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The RD Sharma class 12 solution of area of bounded region exercise 20.3 is one of the easy chapters to start with if any student is thinking of practicing the maths solutions with a less challenging chapter. The RD Sharma class 12th exercise 20.3 will give you an insight of the different types of concepts covered in the RD Sharma solutions chapter and hence improve your skills and enhance solving qualities to help you Ace the maths exam. The class 12 RD Sharma chapter 20 exercise 20.3 solution guides you so that you can get a grip on the subject of maths even if you think that you are not so good at the subject.

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Areas of Bounded Region Exercise 20.3 Question 1

12 sq.units.

The intersecting points of the given parabolas are obtained by solving these equations for x and y , which are 0(0,0)and (6,6).

The given equations are

...…(i)

And

(ii)

Putting x value on x = 6

When y =0 then x=6

And when y =6 then x = 6

On solving these two equations ,we get point of intersections

The points are O(0,0) and A(6,6). These are shown in the graph below

Bounded Area ,A=[Area between the curve (i) and x -axis from 0 to 6]-[Area between the curve (ii) and x -axis from 0 to 6)

On integration the above definite integration

Areas of Bounded Region Exercise 20.3 Question 2

the bounded area ,A=[Area between the curve (1) and x -axis from 0 to 4]-[Area between the curve (2)and x -axis from 0 to 4]

The given eqation are,

.........(1)

And,

Equating (1)and (2)

When we put x =4 in eqation (I)

Then y =3 ,

When we put x =0 in eqation (I)

Then y = 3 ,

On solving these two equations,we get the point of intersections

The points are shown in the graph below

Now the bounded area ,

A= [Area between the curve (1) and x- axis from 0 to 4] - [ Area between the curve (2) and x - axis from 0 to 4]

On integrating the above definate integration ,

The required area

Areas of Bounded Region Exercise 20.3 Question 3

sq.units

and

Bounded Area ,

A = [Area between curve 1 and axis from 0 to 1] - [Area between the curve 2 and axis from 0 to 1]

The given eqation are,

(i)and (ii)

Solving eqation (1) and( 2)

'

So,

Or y = 1 and x = 0 or x = 1

On solving these two equations we get the points of intersection .

The point are O(0,0)and A(1,1) these are shown in the graph below

Now the bounded area is the required area to be calculated,

Hence,Bounded Area ,

A = [Area between curve 1 and axis from 0 to 1] - [Area between the curve 2 and axis from 0 to 1]

On integrating the above definate integration,

= sq.units

Areas of Bounded Region Exercise 20.3 Question 4

Bounded area ,A = 2 times [area between the equation 1 and y axis from y =0 to y = 3 ]

The given equation are ,

…(1)

..(2)

And …(3)

Equation 1 represents a parabola with vertex (0,4) and passes through (0,2),(0,2)

Equation 2 is x - axis and cutting the parabola at C(2,0) and D(-2,0)

Equatin 3 is a line parellel to x - axis cutting the parabola at A(3,1)and B(-3,1)

On solving these equations ,we get pont of intersection of a parabola with the other two lines are A(3,1) B(-3,1),c(2,0)and D(-2,0) these are shown in the graph below

Now the bounded area is the required area to be calculated,

Hence ,bounded area ,A = 2 times [area between the equation 1 and y axis from y = 0 to y =3 ]

On intregrating the above definate integration,

, as area in not negative.

= sq units

The area bounded by the curved and the lined y=0, y=3 is sq units

Areas of Bounded Region Exercise 20.3 Question 5

considered ,this represent an ellipse ,symmetrical about both axis and cutting x-axis at A(a,0)and A`(-a,0) and y-axis at B(0,b),B`(0,-b)

Represents the area inside the ellipse

Represent a straight line cutting x-axis at A(a,0) and y- axis at B(0,b)

Represent the area above the straight line

Represent the smaller shaded arera bounded by the line and the ellips in the shaded region, consider a vertical strip with length= and width=dx such that lies on ellipse and lies on the straight line area of approximating rectangle=

The approximating rectangle move from x = 0 to x = a

Area of the shaded region

Areas of Bounded Region Exercise 20.3 Question 6

4 square units

A(2,1),B(3,4)and C(5,2)

