Solution

(b) According to the given condition,

Area of circle $=$ Area of first circle + Area of second circle

$ \begin{matrix} \therefore & \pi R^{2}=\pi R_1^{2}+\pi R_2^{2} \\ \Rightarrow & R^{2}=R_1^{2}+R_2^{2} \end{matrix} $

Solution

(a) According to the given condition,

Circumference of circle $=$ Circumference of first circle + Circumference of second circle

$ \begin{aligned} \therefore & 2 \pi R = 2 \pi R_1+2 \pi R_2 \\ \Rightarrow & R = R_1+R_2 \end{aligned} $

Solution

(b) According to the given condition,

Circumference of a circle $=$ Perimeter of square

$ 2 \pi r=4 a $

[where, $r$ and $a$ are radius of circle and side of square respectively]

$\Rightarrow \quad \dfrac{22}{7} r=2 a \Rightarrow 11 r=7 a$

$\Rightarrow \quad a=\dfrac{11}{7} r \Rightarrow r=\dfrac{7 a}{11}$

Now, $\quad$ area of circle, $A_1=\pi r^{2}$

$ \begin{aligned} & =\pi \dfrac{7 a}11^{2}=\dfrac{22}{7} \times \dfrac{49 a^{2}}{121} \\ & =\dfrac{14 a^{2}}{11} \end{aligned} $

and area of square, $A_2=(a)^{2}$

From Eqs. (ii) and (iii), $\quad A_1=\dfrac{14}{11} A_2$

$\therefore \quad A_1>A_2$

Hence, Area of the circle $>$ Area of the square.

Solution

(a) Take a point $C$ on the circumference of the semi-circle and join it by the end points of diameter $A$ and $B$.

$\therefore \quad \angle C=90^{\circ}$

So, $\triangle A B C$ is right angled triangle.

$\therefore$ Area of largest $\triangle A B C=\dfrac{1}{2} \times A B \times C D$

$ \begin{aligned} & =\dfrac{1}{2} \times 2 r \times r \\ & =r^{2} \text{ sq units } \end{aligned} $

[by property of circle] [angle in a semi-circle are right angle]

Solution

(b) Let radius of circle be $r$ and side of a square be $a$. According to the given condition,

Perimeter of a circle $=$ Perimeter of a square

$$ \begin{align*} & \therefore \quad \begin{align*} 2 \pi r = 4 a \Rightarrow a=\dfrac{\pi r}{2} \\ \text{ Now, } \quad \dfrac{\text{ Area of circle }}{\text{ Area of square }} = \dfrac{\pi^{2}}{(a)^{2}}=\dfrac{\pi r^{2}}{\dfrac{\pi r}{2}} \\ & =\dfrac{\pi^{2}}{\pi^{2} r^{2} / 4}=\dfrac{4}{\pi}=\dfrac{4}{22 / 7}=\dfrac{28}{22}=\dfrac{14}{11} \end{align*} \tag{i} \end{align*} $$

Solution

(a) Area of first circular park, whose diameter is $16 m$

$ =\pi^{2}=\pi \dfrac{16}2^{2}=64 \pi m^{2} $

$ \because r=\dfrac{d}{2}=\dfrac{16}{2}=8 m $

Area of second circular park, whose diameter is $12 m$

$ =\pi \dfrac{12}2^{2}=\pi(6)^{2}=36 \pi m^{2} \quad \because r=\dfrac{d}{2}=\dfrac{12}{2}=6 m $

According to the given condition,

Area of single circular park $=$ Area of first circular park + Area of second circular park

$ \pi R^{2}=64 \pi+36 \pi \quad[\because R \text{ be the radius of single circular park }] $

$ \begin{matrix} \Rightarrow & \pi R^{2}=100 \pi \Rightarrow R^{2}=100 \\ \therefore & R=10 m \end{matrix} $

Solution

(d) Given, $\quad$ side of square $=6 cm$

$\therefore$ Diameter of a circle, $(d)=$ Side of square $=6 cm$

$\therefore \quad$ Radius of a circle $(r)=\dfrac{d}{2}=\dfrac{6}{2}=3 cm$

$\therefore \quad$ Area of circle $=\pi(r)^{2}$

$ =\pi(3)^{2}=9 \pi cm^{2} $

Solution

(b) Given, radius of circle, $r=O C=8 cm$.

$\therefore$ Diameter of the circle $=A C=2 \times O C=2 \times 8=16 cm$

which is equal to the diagonal of a square.

Let side of square be $x$.

In right angled $\triangle A B C, \quad A C^{2}=A B^{2}+B C^{2}$

[by Pythagoras theorem]

$ \begin{aligned} & \Rightarrow \quad(16)^{2}=x^{2}+x^{2} \\ & \Rightarrow \quad 256=2 x^{2} \\ & \Rightarrow \quad x^{2}=128 \\ & \therefore \quad \text{ Area of square }=x^{2}=128 cm^{2} \end{aligned} $

Alternate Method

Radius of circle $(r)=8 cm$

Diameter of circle $(d)=2 r=2 \times 8=16 cm$

Since, square inscribed in circle.

$\therefore$ Diagonal of the squre $=$ Diameter of circle

Now, $\quad$ Area of square $=\dfrac{(\text{ Diagonal })^{2}}{2}=\dfrac{(16)^{2}}{2}=\dfrac{256}{2}=128 cm^{2}$

Solution

(c) $\because$ Circumference of first circle $=2 \pi r=\pi d_1=36 \pi cm$

[given, $d_1=36 cm$ ] and circumference of second circle $=\pi d_2=20 \pi cm$ According to the given condition,

[given, $d_2=20 cm$ ]

Circumference of circle $=$ Circumference of first circle + Circumference of second circle

$\Rightarrow \quad \pi D=36 \pi+20 \pi \quad$ [where, $D$ is diameter of a circle]

$\Rightarrow \quad D=56 cm$

So, diameter of a circle is $56 cm$.

$\therefore \quad$ Required radius of circle $=\dfrac{56}{2}=28 cm$

Solution

(d) Let $r_1=24 cm$ and $r_2=7 cm$

$\therefore \quad$ Area of first circle $=\pi_1^{2}=\pi(24)^{2}=576 \pi cm^{2}$

and area of second circle $=\pi r_2^{2}=\pi(7)^{2}=49 \pi cm^{2}$

According to the given condition,

Area of circle $=$ Area of first circle + Area of second circle

$\begin{array}{lll}\therefore & \pi R^2 = 576 \pi+49 \pi \\ \Rightarrow & R^2 = 625 \Rightarrow R=25 \mathrm{~cm} \\ \therefore & \text { Diameter of a circle } = 2 R=2 \times 25=50 \mathrm{~cm}\end{array}$

Solution

False

Let $A B C D$ be a square of side a.

