Class 9 Chapter 9 (Areas of parallelogram and triangles) Notes

Notes of Chapter 9 Area of Parallelograms and Triangles

Topics in the Chapter

• Area of Plane Figures
• Fundamentals
• Theorems

Areas of Plane Figures

• Two congruent figures have equal areas.
• A diagonal of a parallelogram divides it into two triangles of equal area.
• Parallelograms on the same base (or equal base) and between the same parallels are equal in area.
• Triangles on the same base (or equal bases) and between the same parallels are equal in area.
• Area of a parallelogram = base × height.
• Area of a triangle = 1/2 ×base × height
• A median of a triangle divides it into two triangles of equal area.
• Diagonals of a parallelogram divides it into four triangles of equal area.

Fundamentals

• Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. 

Examples:

 

 

• Area of triangle is half the product of its base (or any side) and the corresponding altitude (or height).
• Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
• A median of a triangle divides it into two triangles of equal areas. 

Theorems:

(1) Prove that a diagonal of a parallelogram divides it into two triangles of equal area.

Given:  A parallelogram ABCD in which BD is one of the diagonals.

To prove:   

Proof: Since two congruent geometrical figures have equal area. Therefore, in order to prove that it is sufficient to show that

ΔABD ≅ ΔCDB

In Δs ABD and CDB, we have

AB = CD

AD = CB

And, BD = DB

So, by SSS criterion of congruence, we have

ΔABD ≅ ΔCDB

Hence, ar(ΔABD) = ar(ΔCDB)

(2) Prove that parallelograms on the same base and between the same parallels are equal in area.

Given: Two parallelograms ABCD and ABEF, which have the same base AB and which are between the same parallel lines AB and FC.

To prove: ar(||gm ABCD) = ar(||gm ABCD)

Proof: In Δs ADF and BCE, we have

AD = BC

AF = BE

And, ∠DAF = ∠CBE         [ ⸪ AD ∥ BC and AF ∥ BE]

So, by SAS criterion of congruence, we have

ΔADF ≅ ΔBCE

ar(ΔADF) = ar(ΔBCE)   …..(i)

Now, ar(||gm ABCD) = ar(square ABED) + ar(ΔBCE)

ar(||gm ABCD) = ar(square ABED) + ar(ΔADF)    [Using(i)]

ar(||gm ABCD) = ar(||gm ABEF)

Hence, ar(||gm ABCD) = ar(||gm ABEF)

(3) Prove that the area of a parallelogram is the product of its base and the corresponding altitude.

Given: A parallelogram ABCD in which AB is the base and AL the corresponding altitude.

To prove: ar(parallelogram ABCD) = A B × AL

Construction: Complete the rectangle ALMB by drawing BM⊥CD.

Proof: Since ar(parallelogram ABCD) and rectangle ALMB are on the same base and between the same parallels.

ar(parallelogram ABCD)

= ar(rect.ALMB)

=AB × AL    [By rect. Area axiom area of a rectangle = Base \(\times\) Height]

Hence, ar(parallelogram ABCD) = AB×AL

(4) Prove that parallelograms on equal bases and between the same parallels are equal in area.

Given: Two parallelograms ABCD and PQRS with equal bases AB and PQ and between the same parallels AQ and DR.

To prove: ar(parallelogram ABCD) = ar(parallelogram PQRS)

Construction: Draw AL ⊥ DR and PM ⊥ DR

Proof: Since AB ⊥ DR, AL ⊥ DR and PM ⊥ DR

AL = PM

Now, ar(parallelogram ABCD) = AB × AL

ar(parallelogram ABCD) = PQ × PM  [AB = PQ and AL = PM]

ar(parallelogram ABCD) = ar(parallelogram PQRS)

(5) Prove that triangles on the same bases and between the same parallels are equal in area.

Proof: We have,

BD∥CA

And, BC∥DA

sq.BCAD is a parallelogram.

Similarly, sq.BCQP is a parallelogram.