The equation of AB,

The equation of BC,

The equation of AC,

Now the required area (A)=[(Area between line AB and x-axis)-(Area between line AC and x-axis)from x=2 to x=3]

+[(Area between line BC and x-Axis )-(Area between line AC and x-Axis )from x=3 to x=5]

=4 square unit

The area of the region bounded by the triangle whose vertices are (2,1),(3,4) and (5,2) is 4 sq. units

Areas of Bounded Region Exercise 20.3 Question 7

sq. Units

We have to find the area of the triangle whose vertices are A(-1,1),B(0,5),C(3,2) as shown below

The equation of AB

The equation of BC,

The equation oa AC,

Now the required Area (A)=[(Area between line BC and x-Axis )-(Area between line AC and x-Axis) from x=0 to x=3]

Say ,Area

So the enclosed area of the triangle is square units

Areas of Bounded Region Exercise 20.3 Question 8

8 sq units

y=2x+1,y=3x+1 and x=4

The given lines are

y=2x+1…..(1)

y=3x+1…..(2)

x=4…….(3)

For intersection points of (1) and( 3)

Y=2x4+1=9

Coordinates of intersecting point of 1 and 3 is (4,9) for intersection point of (2) and (3)

Y=3x4+1=13

i.e, coordinates of intersection point of (2) and (3) is (4, 3)

For intersection point of (1) and (2)

2x+1=3x+1=>x=0

Y=1

i.e., coordinates of intersection point of (1) and (2) is (0, 1)

Shaded region is required triangle region.

Required Area =Area of trapezium OABD-Area of trapezium OACD

=8 sq units

Areas of Bounded Region Exercise 20.3 Question 9

square units

To find area

On solving the equation (1) and (2)

And when x=1 then y=

Equation (1) represents a parabola with vertex (0,0) and axis as x-axis equation (2) represents a circle with centre (0,0) and radius 3 units, so it meets area at

Point of intersection of parabola and circle is and

The sketch of the curves is as below

Or, required area=(region ODCO+REGION DBCD)

Hence, the required area is square unit

Areas of Bounded Region Exercise 20.3 Question 10

Square unit

…(1)

Equation of circle with centre (0,0) and radius 4

Equation of parabola:

….(2)

Intersecting piont of (1) and (2)

So points are and

Area of shaded region

square unit

Areas of Bounded Region Exercise 20.3 question 11.

.

........(I)

.........(II)

........(I)

.........(II)

Clearly represents a circle with center (0, 0) and radius 2. Also, represents a circle with centre (2, 0) and radius 2. To find the point of intersection of the given curves, we solve (i) and (ii). Simultaneously, we find the two curves intersect at and .

Since both the curves are symmetrical about x-axis, So, the required area = 2(Area OABCO) Now, we slice the area OABCO into vertical strips. We observe that the vertical strips change their character at A(1,√3). So, Area OABCO = Area OACO + Area CABC.

When area OACO is sliced in the vertical strips, we find that each strip has its upper end on the circle (x - 2)

Similarly, approximating rectangle in the region CABC has length = y2 , width = and area = y2

As it can move from x =1 to x =2

Hence,required area A is given by

Areas of Bounded Region Exercise 20.3 question 12.

.

..........(I)

........(II)

..........(I)

........(II)

Solving (I) and (II) ,we have

We have to determine the area of shaded region.

Required Area

Areas of Bounded Region Exercise 20.3 question 13.

Here we know that

We can write it as

Here which represents the region inside the parabola, y

represents the interior of circle having (0, 0) as centre and as radius

So the region R which is intersection of points R1 and R2 is shaded in the figure

…… (1)

…… (2)

Solving both the equations

3x

So we get

Here is the rejecting negative value

Substituting y = 0 in equation (1)

We know that the circle (1) cuts x-axis at P and P’ (

So the required area can be written as

Required area = 2 [area of ODPAO] = 2 [area of ODAO + area of ADPA] .

Intergrating w.r.t

Substituting the value of x

Areas of Bounded Region Exercise 20.3 question 15.