$\therefore$ Diameter of circle $=$ Side of square $=a$

$\therefore \quad$ Radius of circle $=\dfrac{a}{2}$

$\therefore \quad$ Area of circle $=\pi(\text{ Radius })^{2}=\pi \quad \dfrac{a}2^{2}=\dfrac{\pi a^{2}}{4}$

Hence, area of the circle is $\dfrac{\pi a^{2}}{4} cm^{2}$.

Solution

True

Given, radius of circle, $r=a cm$

$\therefore$ Diameter of circle, $d=2 \times$ Radius $=2 acm$

$\therefore \quad$ Side of a square $=$ Diameter of circle

$ =2 acm $

$\therefore$ Perimeter of a square $=4 \times($ Side $)=4 \times 2 a$

$ =8 acm $

Solution

False

Given diameter of circle is $d$.

$\therefore$ Diagonal of inner square $=$ Diameter of circle $=d$

Let side of inner square $E F G H$ be $x$.

$\therefore$ In right angled $\triangle E F G$,

$ \begin{aligned} & \qquad \quad E G^{2}=E F^{2}+F G^{2} \quad \text{ [by Pythagoras theorem] } \\ & \Rightarrow \quad d^{2}=x^{2}+x^{2} \\ & \therefore \quad d^{2}=2 x^{2} \Rightarrow x^{2}=\dfrac{d^{2}}{2} \\ & \text{ But side of the outer square } A B C S=\text{ Diameter of circle }=d \\ & \therefore \quad \text{ Area of outer square }=d^{2} \\ & \text{ Hence, area of outer square is not equal to four times the area of the inner square. } \end{aligned} $

Solution

False

It is true only in the case of minor segment. But in case of major segment area is always greater than the area of sector.

Solution

False

Because the distance travelled by the wheel in one revolution is equal to its circumference i.e., $\pi d$.

i.e., $\quad \pi(2 r)=2 \pi r=$ Circumference of wheel $\quad[\because d=2 r]$

Solution

True

The distance covered in one revolution is $2 \pi$ r. i.e., its circumference.

Solution

False

If $0<r<2$, then numerical value of circumference is greater than numerical value of area of circle and if $r>2$, area is greater than circumference.

Solution

False

Let two circles $C_1$ and $C_2$ of radius $r$ and $2 r$ with centres $O$ and $O^{\prime}$, respectively.

It is given that, the arc length $\widehat{A B}$ of $C_1$ is equal to arc length $\widehat{C D}$ of $C_2$ i.e., $\widehat{A B}=\widehat{C D}=l$ (say)

Now, let $\theta _1$ be the angle subtended by arc $ \widehat{A B} $ of $\theta _2$ be the angle subtended by arc $\widehat{C D}$ at the centre.

$ \begin{aligned} & \widehat{A B}=l=\dfrac{Q_1}{360} \times 2 \pi r \\ & \widehat{C D}=l=\dfrac{\theta _2}{360} \times 2 \pi(2 r)=\dfrac{\theta _2}{360} \times 4 \pi r \end{aligned} $

From Eqs. (i) and (ii),

$ \begin{aligned} \Rightarrow \quad \dfrac{\theta _1}{360} \times 2 \pi r = \dfrac{\theta _2}{360} \times 4 \pi r \\ \theta _1 = 2 \theta _2 \end{aligned} $

i.e., angle of the corresponding sector of $C_1$ is double the angle of the corresponding sector of $C_2$.

It is true statement.

Solution

False

It is true for arcs of the same circle. But in different circle, it is not possible.

Solution

False

It is true for arcs of the same circle. But in different circle, it is not possible.

Solution

False

The area of the largest circle that can be drawn inside a rectangle is $\pi \dfrac{b}2^{2} cm$, where $\dfrac{b}{2}$ is the radius of the circle and it is possible when rectangle becomes a square.

Solution

True

If circumference of two circles are equal, then their corresponding radii are equal. So, their areas will be equal.

Solution

True

If areas of two circles are equal, then their corresponding radii are equal. So, their circumference will be equal.

Solution

True

When the square is inscribed in the circle, the diameter of a circle is equal to the diagonal of a square but not the side of the square.

Solution

Let the radius of a circle be $r$.

$\therefore \quad$ Circumference of a circle $=2 \pi r$

Let the radii of two circles are $r_1$ and $r_2$ whose values are $15 cm$ and $18 cm$ respectively.

i.e. $\quad r_1=15 cm \text{ and } r_2=18 cm$

Now, by given condition,

Circumference of circle $=$ Circumference of first circle + Circumference of second circle

$\Rightarrow \quad 2 \pi r=2 \pi r_1+2 \pi r_2$

$\Rightarrow \quad r=r_1+r_2$

$\Rightarrow \quad r=15+18$

$\therefore \quad r=33 cm$

Hence, required radius of a circle is $33 cm$.

Solution

Let the side of a square be $a$ and the radius of circle be $r$.

Given that, length of diagonal of square $=8 cm$

$\Rightarrow$ $a \sqrt{2}=8$

$\Rightarrow$ $a=4 \sqrt{2} cm$

Now, $\quad$ Diagonal of a square $=$ Diameter of a circle

$\Rightarrow$ Diameter of circle $=8$

$\Rightarrow$ Radius of circle $=r=\dfrac{\text{ Diameter }}{2}$

$\Rightarrow \quad r=\dfrac{8}{2}=4 cm$

$\therefore \quad$ Area of circle $=\pi r^{2}=\pi(4)^{2}$

$ =16 \pi \times cm^{2} $

and Area of square $=a^{2}=(4 \sqrt{2})^{2}$

$ =32 cm^{2} $

So, the area of the shaded region $=$ Area of circle - Area of square

$ =(16 \pi-32) cm^{2} $

Hence, the required area of the shaded region is $(16 \pi-32) cm^{2}$.

Solution

Given that, Radius of a circle, $r=28 cm$

and measure of central angle $\theta=45^{\circ}$

$\therefore \quad$ Area of a sector of a circle $=\dfrac{\pi^{2}}{360^{\circ}} \times \theta$

$ \begin{aligned} & =\dfrac{22}{7} \times \dfrac{(28)^{2}}{360} \times 45^{\circ} \\ & =\dfrac{22 \times 28 \times 28}{7} \times \dfrac{45^{\circ}}{360^{\circ}} \\ & =22 \times 4 \times 28 \times \dfrac{1}{8} \\ & =22 \times 14 \\ & =308 cm^{2} \end{aligned} $

Hence, the required area of a sector of a circle is $308 cm^{2}$.