Now, parallelograms ECQP and BCAD are on the same base BC, and between the same parallels.

ar(parallelogram BCQP) = ar(parallelogram BCAD)    ….(i)

We know that the diagonals of a parallelogram divides it into two triangles of equal area.

ar(ΔPBC) = \(\dfrac{1}{2}\)ar(parallelogram BCQP)   …..(ii)

And, ar(ΔABC) = \(\dfrac{1}{2}\)ar(parallelogram BCAQ)  ....(iii)

Now, ar(parallelogram BCQP) = ar(parallelogram BCAD) [ From (i)]

\(\dfrac{1}{2}\)ar(parallelogram BCQP) = \(\dfrac{1}{2}\)ar(parallelogram BCAD)

ar(ΔABC) = ar(ΔPBC)    [From (ii) and (iii)]

Hence, ar(ΔABC) = ar(ΔPBC)

(6) Prove that the area of a triangle is half the product of any of its sides and the corresponding altitude.

Given: A ΔABC in which AL is the altitude to the side BC.

To prove:

Construction: Through C and A draw CD ∥ BA and AD ∥ BC respectively, intersecting each other at D.

Proof: We have,

BA∥CD and AD ∥ BC

So, BCDA is a parallelogram.

Since AC is a diagonal of parallelogram BCDA.

[BC is the base and AL is the corresponding altitude of parallelogram BCDA]

(7) Prove that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram.

Given: A ΔABC and a parallelogram BCDE on the same base BC and between the same parallel BC and AD.

To prove: ar(ΔABC) = \(\dfrac{1}{2}\)ar(||gm BCDE)

Construction: Draw AL ⊥ BC and DM ⊥ BC, meeting BC produced in M.

Proof: Since A, E and D are collinear and BC ∥ AD

AL = DM    …..(i)

Now,

ar(ΔABC) = \(\dfrac{1}{2}\)(BC × AL)

ar(ΔABC) = \(\dfrac{1}{2}\)(BC × DM)   [AL = DM (from (i)]

ar(ΔABC) = \(\dfrac{1}{2}\)ar(parallelogram BCDE)

 

(8) Prove that the area of a trapezium is half the product of its height and the sum of parallel sides.

Given: A trapezium ABCD in which AB ∥ DC; AB = a, DC = b and AL = CM = h, where AL⊥DC and CM ⊥ AB.

To prove: ar(trap. ABCD) = \(\dfrac{1}{2}\)h × (a + b)

Construction: Join AC

Proof: We have,

ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD)

ar(trap. ABCD) = \(\dfrac{1}{2}\)(AB × CM) + \(\dfrac{1}{2}\)(DC × AL)

ar(trap. ABCD) = \(\dfrac{1}{2}\)ah + \(\dfrac{1}{2}\)bh  [AB = a and DC = b]

ar(trap. ABCD) = \(\dfrac{1}{2}\)h×(a+b)

(9) Prove that 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.

Given: Two triangles ABC and PQR such that:

ar(ΔABC) = ar(ΔPQR)

AB = PQ

CN and RT are the altitudes corresponding to AB and PQ respectively of the two triangles.

To prove: CN = RT

Proof: In ΔABC, CN is the altitude corresponding to side AB.

ar(ΔABC) = \(\dfrac{1}{2}\)(AB × CN)    ….(i)

Similarly, we have,

ar(ΔPQR) = \(\dfrac{1}{2}\)(PQ × RT)   …..(ii)

Now, ar(ΔABC) = ar(ΔPQR)

\(\dfrac{1}{2}\)(AB × CN) = \(\dfrac{1}{2}\)(PQ × RT)

(AB × CN) = (PQ × RT)

(PQ × CN) = (PQ × RT)    [ AB = PQ (Given)]

CN = RT

(10) Prove that if each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram.

Given: A quadrilateral ABCD such that its diagonals AC and BD are such that

ar(ΔABD) = ar(ΔCDB) and ar(ΔABC) = ar(ΔACD)

To prove: Quadrilateral ABCD is a parallelogram.