First we need to find the point of intersection of the given curve

Equation of parabola ---(i)

Equation of line ---(ii)

Putting the value of y in equation (ii)

When then

When then

So the points of the interaction of the given curves is (1 , -3 ) and (2 , 12)

Hence , area of the region is .

Areas of Bounded Region Exercise 20.3 question 16

.

Use basic concepts.

and .

..........(1)

.........(2)

Equation (1) represents a parabola with vertex (0,0) and axis as x-axis,

Equation (2) represents a parabola with vertex (0,0), and axis as y-axis,

Point of intersection of parabolas are (0, 0) and

A rough sketch is given as:

The shaded region is required area and it is sliced into rectangle of width= and length

Area of rectangle=

This approximation rectangle slides from to, , so

Required area = Region OQAPO

Hence the required area is .

Areas of Bounded Region Exercise 20.3 question 17.

use basic concepts.

.

x | 0 | 1 |

y | 0 | 0.58 |

For intersecting point of given circle and line

Putting in we get

Intersecting points are

Shaded region is required region.

Now required area

Areas of Bounded Region Exercise 20.3 question 18

9 sq. units.

he given equations of the curves are

,

,

Solving we get,

The graph of function is part of parabola lying above x –axis.

The graph is as shown in the figure.

From the figure, area of shaded region,

Areas of Bounded Region Exercise 20.3 question 19

.

Given curves and .

The point of intersection of circle

And parabola, is

The required common area ,

Areas of Bounded Region Exercise 20.3 question 20.

.

Given curves are and .

We have given equations

...............(1)

........(2)

Equation (1) can be written as

So equation (i) represents a circle with centre (4, 0) and radius 4.

Again, clearly equation (ii) represents parabola with vertex (0, 0) and axis as x -axis. The curve (i) and (ii) are shown infigure and the required region is shaded. On solving equation (i) and (ii) we have points of intersection 0(0, 0) and A(4, 4), C(4, - 4)

Now, we have to find the area of region bounded.

By(i) and (ii) & above x - axis.

So requires region is OBAO.

Areas of Bounded Regions exercise 20.3 question 21

.

Given curve and .

Represents a parabola with vetex(0,0) and opening upward , symmetrical about +ve axis

Represents the wider parabola, with vertex at C (0,9)

To find point of intersection, solve the two equations

Thus and ’ are point of intersection of two parabolas.

Shaded area A’OA=2 x area (OCAO)

Consider a vertical stip of length and width dx

Area of approximating rectangle

The approximating rectangle moves from to

Shaded area B’A’B=2 area OCAO=

Areas of Bounded Regions exercise 20.3 question 22.

sq. Units

Use concept.

The given equations are and

To find the area enclosed by,

........(i)

........(ii)

On solving the equation (i) and (ii),

or

or

Equation (1) represents a parabola with vertex (0, 0) and axis as y - axis.

Equation (2) represents a parabola with vertex (0,4) and axis as the y - axis.

Points of intersection of parabolas are A(2, 8) and B(– 2, 8).

These are shown in the graph below:

Required area=Region AOBCA

=2(RegionAOCA)

Areas of Bounded Regions exercise 20.3 question 23.

4 sq. units

The vertices of triangle are (-1,2), (1,5) and (3,4).

Equation of,

AB is:

BC is:

AC is:

Required area

(ii)

Equation of,

AB is:

BC is:

AC is:

Required area

(III)

Equation of ,

AB is:

BC is :

AC is:

Required area

Areas of Bounded Regions exercise 20.3 question 24

Areas of Bounded Regions exercise 20.3 question 25

.

The given equations are and .

Represents a circle with centreo(0,0) and cutting the x axis at A(4,0)

Represents straight passing through o(0,0) .

Point of intersection is obtained by solving the two equations.

and

and are points of intersection of circle and straight line

Shaded area (OBQAO) =area(OBPO)+area(PBQAP)

Areas of Bounded Regions exercise 20.3 question 26

.

Use concept.

The given equations are and

To find area bounded by

...........(1)

........(2)

On solving the equation (i) and (ii)

Equation (i) is a parabola with vertex (−12,0) and passes through (0, 1), A (0, – 1)

Equation (ii) is a line passing through (1, 0) and (0, – 1).