Solution

Given, radius of wheel, $r=35 cm$

Circumference of the wheel $=2 \pi$

$ \begin{aligned} & =2 \times \dfrac{22}{7} \times 35 \\ & =220 cm \end{aligned} $

But speed of the wheel $=66 kmh^{-1}=\dfrac{66 \times 1000}{60} m / min$

$ \begin{aligned} & =1100 \times 100 cmmin^{-1} \\ & =110000 cmmin^{-1} \end{aligned} $

$\therefore$ Number of revolutions in $1 min=\dfrac{110000}{220}=500$ revolution

Hence, required number of revolutions per minute is 500 .

Solution

Let $A B C D$ be a rectangular field of dimensions $20 m \times 16 m$. Suppose, a cow is tied at a point $A$. Let length of rope be $A E=14 m=r$ (say).

$\therefore$ Area of the field in which the cow graze $=$ Area of sector $A F E G=\dfrac{\theta}{360^{\circ}} \times \pi r^{2}$

$ =\dfrac{90}{360} \times \pi(14)^{2} $

[so, the angle between two adjacent sides of a rectangle is $90^{\circ}$ ]

$ \begin{aligned} & =\dfrac{1}{4} \times \dfrac{22}{7} \times 196 \\ & =154 m^{2} \end{aligned} $

Solution

Length and breadth of a circular bed are $38 cm$ and $10 cm$.

$\therefore \quad$ Area of rectangle $A C D F=$ Length $\times$ Breadth $=38 \times 10=380 cm^{2}$

Both ends of flower bed are semi-circles.

$\therefore \quad$ Radius of semi-circle $=\dfrac{D F}{2}=\dfrac{10}{2}=5 cm$

$\therefore \quad$ Area of one semi-circles $=\dfrac{\pi r^{2}}{2}=\dfrac{\pi}{2}(5)^{2}=\dfrac{25 \pi}{2} cm^{2}$

$\therefore \quad$ Area of two semi-circles $=2 \times \dfrac{25}{2} \pi=25 \pi cm^{2}$

$\therefore \quad$ Total area of flower bed $=$ Area of rectangle ACDF + Area of two semi-circles

$ =(380+25 \pi) cm^{2} $

Solution

Given,

$A C=6 cm$ and $B C=8 cm$

We know that, triangle in a semi-circle with hypotenuse as diameter is right angled triangle.

$\therefore$ $\angle C=90^{\circ}$

In right angled $\triangle A C B$, use Pythagoras theorem,

$ \begin{matrix} \therefore & & A B^{2} = A C^{2}+C B^{2} \\ \Rightarrow & & A B^{2} = 6^{2}+8^{2}=36+64 \\ \Rightarrow & & A B^{2} = 100 \\ \Rightarrow & & A B = 10 cm \quad \text{ [since, side cannot be negative] } \\ \therefore & & \text{ Area of } \triangle A B C=\dfrac{1}{2} \times B C \times A C = \dfrac{1}{2} \times 8 \times 6=24 cm^{2} \end{matrix} $

Here, diameter of circle, $A B=10 cm$

$\therefore \quad$ Radius of circle, $r=\dfrac{10}{2}=5 cm$

$ \begin{aligned} \text{ Area of circle } = \pi^{2}=3.14 \times(5)^{2} \\ & =3.14 \times 25=78.5 cm^{2} \end{aligned} $

$\therefore \quad$ Area of the shaded region $=$ Area of circle - Area of $\triangle A B C$

$ =78.5-24=54.5 cm^{2} $

Solution

In a figure, join $E D$

From figure, radius of semi-circle DFE, $r=6-4=2 m$

Now, $\quad$ area of rectangle $A B C D=B C \times A B=8 \times 4=32 m^{2}$

and $\quad$ area of semi-circle DFE $=\dfrac{\pi^{2}}{2}=\dfrac{\pi}{2}(2)^{2}=2 \pi m^{2}$

$\therefore$ Area of shaded region $=$ Area of rectangle $A B C D+$ Area of semi-circle DFE $=(32+2 \pi) m^{2}$

Solution

Join $G H$ and $F E$

Here, $\quad$ breadth of the rectangle $B C=12 m$

$\therefore \quad$ Breadth of the inner rectangle $E F G H=12-(4+4)=4 cm$

which is equal to the diameter of the semi-circle $E J F, d=4 m$

$\therefore \quad$ Radius of semi-circle EJF, $r=2 m$

$\therefore \quad$ Length of inner rectangle $E F G H=26-(5+5)=16 m$

$\therefore$ Area of two semi-circles EJF and HIG $=2 \quad \dfrac{\pi r^{2}}{2}=2 \times \pi \dfrac{(2)^{2}}{2}=4 \pi m$

Now, area of inner rectangle $E F G H=E H \times F G=16 \times 4=64 m^{2}$

and $\quad$ area of outer rectangle $A B C D=26 \times 12=312 m^{2}$

$\therefore \quad$ Area of shaded region $=$ Area of outer rectangle - (Area of two semi-circles

$ =312-(64+4 \pi)=(248-4 \pi) m^{2} $

(Area of inner rectangle)

Solution

Given that, radius of circle $(r)=14 cm$

and angle of the corresponding sector i.e., central angle $(\theta)=60^{\circ}$

Since, in $\triangle A O B, O A=O B=$ Radius of circle i.e., $\triangle A O B$ is isosceles.

$\Rightarrow \quad \angle O A B=\angle O B A=\theta$

Now, in $\triangle O A B \quad \angle A O B+\angle O A B+\angle O B A=180^{\circ}$

[since,sum of interior angles of any triangle is $180^{\circ}$ ]

$\Rightarrow$ $60^{\circ}+\theta+\theta=180^{\circ}$

$\Rightarrow$ $2 \theta=120^{\circ}$

$\Rightarrow$ $\theta=60^{\circ}$

i.e. $\quad \angle O A B=\angle O B A=60^{\circ}=\angle A O B$

Since, all angles of $\triangle A O B$ are equal to $60^{\circ}$ i.e., $\triangle A O B$ is an equilateral triangle.