Proof: Since diagonal AC of the quadrilateral ABCD separates it into two triangles of equal area. Therefore,

ar(ΔABC) = ar(ΔACD)    …..(i)

But, ar(ΔABC) + ar(ΔACD) = ar(ABCD)

2ar(ΔABC) = ar(ABCD)    [Using (i)]

ar(ΔABC) = \(\dfrac{1}{2}\)ar(ABCD) ….(ii)

Since diagonal BD of the quadrilateral ABCD separates it into triangles of equal area.

ar(ΔABD) = ar(ΔBCD)   ….(iii)

But, ar(ΔABD) + ar(ΔBCD) = ar(ABCD)

2ar(ΔABD) = ar(ABCD)    [Using(iii)]

ar(ΔABD) = \(\dfrac{1}{2}\)ar(ABCD)   …..(iv)

From (ii) and (iv), we get

ar(ΔABC) = ar(ΔABD)

Since Δs ABC and ABD are on the same base AB. Therefore they must have equal corresponding altitudes.

i.e. Altitude from C of ΔABC = Altitude from D of ΔABD

DC∥AB

Similarly, AD∥BC

Hence, quadrilateral ABCD is a parallelogram.

(11) Prove that the area of a rhombus is half the product of the lengths of its diagonals.

Given: A rhombus ABCD whose diagonals AC and BD intersect at O.

To prove: ar(rhombus ABCD) = \(\dfrac{1}{2}\)(AC×BD)

Proof: Since the diagonals of a rhombus intersect at right angles. Therefore,

OB ⊥ AC and OD ⊥ AC

ar(rhombus) = ar(ΔABC) + ar(ΔADC)

ar(rhombus) = \(\dfrac{1}{2}\)(AC × BO) + \(\dfrac{1}{2}\)(AC × DO)

ar(rhombus) = \(\dfrac{1}{2}\)(AC × (BO + DO))

ar(rhombus) =\(\dfrac{1}{2}\)(AC×BD)

(12) Prove that diagonals of a parallelogram divide it into four triangles of equal area.

Given: A parallelogram ABCD. The diagonals AC and BD intersect at O.

To prove: ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

Proof: Since the diagonals of a parallelogram bisect each other at the point of intersection.

OA = OC and OB = OD

Also, the median of a triangle divides it into two equal parts.

Now, in ΔABC, BO is the median.

ar(ΔOAB) = ar(ΔOBC)  ….(i)

In ΔBCD, CO is the median

ar(ΔOBC) = ar(ΔOCD)    …..(ii)

In ΔACD, DO is the median

ar(ΔOCD) = ar(ΔAOD)   ….(iii)

From (i), (ii) and (iii), we get

ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

(13) Prove that if the diagonals AC and BD of a quadrilateral ABCD, intersect at O and separate the quadrilateral into four triangles of equal area, then the quadrilateral ABCD is parallelogram.

Given: A quadrilateral ABCD such that its diagonals AC and BD intersect at O and separate it into four parts such that

ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

To prove: Quadrilateral ABCD is a parallelogram.

Proof: We have,

ar(ΔAOD) = ar(ΔBOC)

ar(ΔAOD) + ar(ΔAOB) = ar(ΔBOC) + ar(ΔAOB)

ar(ΔABD) = ar(ΔABC)

Thus, Δs ABD and ABC have the same base AB and have equal areas. So, their corresponding altitudes must be equal.

Altitude from ΔABD Altitude from C of ΔABC

DC ∥ AB

Similarly, we have, AD ∥ BC.

Hence, quadrilateral ABCD is a parallelogram.

(14) Prove that a median of a triangle divides it into two triangles of equal area.

Given: A ΔABC in which AD is the median.

To prove: ar(ΔABD) = ar(ΔADC)

Construction: Draw AL ⊥ BC.

Proof: Since AD is the median of ΔABC. Therefore, D is the mid point of BC.

BD = DC

BD × AL = DC × AL  [ Multiplying both sides by AL]

\(\dfrac{1}{2}\)(BD × AL) = \(\dfrac{1}{2}\)(DC × AL)

ar(ΔABD) = ar(ΔADC)

ALITER, Since Δs ABD and ADC have equal bases and the same altitude AL.

ar(ΔABD) = ar(ΔADC)

Share:

Read More

Class 10 Chapter 7 (Coordinate Geometry) Class Notes

 Revision Notes on Coordinate Geometry

Cartesian Coordinate System

In the Cartesian coordinate system, there is a Cartesian plane which is made up of two  number lines which are perpendicular to each other, i.e. x-axis (horizontal) and y-axis (vertical) which represents the two variables. These two perpendicular lines are called the coordinate axis.

