Areas of Parallelograms and Triangles Class 9 Notes CBSE Maths Chapter 9 (Free PDF Download)

Definition of Triangle:

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A figure in two dimensions, constituting three lines and corners is called a triangle.

Properties of a Triangle:

• A triangle is the polygon having least number of sides.

• It is a closed figure in two dimensions formed by three line segments and corners thus, it occupies area bounded within the sides.

As triangle is the simplest polygon, the area of other polygons can be defined by using area of finite sets of triangles. For example: A hexagon is made up of two triangular parts thus, the area of a hexagon is the union of two triangular regions.

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Definition of Unit Area:

Area enclosed by a figure having sides of unit length is called unit area. It is generally represented as square units and is a positive real number.

Notation of Area of a Polygon:

The area of a polygonal figure A is denoted by $ar(A)$. And in meters, it is denoted by ${{m}^{2}}$.

Area Axioms

1. Congruent Area Axiom: 

If $\vartriangle ABC\cong \vartriangle PQR$ then area of triangle ABC = Area of triangle PQR.

2. Area Monotone Axiom: 

If ${{R}_{1}}$ and ${{R}_{2}}$ two polygonal regions such that ${{R}_{1}}\subset {{R}_{2}}$ area of ${{R}_{1}}\le {{R}_{2}}$.

3. Area Addition Axiom

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If ${{R}_{1}}$ and ${{R}_{2}}$ are two polygonal whose intersection is either a finite number of line segments or a single point and \[R\text{ }=\text{ }{{R}_{1}}+\text{ }{{R}_{2}}\]then \[ar\left( {{R}_{0}} \right)\text{ }=\text{ }ar\text{ }\left( {{R}_{1}} \right)\text{ }+\text{ }ar\text{ }\left( {{R}_{2}} \right).\]In figs (i) the region is divided into two regions R1 and R2.

Area of a Rectangular Region

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Given that \[AB\text{ }=\text{ }a\] metres and \[AD\text{ }=\text{ }b\] metres, hence \[ar\text{ }\left( ABCD \right)\text{ }=\text{ }ab\]sq. m.  (Using addition area axiom)

Theorem 1

Statement:

Diagonals of a parallelogram divides it into two triangles of equal area.

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Given:

ABCD is a parallelogram. AC is one of the diagonals of the parallelogram ABCD.

To Prove:

$ar(ABC)=ar(DBC)$

Proof:

In triangles ABD and DBC,

\[AB\text{ }=\text{ }DC\] (Opposite sides of parallelogram)

\[AD\text{ }=\text{ }BC\] (Opposite sides of parallelogram)

\[BD\text{ }=\text{ }BD\] (Common side)

(Area congruency axiom)

Hence, $ABC\cong DBC$ (SSS congruency)

$ar(ABC)=ar(DBC)$  (Using congruent area axiom)

Theorem 2

Statement:

Parallelograms on the same base and between the same parallel lines are equal in area.

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Given:

ABCD and ABEF are two parallelograms having same base AB and same parallels AB and CF.

To Prove:

Area of parallelogram \[ABCD\text{ }=\text{ }ABEF\]

Proof:

$ar({{\parallel }^{m}}ABCD)=ar(ABED)+ar(EBC)$   …. (1) (area addition axiom) 

$ar({{\parallel }^{m}}ABEF)=ar(ABED)+ar(AFD)$     …. (2) (area addition axiom)

Now in triangles EBC and AFD,

\[AF\text{ }=\text{ }BE\] (Opposite sides of a parallelogram)

\[AD\text{ }=\text{ }BC\] (Opposite sides of a parallelogram)

Angle $AFD=BEC$ (\[AB\text{ }||\text{ }BE\]and \[FC\]is a transversal)

Hence are corresponding angles.

\[EF\text{ }=\text{ }AB\text{ }=\text{ }CD\]

\[EF\text{ }-\text{ }DE\text{ }=\text{ }CD\text{ }-\text{ }DE\]i.e., FD = EC

Triangle $EBC\cong AFD$  (SAS congruency condition)

$ar(EBC)=ar(AFD)$  (Area congruency condition)

\[ar({{\parallel }^{m}}ABCD)=ar({{\parallel }^{m}}ABCD)\]  From (1), (2) and (3),

Corollary Statement: 

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Parallelograms on equal bases and between the same parallels are equal in area.

Given: \[|{{|}^{m}}ABCD\]and \[|{{|}^{m}}PQRS\] are between the same parallels l and m such that \[AB\text{ }=\text{ }PQ\](equal bases).