Also,

$O A=O B=A B=14 cm$

So, $\quad$ Area of $\triangle O A B=\dfrac{\sqrt{3}}{4}(\text{ side })^{2}$

$ \begin{aligned} & =\dfrac{\sqrt{3}}{4} \times(14)^{2} \quad[\because \text{ area of an equilateral triangle }=\dfrac{\sqrt{3}}{4}(\text{ side })^{2}] \\ & =\dfrac{\sqrt{3}}{4} \times 196=49 \sqrt{3} cm^{2} \end{aligned} $

and $\quad$ area of sector $O B A O=\dfrac{\pi^{2}}{360^{2}} \times \theta$

$ \begin{aligned} & =\dfrac{22}{7} \times \dfrac{14 \times 14}{360} \times 60^{\circ} \\ & =\dfrac{22 \times 2 \times 14}{6}=\dfrac{22 \times 14}{3}=\dfrac{308}{3} cm^{2} \end{aligned} $

$\therefore \quad$ Area of minor segment $=$ Area of sector OBAO - Area of $\triangle O A B$

$ =\dfrac{308}{3}-49 \sqrt{3} cm^{2} $

Hence, the required area of the minor segment is $\dfrac{308}{3}-49 \sqrt{3} cm^{2}$.

Solution

Given, side of a square $B C=12 cm$

Since, $Q$ is a mid-point of $B C$.

$\therefore \quad$ Radius $=B Q=\dfrac{12}{2}=6 cm$

Now,

Now, area of quadrant $B P Q=\dfrac{\pi^{2}}{4}=\dfrac{3.14 \times(6)^{2}}{4}=\dfrac{113.04}{4} cm^{2}$ Area of four quadrants $=\dfrac{4 \times 113.04}{4}=1123.04 cm^{2}$

$\therefore \quad$ Area of the shaded region $=$ Area of square - Area of four quadrants

$ =144-113.04=30.96 cm^{2} $

Solution

Since, $A B C$ is an equilateral triangle. $\therefore$ $\angle A=\angle B=\angle C=60^{\circ}$ and $A B=B C=A C=10 cm$

So, $E, F$ and $D$ are mid-points of the sides.

$ \begin{aligned} \therefore \quad A E = E C=C D=B D=B F=F A=5 cm \\ \text{ Now, } \quad \text{ area of sector } C D E = \dfrac{\theta \pi r^{2}}{360}=\dfrac{60 \times 3.14}{360}(5)^{2} \\ =\dfrac{3.14 \times 25}{6}=\dfrac{78.5}{6}=13.0833 cm^{2} \\ \therefore \quad \text{ Area of shaded region } = 3(\text{ Area of sector } C D E) \\ =3 \times 13.0833 \\ =39.25 cm^{2} \end{aligned} $

Solution

Given that, radii of each $arc(r)=14 cm$

Now, area of the sector with central $\angle P=\dfrac{\angle P}{360^{\circ}} \times \pi^{2}$

$ =\dfrac{\angle P}{360^{\circ}} \times \pi \times(14)^{2} cm^{2} $

$[\because.$ area of any sector with central angle $\theta$ and radius $.r=\dfrac{\pi r^{2}}{360^{\circ}} \times \theta]$

Area of the sector with central angle $=\dfrac{\angle Q}{360^{\circ}} \times \pi r^{2}=\dfrac{\angle Q}{360^{\circ}} \times \pi \times(14)^{2} cm^{2}$

and area of the sector with central angle $R=\dfrac{\angle R}{360^{\circ}} \times \pi^{2}=\dfrac{\angle R}{360^{\circ}} \times \pi \times(14)^{2} cm^{2}$

Therefore, sum of the areas (in $cm^{2}$ ) of three sectors

$ \begin{aligned} & =\dfrac{\angle P}{360^{\circ}} \times \pi \times(14)^{2}+\dfrac{\angle \theta}{360^{\circ}} \times \pi \times(14)^{2}+\dfrac{\angle R}{360^{\circ}} \times \pi \times(14)^{2} \\ & =\dfrac{\angle P+\angle Q+\angle R}{360} \times 196 \times \pi=\dfrac{180^{\circ}}{360^{\circ}} \times 196 \pi cm^{2} \end{aligned} $

[since, sum of all interior angles in any triangle is $180^{\circ}$ ]

$=98 \pi cm^{2}=98 \times \dfrac{22}{7}$

$=14 \times 22=308 cm^{2}$

Hence, the required area of the shaded region is $308 cm^{2}$.

Solution

Given that, a circular park is surrounded by a road.

Width of the road $=21 m$

Radius of the park $(r_i)=105 m$

$\therefore$ Radius of whole circular portion (park + road),

$ r_e=105+21=126 m $

Now, area of road $=$ Area of whole circular portion

$ \begin{aligned} \text{-Area of circular park } & \longrightarrow \\ & =\pi r_e^{2}-\pi_i^{2} \\ & =\pi(r_e^{2}-r_i^{2}) \\ & =\pi{(126^{2}-(105)^{2}}. \\ & =\dfrac{22}{7} \times(126+105)(126-105) \\ & =\dfrac{22}{7} \times 231 \times 21 \quad[\because \text{ area of circle }=\pi r^{2}] \\ & =66 \times 231 \\ & =15246 cm^{2} \end{aligned} $

Hence, the required area of the road is $15246 cm^{2}$.

Solution

Given that, radius of each $arc(r)=21 cm$

Area of sector with $\angle A=\dfrac{\angle A}{360^{\circ}} \times \pi r^{2}=\dfrac{\angle A}{360^{\circ}} \times \pi \times(21)^{2} cm^{2}$

$[\because.$ area of any sector with central angle $\theta$ and radius $.r=\dfrac{\pi^{2}}{360^{\circ}} \times \theta]$

Area of sector with $\angle B=\dfrac{\angle B}{360^{\circ}} \times \pi r^{2}=\dfrac{\angle B}{360^{\circ}} \times \pi \times(21)^{2} cm^{2}$

Area of sector with $\angle C=\dfrac{\angle C}{360^{\circ}} \times \pi r^{2}=\dfrac{\angle C}{360^{\circ}} \times \pi \times(21)^{2} cm^{2}$

and area of sector with $\angle D=\dfrac{\angle D}{360^{\circ}} \times \pi^{2}=\dfrac{\angle D}{360^{\circ}} \times \pi \times(21)^{2} cm^{2}$

Therefore, sum of the areas (in $cm^{2}$ ) of the four sectors

$ \begin{aligned} & =\dfrac{\angle A}{360^{\circ}} \times \pi \times(21)^{2}+\dfrac{\angle B}{360^{\circ}} \times \pi \times(21)^{2}+\dfrac{\angle C}{360^{\circ}} \times \pi \times(21)^{2}+\dfrac{\angle D}{360^{\circ}} \times \pi \times(21)^{2} \\ & =\dfrac{(\angle A+\angle B+\angle C+\angle D)}{360^{\circ}} \times \pi \times(21)^{2} \\ & \quad[\because \text{ sum of all interior angles in any quadrilateral }=360^{\circ}] \end{aligned} $

Hence, required area of the shaded region is $1386 cm^{2}$.