Points of intersection of parabola and line are B (4, 3) and A (0, – 1)

These are shown in the graph below:

Required area=RegionABCDA

Areas of Bounded Regions exercise 20.3 question 27

use concept.

The given equations are and

We have. and

Consider a horizontal strip of length and width dy where lies on straightline and lies on the parabola.

Area of approximating rectangle= and it moves from y =1 to y =- 7

Areas of Bounded Regions exercise 20.3 question 28

Use concept.

The given equations are and .

Areas of Bounded Regions exercise 20.3 question 29

Use concept.

The given equations are and .

To find area bounded by

......(1)

.......(2)

Equation (1) represents a parabola with vertex (0,2) and downward, meets axes at .

Equation (2) represents a line passing through (0,0) and (2,-2) .The points of intersection of line and parabola are (2,-2) and (1,-1) .

A rough sketch of curve is as follows:

Shaded region is sliced into rectangles with area . It slides from x =-1 to x = 2 so,

Required area =Region ABPCOA

Areas of Bounded Regions exercise 20.3 question 30 sub question 1.

11 sq.units.

Use concept of definite integrals.

The given equations are 3x - y -3 =0 , 2x + y - 1 =0 and x -2y -1 =0.

We have

Area of bounded region =

Areas of Bounded Regions exercise 20.3 question 30 sub question 2

6.5 sq. units

Point off intersection of (i) and (ii) is A(3,5)

Point off intersection of (ii) and (iii) is B(6,3)

Point off intersection of (iii) and (i) is C(1,2)

Area of the region bonded

Areas of Bounded Regions exercise 20.3 question 31

Use concept.

The given equation are and .

and

........(1)

........(2)

Equation(1) is a line passing through (2,2) and (0,0).Equation (2) is a parabola upward with vertex at (0,2).

A rough sketch of curve is as under:

Shaded region is sliced into rectangle of area=. .It slides from to ,so

Required Area=Region OABCO

Areas of Bounded Regions exercise 20.3 question 32

4 sq. Units

Use concept.

Equation of the given curve and

…(i)

And …(ii)

On solving the equation (i) and (ii),

Or

Or

when y =1 then x =1 and when y = -1 then x =1

Equation (i) represents an upward parabola with vertex (0, 0) and axis -y

Equation (ii) represents a parabola with vertex (3, 0) and axis as x-axis

They intersect at A (1, – 1) and C (1, 1)

These are shown in the graph below:

Required area = Region OABCO = 2

= 2 Region OBCO

= 2[Region ODCO + Region BDCB]

= 4 sq.units

Areas of Bounded Regions exercise 20.3 question 33

7sq. Units.

Use conceptual part.

Triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C (8,4).

Solution:

Equation of AB is given by

Equation of BC is given by

Equation of AC is given by

Area of ABC

Areas of Bounded Regions exercise 20.3 question 34

sq. Units.

centre at (0, 0) and? radius? .

Equation of the line is

Consider, y =x -1 and

Eliminating y ,we get

The required area is

= sq.units

Therefore, the area of the region is sq.units.

Areas of Bounded Regions exercise 20.3 question 35

1sq.unit.

Use concept.

The curves and .

To find area bounded by and

A rough sketch of the curve is as under:

Shaded region is the required area. So

Required area = Region ABCA

A = Region ABDA + Region BCDB

A = 1 sq. units.

Areas of Bounded Regions exercise 20.3 question 36

sq. Units

Use concept.

The first quadrant enclosed by x-axis, the line y = x and the circle

Point of intersection, x =4

Area of a shaded region =

Areas of Bounded Regions exercise 20.3 question 37

(not possible cannot be -ve )

Or x=2 (only allowed value)

Therefore, the area of the circle is

Areas of Bounded Regions exercise 20.3 question 38

sq. Units

Use concept .

is a parabola line

They meet at pts (2,4) and (-1,1)

By solving the equation,

Required area

= Area of line - Area of parabola

Areas of Bounded Regions exercise 20.3 question 39

...... (i)

....... (ii)

And .......(iii)

Solving the above three equations to get the intersectons points,

Equation (i) represents a parabola with vertices (3,0) and axis as y - axis

Equation (ii) represents a line passing through (0,3) and

The points of intersection are A(0,3) and B(2,7)

These are shown in the graph below:

Required area =

= sq.units

Therefore, the area of the region is sq.units

Areas of Bounded Regions exercise 20.3 question 40

sq. Units

Use concept of definite integrals.

