# Class 12 RD Sharma Solutions – Chapter 21 Areas of Bounded Regions – Exercise 21.3

Last Updated : 29 Mar, 2023

### Question 1: Calculate the area of the region bounded by the parabolas y2 = 6x, x2 = 6y.

Solution:

Given,

y2 = 6x, x2 = 6y

y2 = 6x

â‡’ y = âˆš(6x)

x2 = 6y

â‡’ y = âˆš(x2/6)

Area of the region bounded by curve = Area under the curve y2=6x along x-axis – Area under the curve x2=6y along the x-axis

Required Area =

Putting the upper and lower limits we get,

Required Area =

= 24 – 12

= 12 sq. units

### Question 2. Find the area of the region common to the parabolas 4y2 = 9x, 3x2 = 16y.

Solution:

Given,

4y2 = 9x, 3x2 = 16y

4y2 = 9x â‡’ y = âˆš(9x/4)

3x2=16y â‡’ y = âˆš(3x2/16)

Area of the region bounded by curve = Area under the curve 4y2=9x along x-axis – Area under the curve 3x2=16y along x-axis

Required area =

[Tex] [/Tex]

Putting the upper and lower limits we get,

Required Area = 43/2-43/16

= 8 – 64/16

= 4 sq. units

### Question 3. Find the area of the region bounded by y=âˆšx and x=y.

Solution:

Given,

y = âˆšx, x = y

Area of the shaded region = Area under the curve along the x-axis – Area under the line along the x-axis

Putting the integration limits, we get:

= 2/3 -1/2

= 1/6 sq. units

### Question 4. Find the area bounded by the curve y=4-x2 and the lines y=0 and y=3.

Solution:

Given,

y = 4-x2, y = 0, y = 3

Area of shaded region = Area of region OCBEO + Area of region ODAEO

As the Y-axis divides the parabola into two symmetrical regions, the area of OCBEO = area of ODAEO

So,

Area of shaded region = 2 Area of ODAEO

Putting the limits, we get

Area of shaded region = 2[4(2)-2^3/3-0]

= 2[8-8/3]

= 32/3 sq. units

### Question 5. Find the area of the region

Solution:

Given

Area of shaded region = Area under the sector OAB – Area of triangle OAB

Putting the upper and lower limits, we get:

### Question 6. Using integration, find the area of the region bounded by the triangle whose vertices are (2,1), (3,4), and (5,2).

Solution:

Given, A(2,1), B(3,4) and C(5,2). We need to find the area of triangle ABC. To calculate the area we need to find the equations of the lines AB, BC, and CA.

The equation of a line can be calculated as:

Equation of AB:

y-1 = 3x-6 â‡’ y=3x-5

Equation of BC:

y-4= -x+3 â‡’ y=-x+7

Equation of CA:

y = x/3 + 1/3 â‡’ 3y = x+1

Required Area = Area of Region I + Area of Region II………………(1)

Area of Region I = Area under AB from x=2 to x=3 – Area under AC from x=2 to x=3

Area of Region I = 4/3 sq. units …………………(2)

Area of Region II = Area under BC from x=3 to x=5 – Area under AC from x=3 to x=5

= 8/3 sq. units

Area of Region II = 8/3 sq. units ……………………….(3)

Putting the value of (2) and (3) in (1), we get:

Required Area = Area of region I + Area of region II

= 4/3 + 8/3 = 4 sq. units

### Question 7. Using integration, find the area of the region bounded by the triangle whose vertices A, B, and C are (-1,1), (0,5), and (3,2).

Solution:

Given, A (2,1), B (3,4) and C (5,2). We need to find the area of triangle ABC. To calculate the area we need to find the equations of the lines AB, BC, and CA.

The equation of a line can be calculated as:

Equation of AB:

y-1 = 4x+4

y = 4x+5

Equation of BC:

y-5= -x

y = -x+5

Equation of CA

y = x/4 + 5/4

4y = x+5

Required Area = Area of Region I + Area of Region II   ………………(1)

Area of Region I = Area under AB from x=-1 to x=0 – Area under AC from x=-1 to x=0

Area of Region I = 15/8 sq. units …………………(2)

Area of Region II = Area under BC from x=0 to x=3 – Area under AC from x=0 to x=3

Area of Region II = 45/8 sq. units ……………………….(3)

Putting the value of (2) and (3) in (1), we get:

Required Area = 15/8+ 45/8

= 15/2 sq. units

### Question 8. Using integration, find the area of triangular the region, the equations of whose sides are y = 2x+1, y = 3x+1, and x = 4.

Solution:

Area of the shaded region in the above graph can be calculated as:

Required area = Area under line y=3x+1 from x=0 to x=4 – Area under the line y=2x+1 from x=0 to x=4

= 8 sq. units

### Question 9. (x – a)2 + (y – b)2 = r2. Find the area of the region {(x, y): y2â‰¤8x, x2+y2â‰¤9}

Solution:

Given,

y2 = 8x â‡’ y = âˆš8x

x2 + y2=9 â‡’ y = âˆš(9-x2)

As the circle is divided by the parabola into two symmetrical halves,

Area of the shaded region = 2(Area under the parabola y2=8x from x=0 to x=1 + Area under the circle x2+y2=9 from x = 1 to x = 3)

Required Area

### Question 10. Find the area of the region bounded by the circle x2+y2=16 and the parabola y2=6x.

Solution:

Given

y2 = 6x â‡’ y = âˆš6x

x2 + y2=16 â‡’ y = âˆš(16-x2)

As the circle is divided by the parabola into two symmetrical halves,

Area of the shaded region = 2(Area under the parabola y2=6x from x=0 to x=2 + Area under the circle x2+y2=16 from x = 2 to x = 4)

Required Area =  sq.units

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