To Prove: \[ar({{\parallel }^{m}}ABCD)=ar({{\parallel }^{m}}PQRS)\].

Construction:

Draw the altitude EF and GH.

Proof:

\[l\text{ }||m\] (From given data)

EF = GH (perpendicular distance between the same parallels) 

\[ar({{\parallel }^{m}}ABCD)=AB\times EF\]  

\[ar({{\parallel }^{m}}PQRS)=PQ\times GH\] (area of a = base x alr)

Since \[AB\text{ }=\text{ }GH\] ( given)

and \[EF\text{ }=\text{ }GH\]( construction)

Hence, \[ar({{\parallel }^{m}}ABCD)=ar({{\parallel }^{m}}PQRS)\]

Theorem 3

Statement:

Triangles on the same base and between the same parallels are equal in area.

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Given:

Triangles ABC and DBC stand on the same BC and between the same parallels l and m.

To prove:

$ar(ABC)=ar(DBC)$

Construction:

CE || AB and BF || CA

Proof:

\[|{{|}^{m}}ABCE\]and \[|{{|}^{m}}DCBF\]  has same base \[BC\] and lies between the same parallels\[l\text{ }and\text{ }m.\]

\[|{{|}^{m}}ABCE\]=\[|{{|}^{m}}DCBF\]  .... (1) 

AC is a diagonal of \[{{\parallel }^{m}}ABCE\] which divides the parallelogram into two triangles of equal areas.

Similarly, we can prove that 

\[ar(BCD)=\dfrac{1}{2}ar(|{{|}^{m}}DCBF)\]

From (1), (2) and (3), we can write

$ar(ABC)=ar(DBC)$

Hence the theorem is proved.

Theorem 4

Statement:

Triangles of equal areas, having one side of one of the either triangles equal to one side of the other, have their corresponding altitudes equal.

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Given:

Two triangles ABC and DEF are such that:

(i) $ar(ABC)=ar(DEF)$

(ii) \[BC\text{ }=\text{ }EF\]

\[AM\text{ }and\text{ }DN\] are altitudes of triangle \[ABC\]and triangle \[DEF\]respectively.

To prove:

\[AM\text{ }=\text{ }DN\]

Proof:

In triangle \[ABC,\text{ }AM\] is the altitude, \[BC\] is the base.

$\Delta ABC=\dfrac{1}{2}\times BC\times AM$

In \[\Delta DEF,\text{ }DN\] is the altitude and\[~EF\] is the base.

$\Delta DEF=\dfrac{1}{2}\times EF\times DN$

$\dfrac{1}{2}\times BC\times AM=\dfrac{1}{2}\times EF\times DN$

Also \[BC\text{ }=\text{ }EF\] (given)

\[{\scriptscriptstyle 1\!/\!{ }_2}\text{ }AM\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }DN\] 

i.e.,\[AM\text{ }=\text{ }DN\].

Hence the theorem is proved.

Areas of Parallelograms and Triangles Class 9 Maths Revision Notes Chapter 9

Area Proposition 

Area of a Closed Shape

The portion of any plane that is enclosed with a closed figure is known as the Planar Area. The measure of this region is referred to as the ‘area’ of that figure. This is represented in the form of numbers used by any unit.

Properties of the Area of a Figure

• If the shape and size of the two figures are the same, they are said to be congruent. And if the two figures are congruent, their area would also be the same.

• However, if the two figures have the same area, they do not need to be congruent.

Area of Parallelogram

• Area of parallelogram = base × height

• Height is perpendicular to the base.

Figures on the Same Base and Between the Same Parallels

If the two figures have the same base and the vertices opposite the base are also on the line parallel to the base, then the two figures are assumed to be on the same base and between the same parallels.

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Parallelograms on the Same Base and Between the Same Parallels

If the two parallelograms have the same base and are between the same parallel lines, the area must be equal to those two parallelograms.

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Triangles on the Same Base and Between the Same Parallels

If the two triangles are on the same base and their opposite vertex is on the parallel axis, their area must be equal.

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Here, ABC and PBC are the two triangles having a common base i.e. BC, and between the two parallel lines i.e. XY and BC.

ar (ABC) = ar (PBC)

Area of Triangle

Area of a rectangle = ½(height)(base)

Median of a Triangle

• The line segment from each vertex of the triangle to the centre of the opposite side is the Median.

• There are three medians of the triangle, and the intersection of the three medians is known as the Centroid.

• The median divides the triangle into two equal sections.