Solution

Length of arc of circle $=20 cm$

Here,

central angle $\theta=60^{\circ}$

$\therefore$ $ \hspace{10 mm} \text{Length of arc } = \dfrac{\theta}{360^{\circ}} \times 2 \pi r $

$\Rightarrow$ $\hspace{10 mm} 20 = \dfrac{60^{\circ}}{360^{\circ}} \times 2 \pi r \Rightarrow \dfrac{20 \times 6}{2 \pi} = r$

$\therefore$ $\hspace{10 mm} r = \dfrac{60}{\pi} cm $

Hence, the radius of circle is $\dfrac{60}{\pi} cm$.

Solution

Given, area of a circular playground $=22176 m^{2}$

$$ \begin{array} \therefore & \pi r^{2}=22176 \quad[\because \text{ area of circle }=\pi r^{2}] \\ \Rightarrow & \dfrac{22}{7} r^{2}=22176 \Rightarrow r^{2}=1008 \times 7 \\ \Rightarrow & r^{2}=7056 \Rightarrow r=84 m \\ \therefore & \text{ Circumference of a circle }=2 \pi r=2 \times \dfrac{22}{7} \times 84 \\ & =44 \times 12=528 m \\ \therefore & \text{ Cost of fencing this ground }=528 \times 50=\text{₹} 26400 \end{array} $$

Solution

Given, diameter of front wheels, $d_1=80 cm$

and diameter of rear wheels, $d_2=2 m=200 cm$

$\therefore \quad$ Radius of front wheel $(r_1)=\dfrac{80}{2}=40 cm$

and radius of rear wheel $(r_2)=\dfrac{200}{2}=100 cm$

$\therefore \quad$ Circumference of the front wheel $=2 \pi r_1=\dfrac{2 \times 22}{7} \times 40=\dfrac{1760}{7}$

$\therefore$ Total distance covered by front wheel $=1400 \times \dfrac{1760}{7}=200 \times 1760$

$ =352000 cm $

Number of revolutions by rear wheel $=\dfrac{\text{ Distance coverd }}{\text{ Circumference }}$

$ =\dfrac{352000}{2 \times \dfrac{22}{7} \times 100}=\dfrac{7 \times 3520}{2 \times 22}=\dfrac{24640}{44}=560 $

Solution

Given that, a triangular field with the three corners of the field a cow, a buffalo and a horse are tied separately with ropes. So, each animal grazed the field in each corner of triangular field as a sectorial form.

Given, radius of each sector $(r)=7 m$

Now, area of sector with $\angle C=\dfrac{\angle C}{360^{\circ}} \times {\pi} r^2=\dfrac{\angle C}{360^{\circ}} \times \pi \times(7)^{2} m^{2}$

Area of the sector with $\angle B=\dfrac{\angle B}{360^{\circ}} \times {\pi} r^2=\dfrac{\angle B}{360^{\circ}} \times \pi \times(7)^{2} m^{2}$

and area of the sector with $\angle H=\dfrac{\angle H}{360^{\circ}} \times \pi r^{2}=\dfrac{\angle H}{360^{\circ}} \times \pi \times(7)^{2} m^{2}$

Therefore, sum of the areas (in $cm^{2}$ ) of the three sectors

$ \begin{aligned} & =\dfrac{\angle C}{360^{\circ}} \times \pi \times(7)^{2}+\dfrac{\angle B}{360^{\circ}} \times \pi \times(7)^{2}+\dfrac{\angle H}{360^{\circ}} \times \pi \times(7)^{2} \\ & =\dfrac{(\angle C+\angle B+\angle H)}{360^{\circ}} \times \pi \times 49 . \\ & =\dfrac{180^{\circ}}{360^{\circ}} \times \dfrac{22}{7} \times 49=11 \times 7=77 m^{2} \end{aligned} $

Given that, sides of triangle are $a=15, b=16$ and $c=17$

Now, semi-perimeter of triangle, $s=\dfrac{a+b+c}{2}$

$ \begin{matrix} \Rightarrow \quad = \dfrac{15+16+17}{2}=\dfrac{48}{2}=24 \\ \therefore \quad \text{ Area of triangular field } = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{24 \cdot 9 \cdot 8 \cdot 7} \\ =\sqrt{64 \cdot 9 \cdot 21} \\ =8 \times 3 \sqrt{21}=24 \sqrt{21} m^{2} \end{matrix} \quad \text{ [by Heron’s formula] } $

So, area of the field which cannot be grazed by the three animals

$ \begin{aligned} & =\text{ Area of triangular field }- \text{ Area of each sectorial field } \\ & =24 \sqrt{21}-77 m^{2} \end{aligned} $

Hence, the required area of the field which can not be grazed by the three animals is $(24 \sqrt{21}-77) m^{2}$.

Solution

Given that, radius of a circle $(r)=12 cm$

and central angle of sector $OBCA(\theta)=60^{\circ}$

$\therefore$ Area of sector $O B C A=\dfrac{\pi r^{2}}{360} \times \theta \quad$ [here, $O B C A=$ sector and $A B C A=$ segment]

$ \begin{aligned} & =\dfrac{3.14 \times 12 \times 12}{360^{\circ}} \times 60^{\circ} \\ & =3.14 \times 2 \times 12 \\ & =3.14 \times 24=75.36 cm^{2} \end{aligned} $

Since, $\triangle O A B$ is an isosceles triangle.

$ \begin{matrix} \text{ Let } & \angle O A B = \angle O B A=\theta _1 \\ \text{ and } & O A = O B=12 cm \\ \therefore & \angle O A B+\angle O B A+\angle A O B = 180^{\circ} \quad[\because \text{ sum of all interior angles of a triangle is } 180^{\circ}] \\ \Rightarrow & \theta _1+\theta _1+60^{\circ} = 180^{\circ} \\ \Rightarrow & 2 \theta _1 = 120^{\circ} \\ \Rightarrow & \theta _1 = 60^{\circ} \\ \therefore & \theta _1 = \theta=60^{\circ} \end{matrix} $

So, the required $\triangle A O B$ is an equilateral triangle.

