpendicular to the base. Intriangle in Example 2, the height is a side of the triangle since it is perpendicular to the base. In the triangles in Examples 1 and 3, the lateral sides are not perpendicular to the base, so a dottedthe triangles in Examples 1 and 3, the lateral sides are not perpendicular to the base, so a dotted line is drawn to represent the height.line is drawn to represent the height.
  • 20.  Definition of a Parallelogram  A parallelogram is a quadrilateral that has two pairs of parallel sides. Properties of Parallelograms  1. The opposite sides are equal in length.  2. The opposite angles are congruent.  3. The diagonals bisect each other.  Conditions to be a Parallelogram  If a quadrilateral has one of the following conditions, then it is a parallelogram.  1. If a quadrilateral has two pairs of parallel sides, then it is a parallelogram.  2. If a quadrilateral has two pairs of opposite sides of the same lengths, then it is a parallelogram. Drag point A.  3. If two pairs of opposite angles of a quadrilateral are the same, then the figure is a parallelogram.  4. If the diagonal of a quadrilateral bisect each other, the figure is a parallelogram. Drag point A and point B. Point C is point symmetry with point A about point O. Point D is point symmetry with point B about point O.  5. If a quadrilateral has one pair of opposite sides which are parallel and equal in length, then the figure is a parallelogram. AD and BC are parallel and equal in length. Drag point A and point B.
  • 21.  We know that Area of a parallelogram is equal to the product of its base and height. The diagonals of a parallelogram bisect each other. A diagonal of the parallelogram divides it into two congruent triangles. Opposite sides and angles of a parallelogram are equal. Parallelograms on the same base and between the same parallels are equal in area. Parallelograms on the same base that have equal areas are between the same parallels. All figures that are on the same base and between same parallels need not have equal areas. Two congruent figures have equal areas but the converse need not be true. 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. If a parallelogram and a triangle are on the same base and they are between the same parallels, then area of the triangle is half the area of the parallelogram.
  • 22.  Two figures are said to be on the same base and between the same parallel lines if they have a common base and the vertices opposite to this common base of each figure lie on a line parallel to the base. Out of these two parallel lines, one should be the line containing the common base and the other is the line passing through the vertices of both the figures opposite to the base. We know that any two congruent figures have same area but the converse is not true. Median of a triangle divides it into two triangles of equal area. Two triangles with same base (or equal bases) and equal areas will have Area of triangle is half the product of the base and the corresponding altitude. Also, Triangles on the same base and between the same parallels are equal in area. equal corresponding altitudes. Triangles on the same base and having equal areas lie between the same parallels.