The curve , line y =x and the positive x-axis

Hence,

represents the upper half of the circle a circle with centre O(0,0) and radius 1 unit.y = x represents equation of a straight line passing through O(0,0)

Point of intersection is obtained by solving two equations

And are two points of intersection between the circles and the straight line.

And is the intersection point of and y =x .

Required area = Shaded area (ODAEO) = Area (ODEO) + Area (EDAE)…..(1)

Now, area (ODEO) =

Area (EDAE) =

Area (ODAEO) = sq.units

Areas of Bounded Region Exercise 20.3 Question 41

sq. units

Use concept.

Lines and

To find area bounded by lines

{Say AB}

{Say BC}

{Say AC}

By solving equation (1) and (2) , we get B(0,5)

By solving equation (2) and (3) , we get C(3,2)

By solving equation (1) and (3) , we get A(-1,1)

A rough sketch of the curve is as under:-

Shaded area ABC is the required area.

Required area = ar( ABD) + ( BDC)

ar( ABD)

AR ( ABD) = sq. units

AR ( BDC) =

AR ( BDC) = sq. units

Using equation (1), (2) and (3),

AR ( ABC) =

ar( ABC) = sq. units

Areas of Bounded Region Exercise 20.3 Question 42

sq. units

Use concept.

The two curves and

To find area enclosed by

……. (1)

…….. (2)

Equation (1) represents a circle with centre (0,0) and meets axis at ,

Equation (2) is a circle with center (3,0) and meets axis at (0,0) , (6,0)

they intersect each other at and . A rough sketch of the curves is as under:

Shaded region is the required area.

Required area = Region OABCO

A = 2(Region OBCO)

= 2(Region ODCO + Region DBCD)

Areas of Bounded Regions Exercise 20.3 Question 43

sq. units

Use concept.

……(1)

…….(2)

Clearly represents a circle and is the equation of a straight line cutting x and y axis at (0,2) and (2,0) respectively.

The smaller region bounded by thesetwo curves is shaded in the following figure.

Length =

Width = x and

Area =

Since the approximating rectangle can move from x=0 to x=2, the required area is given by

Therefore, the area of the region is sq. units

Areas of Bounded Regions Exercise 20.3 Question 44

Use concept.

To find the area of a region

Here,

…..(1)

…..(2)

Equation (1) represents an ellipse with centre at origin and meets axis at , .

Equation (2) is a line that meets axis at (3,0) ,(0,2)

A rough sketch is as under:

Shaded region represents required area. This is sliced into rectangles with area which slides from x=0 to x=3, so

Required area = Region APBQA

Therefore, the area of the region is

Areas of Bounded Regions Exercise 20.3 Question 45

Use concept.

The curve , and the x-axis.

Give, equation of a curves are

……(1)

……(2)

Consider the curve

which represents a circle with centre (0, 0) and radius 2 units.

Now, consider the curve which also represents a circle with centre (2,0) and radius 2 units.

Now Let us sketch the graph of given curves and find their points of intersection.

On substituting the value of y from Eq. (1) in Eq. (2), we get

or or

On substituting x =1 in Eq. (1), we get

Thus, the point of intersection is .

Clearly, required area = Area of shaded region OABO

Areas of Bounded Regions Exercise 20.3 Question 46

.sq. Units.

Use concept.

The curves and

To find area enclosed by

A rough sketch of equation of lines (1),(2),(3),(4) is given as:

Shaded region is the required area.

Required area = Region ABCDA

Required area = Region BDCB + Region ABDA

Region BDCB is sliced into rectangles of area and it slides from to x =1

Region ABDA is sliced into rectangle of area and it slides from x =1 to .

So, using equation (1),

Required area = Region BDCB + Region ABDA

Areas of Bounded Regions Exercise 20.3 Question 47

Use concept.

The curves and .