Now, $\quad$ area of $\triangle A O B=\dfrac{\sqrt{3}}{4}(\text{ side })^{2} \quad[\because.$ area of an equilateral triangle $.=\dfrac{\sqrt{3}}{4}(\text{ side })^{2}]$

$ \begin{aligned} & =\dfrac{\sqrt{3}}{4}(12)^{2} \\ & =\dfrac{\sqrt{3}}{4} \times 12 \times 12=36 \sqrt{3} cm^{2} \end{aligned} $

Now, area of the segment of a circle i.e.,

$A B C A=$ Area of sector $O B C A-$ Area of $\triangle A O B$

$ =(75.36-36 \sqrt{3}) cm^{2} $

Hence, the required area of segment of a circle is $(75.36-36 \sqrt{3}) cm^{2}$.

Solution

Given that, a circular pond is surrounded by a wide path.

The diameter of circular pond $=17.5 m$

$ \begin{matrix} \therefore \quad \text{ Radius of circular pond }(r_i) = \dfrac{\text{ Diameter }}{2} \\ \text{ i.e., } & O A=r_i=\dfrac{17.5}{2}=8.75 m \end{matrix} $

and the width of the path $=2 m$

i.e., $\quad A B=2 m$

Now, length of $O B=O A+A B=r_i+A B$

Let $\quad(r_e)=8.75+2=10.75 m$

So, area of circular path $=$ Area of outer circle i.e., (circular pond + path)

$ \begin{aligned} & =\pi r_e^{2}-\pi r_i^{2} \\ & =\pi(r_e^{2}-r_i^{2}) \\ & =\pi{(10.75)^{2}-(8.75)^{2}} \\ & =\pi{(10.75+8.75)(10.75-8.75)} \\ & =3.14 \times 19.5 \times 2 \\ & =122.46 m^{2} \end{aligned} $

Now, cost of constructing the path per square metre $=\text{₹} 25$

$\therefore$ Cost constructing the path ₹ $122.46 m^{2}=122.46 \times 25$

$=\text{₹} 3061.50$

Hence, required cost of constructing the path at the rate of $ \text{₹} 25 per m^2 $ is ₹ 3061.50.

Solution

Given-

$A B C D$ is a trapezium. $A B | D C$ and $A B=18 \mathrm{~cm}, D C=32 \mathrm{~cm}$.

There are 4 arcs centring the vertices $A, B, C, D$ with radius $r=7 \mathrm{~cm}$.

To find out - area of shaded region

Solution-

Area of trapezium $=\frac{1}{2} \times$ sum of the parallel sides $\times$ distance between the parallel sides.

$=\frac{1}{2} \times(A B+C D) \times 14=\frac{1}{2} \times(18+32) \times 14 \mathrm{~cm}^2=350 \mathrm{~cm}^2 \text {. }$

Let us take the angles of the trapezium as $a$ at $A, b$ at $B, c$ at $C$ and $d$ at $D$.

Now the given arcs form 4 sectors.

Together they form a circle of radius $r=7 \mathrm{~cm}$ as $a+b+c+d=360^{\circ}$.

$\therefore$ The area of the sectors-area of circle with radius $(r=7 \mathrm{~cm})$

$=\pi \times r^2=\frac{22}{7} \times 7^2=154 \mathrm{~cm}^2 \text {. }$

$\therefore$ Area of shaded region $=$ area of trapezium - area of circle

$=(350-154)=196 \mathrm{~cm}^2 \text {. }$

Solution

Given that, three circles are in such a way that each of them touches the other two.

Now, we join centre of all three circles to each other by a line segment. Since, radius of each circle is $3.5 cm$.

So;

$ \begin{aligned} A B = 2 \times \text{ Radius of circle } \\ & =2 \times 3.5=7 cm \\ A C = B C=A B=7 cm \end{aligned} $

which shows that, $\triangle A B C$ is an equilateral triangle with side $7 cm$.

We know that, each angle between two adjacent sides of an equilateral triangle is $60^{\circ}$.

$\therefore$ Area of sector with angle $\angle A=60^{\circ}$.

$ =\dfrac{\angle A}{360^{\circ}} \times \pi r^{2}=\dfrac{60^{\circ}}{360^{\circ}} \times \pi \times(3.5)^{2} $

So,

$ \text{ area of each sector }=3 \times \text{ Area of sector with angle } A \text{. } $

$ \begin{aligned} & =3 \times \dfrac{60^{\circ}}{360^{\circ}} \times \pi \times(3.5)^{2} \\ & =\dfrac{1}{2} \times \dfrac{22}{7} \times 3.5 \times 3.5 \\ & =11 \times \dfrac{5}{10} \times \dfrac{35}{10}=\dfrac{11}{2} \times \dfrac{7}{2} \\ & =\dfrac{77}{4}=19.25 cm^{2} \end{aligned} $

and

$ \begin{aligned} \text{ Area of } \triangle A B C = \dfrac{\sqrt{3}}{4} \times(7)^{2} \\ [\because \text{ area of an equilateral triangle }=\dfrac{\sqrt{3}}{4}(\text{ side })^{2}] \\ =49 \dfrac{\sqrt{3}}{4} cm^{2} \end{aligned} $

$\therefore$ Area of shaded region enclosed between these circles $=$ Area of $\triangle A B C$

-Area of each sector

$ \begin{aligned} & =49 \dfrac{\sqrt{3}}{4}-19.25=12.25 \times \sqrt{3}-19.25 \\ & =21.2176-19.25=1.9676 cm^{2} \end{aligned} $

Hence, the required area enclosed between these circles is $1.967 cm^{2}$ (approx).

Solution

Let the central angle of the sector be $\theta$.

Given that, radius of the sector of a circle $(r)=5 cm$ and arc length $(l)=3.5 cm$

$\therefore$ Central angle of the sector, $\theta=\dfrac{\text{ arc length }(l)}{\text{ radius }}$

$ \begin{matrix} \Rightarrow & \theta=\dfrac{3.5}{5}=0.7 R & \because \theta=\dfrac{l}{r} \\ \Rightarrow & \theta=0.7 \times \dfrac{180}{\pi} & \because 1 R=\dfrac{180^{\circ}}{\pi} D^{\circ} \end{matrix} $

Now, area of sector with angle $\theta=0.7 \times \dfrac{180}{\pi}$

$ \begin{aligned} & =\dfrac{{\pi} r^2}{360^{\circ}} \times(0.7) \times \dfrac{180^{\circ}}{\pi} \\ & =\dfrac{(5)^{2}}{2} \times 0.7=\dfrac{25 \times 7}{2 \times 10}=\dfrac{175}{20}=8.75 cm^{2} \end{aligned} $

Hence, required area of the sector of a circle is $8.75 cm^{2}$.

