# Class 12 RD Sharma Solutions – Chapter 21 Areas of Bounded Regions – Exercise 21.1 | Set 3

**Question 21. Draw a rough sketch of the curve **** and find the area between x-axis, the curve and the ordinates x = 0, x = ****π**

**Solution:**

Here, we have to find the bounded by

x-axis, x = 0 and x = π

Here is the table for values of

x 0 π 1.57 2.07 2.57 3.07 3.57 3.07 2.57 2.07 1.57 Here is the rough sketch,

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABCDO

Required area = square units

**Question 22. Draw a rough sketch of the curve **** and find the area between the x-axis, the curve and the ordinates x = 0, x = **π**.**

**Solution:**

Here, we have the area between y-axis,

x = 0,

x = π

and

Thus, the table for equation (1) is

x 0 π y 0 0.66 1.25 1.88 2.5 1.88 1.25 0.66 0 Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABOA

Required area = square units

**Question 23. Find the area bounded by the curve y = cos x between x = 0 and x = 2****π**

**Solution:**

Here from the figure we can see that

The required area = area of the region OABO + area of the region BCDB + area of the region DEFD

Therefore,

The required area =

**Question 24. Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = **** are the ratio 2:3.**

**Solution:**

We have to find the area under the curve

y = sin x ……..(1)

and

y = sin 2x …………(2)

Between x = 0 and x =

x

y = sin x

y = sin 2x 0 0.8 1 0.8 0 Here is the rough sketch

Area under curve y = sin 2x

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y1

Area of rectangle = y1△x

The approx rectangles slide from x = 0 to x = ,

Thus,

Required area = Region OPACO

We slice it into approximation rectangle of

Width = △x

Length = y2

Area of rectangle = y2△x

The approx rectangles slide from x = 0 to x = ,

Thus,

Required area = Region OQACO

Thus,

**Question 25. Compare the area under the curves y = cos**^{2}x and y = sin^{2}x between x = 0 and x = **π**

^{2}x and y = sin

^{2}x between x = 0 and x =

**Solution:**

Here to compare area under curves

y = cos

^{2}xand

y = sin

^{2}xBetween x = 0 and x = π

This is the table for y = cos

^{2}x and y = sin^{2}x

x

y = cos

^{2}x0

1

y = sin ^{2}x0 0.25 0.5 0.75 1 0.75 0.5 0.25 0 Area of region enclosed by

y = cos

^{2}x and axisA

_{1}= Region OABO + Region BCDB= 2(Region BCDB)

Area of region enclosed by y = sin

^{2}x and axisA

_{2}= Region OEDOFrom equation (1) and (2),

A

_{1}= A_{2}Thus,

Area enclosed by y = cos

^{2}x = Area enclosed by y = sin^{2}x.

**Question 26. Find the area bounded by the ellipse **** and the ordinates x = 0 and x = ae, where, b**^{2} = a^{2}(1 – e ^{2}) and e < 1.

^{2}= a

^{2}(1 – e

^{2}) and e < 1.

**Solution:**

Thus, the required area in the figure below of the region BOB’RFSB is enclosed by the ellipse and the lines x = 0 and x = ae

Here is the area of the region BOB’RFSB

**Question 27. Find the area of the minor segment of the circle x**^{2} + y^{2} = a^{2} cut-off by the line x = ** .**

^{2}+ y

^{2}= a

^{2}cut-off by the line x =

**Solution:**

Area of the mirror segment of the circle

**Question 28. Find the area of the region bounded by the curve x = at, y = 2at between the ordinates corresponding t = 1 and t = 2.**

**Solution:**

Area of the bounded region

**Question 29. Find the area enclosed by the curve x = 3 cos t, y = 2 sin t.**

**Solution:**

Area of the bounded region

= -8 [0 – 1]

= 8 square units