To find area enclosed by

……. (1)

And

Equation (1) represents a downward parabola with vertex and equation (2) represents lines. A rough sketch of curves is given as:

Required area = Region ABECDA

A = Region ABEA + Region AECDA

Areas of Bounded Regions Exercise 20.3 Question 48

sq. units

Use concept.

The parabola and

To area enclosed by

Equation (1) represents a parabola downward with vertex at (2,4) and meets axis at (4,0), (0,0). Equation (2) represents a parabola upward whose vertex is and meets axis at (1,0), (0,0). Points of intersection of parabola are (0,0) and .

A rough sketch of the curves is as under:

Shaded region is required area it is sliced into rectangles with area = . It slides from x =0 to , so

Required area = Region OQAP

Therefore, the area enclosed by the parabola and is sq. units.

Areas of Bounded Regions Exercise 20.3 Question 49

121:4

Use basic concepts.

The parabola and .

The curves are

Equation (1) represents a parabola downward with vertex at (2,4) and meets axis at (4,0), (0,0). Equation (2) represents a parabola upward whose vertex is and meets axis at (1,0), (0,0) and . A rough sketch of the curves is as under:

Area of the region above x-axis

= Area of region OBACO

= Region OBCO + Region BACB

Area of the region below x-axis

= Area of region OPBO

= Region OBCO + Region BACB

Areas of Bounded Regions Exercise 20.3 Question 50

4 sq. units.

Use concept of definite integrals.

The curves and

To find area bounded by the curve

And

Drawing the rough sketch of lines (1), (2), (3) and (4) as under:

Shaded region is the required area

Required area = Region ABCDA

A = Region ABFA + Region AFCEA + Region CDEC

A = 4 sq. units.

Therefore, the area bounded by the curve is 4 sq. units.

Areas of Bounded Region Exercise 20.3 question 51.

M=2.

Use concept of definite integrals

The parabola and the line y = mx is .

Area of the bounded region =

Therefore, the value of m is 2.

Areas of Bounded Region Exercise 20.3 question 52.

Area of the bounded region =

**The RD Sharma class 12th exercise 20.3 consist a total of 55 questions that covers up all the important factors of the chapter that is mentioned below:**

Finding Area using vertical stripes and horizontal stripes

Finding the Area between two curves

Finding the area between two curves using vertical stripes and horizontal stripes

Area of the bounded region

**The benefits of practicing from the RD Sharma class 12th exercise 20.3 Is mentioned below:-**

The RD Sharma class 12 solution chapter 20 exercise 20.3 Is Designed by experts to provide the best solutions to the questions and helpful tips to make it easy for them to solve mathematics.

Teachers are very fond of the solutions of the RD Sharma and most of the questions which they assign for homework are taken in from the RD Sharma class 12th exercise 20.3.

The RD Sharma class 12 chapter 20 exercise 20.3 Is important to practice thoroughly as it has been observed that most of the questions that are asked in the board exams are the similar pattern that has been provided in the RD Sharma solutions.

The RD Sharma solutions are of the similar pattern of the NCERT, therefore students who are appearing for public examinations can also prepare using these solutions.

The RD Sharma class 12th exercise 20.3 Is recommended because it can be easily accessed by downloading the study material from the careers360 website and that is also free of cost.

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1. Is a solution useful for class 12 and 11 students?

Yes, the RD Sharma solutions are completely beneficial for any student who is appearing for board exams for will be appearing in the coming year.

2. How do you find the area of a bounded region?

The area under a cover between two points can be found by doing a definite integral between the two points.

3. Can teachers also use the RD Sharma solutions?

Teachers make the most use of R D Sharma solutions by preparing question papers and providing assignments to the students through the usage of the solutions.

4. Does the careers360 website contain the updated version of R D Sharma solutions?

Yes, The careers360 website contains all the updated version RD Sharma solutions for every class in every exercise.

5. Does the RD Sharma class 12th exercise 20.3 contain all important questions?

Yes, RD Sharma class 12 exercise 20.3 contains all essential questions that cover the concepts important for the exams.

Application Date:05 September,2024 - 20 September,2024

Admit Card Date:13 September,2024 - 07 October,2024

Admit Card Date:13 September,2024 - 07 October,2024

Application Date:17 September,2024 - 30 September,2024

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