Solution

Given that, four circular cardboard pieces arc placed on a paper in such a way that each piece touches other two pieces.

Now, we join centre of all four circles to each other by a line segment. Since, radius of each circle is $7 cm$.

So,

$ \begin{aligned} A B = 2 \times \text{ Radius of circle } \\ & =2 \times 7=14 cm \\ \Rightarrow & A B = B C=C D=A D=14 cm \end{aligned} $

which shows that, quadrilateral $A B C D$ is a square with each of its side is $14 cm$.

We know that, each angle between two adjacent sides of a square is $90^{\circ}$.

$\therefore$ Area of sector with $\angle A=90^{\circ}$

$ \begin{aligned} & =\dfrac{\angle A}{360^{\circ}} \times \pi r^{2}=\dfrac{90^{\circ}}{360^{\circ}} \times \pi \times(7)^{2} \\ & =\dfrac{1}{4} \times \dfrac{22}{7} \times 49=\dfrac{154}{4}=\dfrac{77}{2} \\ & =38.5 cm^{2} \end{aligned} $

$\therefore \quad$ Area of each sector $=4 \times$ Area of sector with $\angle A$

$ \begin{aligned} & =4 \times 38.5 \\ & =154 cm^{2} \end{aligned} $

and area of square $A B C D=(\text{ side of square })^{2}$

$ =(14)^{2}=196 cm^{2} \quad[\therefore \text{ area of square }=(\text{ side })^{2}] $

So, area of shaded region enclosed between these pieces $=$ Area of square $A B C D$ - Area of each sector

$ \begin{aligned} & =196-154 \\ & =42 cm^{2} \end{aligned} $

Hence, required area of the portion enclosed between these pieces is $42 cm^{2}$.

Solution

$\because$

Area of square $=784$

$ \begin{matrix} \therefore & (\text{ Side })^{2} = (28)^{2} \\ \Rightarrow & \text{ Side } = 28 cm \end{matrix} $

Since, all four are congruent circular plates.

$\therefore$ Diameter of each circular plate $=14 cm$

$\therefore \quad$ Radius of each circular plate $=7 cm$

Now, area of one circular plate $=\pi r^{2}=\dfrac{22}{7}(7)^{2}$

$ =154 cm^{2} $

$\therefore \quad$ Area of four circular plates $=4 \times 154=616 cm^{2}$

$\therefore$ Area of the square sheet not covered by the circular plates $=784-616=168 cm^{2}$

Solution

Given, floor of a room is covered with circular tiles.

Length of a floor of a room $(l)=5 m$

and breadth of floor of a room $(b)=4 m$

$\therefore$ Area of floor of a room $=l \times b$

$ =5 \times 4=20 m^{2} $

Diameter of each circular tile $=50 cm$

$\Rightarrow$ Radius of each circular tile $=\dfrac{50}{2}=25 cm$

$ =\dfrac{25}{100} m=\dfrac{1}{4} m \quad[\because \text{ diameter }=2 \times \text{ radius }] $

Now, area of a circular tile $=\pi{\text{ (radius })^{2}}$

$ =3.14 \times \dfrac{1}4^{2}=\dfrac{3.14}{16} m^{2} $

$\therefore$ Area of 80 circular tiles $=80 \times \dfrac{3.14}{16}=5 \times 3.14=15.7 m^{2}$

$[\because 80$ congruent circular tiles covering the floor of a room $]$ So, area of floor that remains uncovered with tiles $=$ Area of floor of a room - Area of 80 circular tiles

$ =20-15.7=4.3 m^{2} $

Hence, the required area of floor that remains uncovered with tiles is $4.3 m^{2}$.

Solution

Let the radius of the circle be $r$.

Given that,

$\pi r^2 = 1256$

$ \begin{matrix} \Rightarrow & r^{2}=\dfrac{1256}{\pi}=\dfrac{1256}{3.14}=400 \\ \Rightarrow & r^{2}=(20)^{2} \\ \Rightarrow & r=20 cm \end{matrix} $

$\therefore$ So, the radius of circle is $20 cm$.

$\Rightarrow \quad$ Diameter of circle $=2 \times$ Radius

$ \begin{aligned} & =2 \times 20 \\ & =40 cm \end{aligned} $

Since, all the vertices of a rhombus lie on a circle that means each diagonal of a rhombus must pass through the centre of a circle that is why both diagonals are equal and same as the diameter of the given circle.

Let $d_1$ and $d_2$ be the diagonals of the rhombus.

$\therefore \quad d_1=d_2=$ Diameter of circle $=40 cm$

So, $\quad$ Area of rhombus $=\dfrac{1}{2} \times d_1 \times d_2$

$ \begin{aligned} & =\dfrac{1}{2} \times 40 \times 40 \\ & =20 \times 40=800 cm^{2} \end{aligned} $

Hence, the required area of rhombus is $800 cm^{2}$.

Solution

Let the diameters of concentric circles be $k, 2 k$ and $3 k$.

$\therefore$ Radius of concentric circles are $\dfrac{k}{2}, k$ and $\dfrac{3 k}{2}$.

$\therefore \quad$ Area of inner circle, $A_1=\pi \dfrac{k}2^{2}=\dfrac{k^{2} \pi}{4}$

$\therefore$ Area of middle region, $A_2=\pi(k)^{2}-\dfrac{k^{2} \pi}{4}=\dfrac{3 k^{2} \pi}{4}$

$[\because.$ area of ring $=\pi(R^{2}-r^{2})$, where $R$ is radius of outer ring and $r$ is radius of inner ring $]$

and area of outer region, $A_3=\pi \dfrac{3 k}2^{2}-\pi k^{2}$

$ \begin{aligned} & =\dfrac{9 \pi k^{2}}{4}-\pi k^{2}=\dfrac{5 \pi k^{2}}{4} \\ \therefore \quad \text{ Required ratio } = A_1: A_2: A_3 \\ & =\dfrac{k^{2} \pi}{4}: \dfrac{3 k^{2} \pi}{4}: \dfrac{5 \pi k^{2}}{4}=1: 3: 5 \end{aligned} $

Solution

We know that, in $60 min$, minute hand revolving $=360^{\circ}$

In $1 min$, minute hand revolving $=\dfrac{360^{\circ}}{60^{\circ}}$

$ \begin{aligned} \therefore \quad \text{ In }(6: 05 \text{ am to } 6: 40 am) = 35 min, \\ \quad \text{ minute hand revolving } = \dfrac{360^{\circ}}{60^{\circ}} \times 35=6 \times 35 \end{aligned} $

Given that, length of minute hand $(r)=5 cm$.