• The intersection point of these two lines is known as the center or the origin of the coordinate plane. Its coordinates are (0, 0).

• Any point on this coordinate plane is represented by the ordered pair of numbers. Let (a, b) is an ordered pair then a is the x-coordinate and b is the y-coordinate.

• The distance of any point from the y-axis is called its x-coordinate or abscissa and the distance of any point from the x-axis is called its y-coordinate or ordinate.

• The Cartesian plane is divided into four quadrants I, II, III and IV.

Distance formula

The distance between any two points A(x₁,y₁) and B(x₂,y₂) is calculated by

Example

Find the distance between the points D and E, in the given figure.

Solution

This shows that this is same as Pythagoras theorem. As in Pythagoras theorem

Distance from Origin

If we have to find the distance of any point from the origin then, one point is P(x,y) and the other point is the origin itself, which is O(0,0). So according to the above distance formula, it will be

Section formula

If P(x, y) is any point on the line segment AB, which divides AB in the ratio of m: n, then the coordinates of the point P(x, y) will be

Mid-point formula

If P(x, y) is the mid-point of the line segment AB, which divides AB in the ratio of 1:1, then the coordinates of the point P(x, y) will be

Area of a Triangle

Here ABC is a triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). To find the area of the triangle we need to draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Now we can see that ABQP, APRC and BQRC are all trapeziums.

Area of triangle ABC = Area of trapezium ABQP + Area of trapezium APRC – Area of trapezium BQRC.

Therefore,

Remark: If the area of the triangle is zero then the given three points must be collinear.

Example

Let’s see how to find the area of quadrilateral ABCD whose vertices are A (-4,-2), B (-3,-5), C (3,-2) and D (2, 3).

If ABCD is a quadrilateral then we get the two triangles by joining A and C. To find the area of Quadrilateral ABCD we can find the area of ∆ ABC and ∆ ADC and then add them.

Area of a Polygon

Like the triangle, we can easily find the area of any polygon if we know the coordinates of all the vertices of the polygon.

If we have a polygon with n number of vertices, then the formula for the area will be

Where x₁ is the x coordinate of vertex 1 and y_n is the y coordinate of the nth vertex etc.

Example

Find the area of the given quadrilateral.

Solution

To find the area of the given quadrilateral-

• Make a table of x and y coordinates of each vertex. Do it clockwise or anti-clockwise.

• Simplify the first two rows by:

  • Multiplying the first row x by the second row y. (red)

  • Multiplying the first row y by the second row x (blue)

  • Subtract the second product form the first.

• Repeat this for all the other rows.

• Now add these results.

The area of the quadrilateral is 45.5 as area will always be in positive.

Centroid of a Triangle

Centroid of a triangle is the point where all the three medians of the triangle meet with each other.

Here ABC is a triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). The centroid of the triangle is the point with the coordinates (x, y).

The coordinates of the centroid will be calculated as

Remarks

In coordinate geometry, polygons are formed by x and y coordinates of its vertices. So in order to prove that the given figure is a:

No.	Figures made of four points	Prove
1.	Square	Its four sides are equal and the diagonals are also equal.
2.	Rhombus	Its four sides are equal.
3.	Rhombus but not square	Four sides are equal and the diagonals are not equal.
4.	Rectangle	Its opposite sides are equal and the diagonals are equal.
5.	Parallelogram	Its opposite sides are equal.
6.	Parallelogram but not a rectangle	Its opposite sides are equal but the diagonals are not equal.

 

No.	Figures made of three points	Prove
1.	A scalene triangle	If none of its sides are equal.
2.	An Isosceles triangle	If any two sides are equal.
3.	Equilateral triangle	If it’s all the three sides are equal.
4.	Right triangle	If the sum of the squares of any two sides is equal to the square of the third side.

Example

If the coordinates of the centroid of a triangle are (1, 3) and two of its vertices are (- 7, 6) and (8, 5), then what will be the third vertex of the triangle?

Solution

Let the third vertex of the triangle be P(x, y)

Since the centroid of the triangle is (1, 3)

Therefore,

Hence the coordinate of the third vertex are (2, – 2).

cbse class 10th coordinate geometry

Share:

Read More