$\therefore$ Area of sector $A O B A$ with angle $\angle O=\dfrac{\pi^{2}}{360} \times \angle O$

$ \begin{aligned} & =\dfrac{22}{7} \dfrac{(5)^{2}}{360^{\circ}} \times 6 \times 35 \\ & =\dfrac{22}{7} \times \dfrac{5 \times 5}{360^{\circ}} \times 6 \times 35 \\ & =\dfrac{22 \times 5 \times 5 \times 5}{60^{\circ}}=\dfrac{22 \times 5 \times 5}{12} \\ & =\dfrac{11 \times 5 \times 5}{6}=\dfrac{275}{6}=45 \dfrac{5}{6} cm^{2} \end{aligned} $

Hence, the required area swept by the minute land is $45 \dfrac{5}{6} cm^{2}$.

Solution

Let the radius of the sector $A O B A$ be $r$.

Given that, Central angle of sector $A O B A=\theta=200^{\circ}$

and $\quad$ area of the sector $A O B A=770 cm^{2}$

We know that, area of the sector $=\dfrac{\pi r^{2}}{360^{\circ}} \times \theta^{\circ}$

$\therefore \quad$ Area of the sector, $770=\dfrac{\pi^{2}}{360^{\circ}} \times 200$

$ \begin{aligned} \Rightarrow & \dfrac{77 \times 18}{\pi} = r^{2} \\ \Rightarrow & r^{2} = \dfrac{77 \times 18}{22} \times 7 \Rightarrow r^{2}=9 \times 49 \\ \Rightarrow & r = 3 \times 7 \\ \therefore & r = 21 cm \end{aligned} $

So, radius of the sector $A O B A=21 cm$.

Now, the length of the correspoding arc of this sector $=$ Central angle $\times$ Radius $\quad \because \theta=\dfrac{l}{r}$

$ \begin{aligned} & =200 \times 21 \times \dfrac{\pi}{180^{\circ}} \quad \because 1^{\circ}=\dfrac{\pi}{180} R \\ & =\dfrac{20}{18} \times 21 \times \dfrac{22}{7} \\ & =\dfrac{220}{3} cm=73 \dfrac{1}{3} cm \end{aligned} $

Hence, the required length of the corresponding arc is $73 \dfrac{1}{3} cm$.

Solution

Let the lengths of the corresponding arc be $l_1$ and $l_2$.

Given that, radius of sector $P O_1 Q P=7 cm$ and radius of sector $A O_2 B A=21 cm$ Central angle of the sector $P O_1 Q P=120^{\circ}$ and central angle of the sector $AO_2 BA=40^{\circ}$

$\therefore$ Area of the sector with central angle $O_1$

$ \begin{aligned} & =\dfrac{\pi^{2}}{360^{\circ}} \times \theta=\dfrac{\pi(7)^{2}}{360^{\circ}} \times 120^{\circ} \\ & =\dfrac{22}{7} \times \dfrac{7 \times 7}{360^{\circ}} \times 120 \\ & =\dfrac{22 \times 7}{3}=\dfrac{154}{3} cm^{2} \end{aligned} $

and area of the sector with central angle $O_2$

$ \begin{aligned} & =\dfrac{\pi^{2}}{360^{\circ}} \times \theta=\dfrac{\pi(21)^{2}}{360^{\circ}} \times 40^{\circ} \\ & =\dfrac{22}{7} \times \dfrac{21 \times 21}{360^{\circ}} \times 40^{\circ} \\ & =\dfrac{22 \times 3 \times 21}{9}=22 \times 7=154 cm^{2} \end{aligned} $

Now, corresponding arc length of the sector $P O_1 Q P$

$ \begin{aligned} & =\text{ Central angle } \times \text{ Radius of the sector } \\ & =120^{\circ} \times 7 \times \dfrac{\pi}{180^{\circ}} \\ & =\dfrac{2}{3} \times 7 \times \dfrac{22}{7} \\ & =\dfrac{44}{3} cm \end{aligned} $

and corresponding arc length of the sector $AO_2 B A$

$ \begin{aligned} & =\text{ Central angle } \times \text{ Radius of the sector } \\ & =40^{\circ} \times 21 \times \dfrac{\pi}{180^{\circ}} \\ & =\dfrac{2}{9} \times 21 \times \dfrac{22}{7} \\ & =\dfrac{2}{3} \times 22=\dfrac{44}{3} cm \end{aligned} $

Hence, we observe that arc lengths of two sectors of two different circles may be equal but their area need not be equal.

Solution

Join $J K, K L, L M$ and $M J$.

Their are four equally semi-circles and $L M J K$ formed a square.

$ \therefore \quad F H=14-(3+3)=8 cm $

So, the side of square should be $4 cm$ and radius of semi-circle of both ends are $2 cm$ each.

$ \begin{aligned} & \therefore \quad \text{ Area of square } J K L M=(4)^{2}=16 cm^{2} \\ & \text{ Area of semi-circle } H J M=\dfrac{\pi r^{2}}{2} \\ & =\dfrac{\pi \times(2)^{2}}{2}=2 \pi cm^{2} \\ & \therefore \quad \text{ Area of four semi-circle }=4 \times 6.28=25.12 cm^{2} \\ & \text{ Now, } \quad \text{ area of square } A B C D=(14)^{2}=196 cm^{2} \\ & \therefore \quad \text{ Area of shaded region }=\text{ Area of square } A B C D \\ & \text{ - [Area of four semi-circle }+ \text{ Area of square JKLM] } \\ & =196-[8 \pi+16]=196-16-8 \pi \\ & =(180-8 \pi) cm^{2} \end{aligned} $

Hence, the required of the shaded region is $(180-8 \pi) cm^{2}$.

Solution

Let the number of revolutions made by a circular wheel be $n$ and the radius of circular wheel ber.

Given that, $\quad$ area of circular wheel $=1.54 m^{2}$

$ \begin{matrix} \Rightarrow & \pi r^{2} = 1.54 \\ \Rightarrow & r^{2} = \dfrac{1.54}{22} \times 7 \Rightarrow r^{2}=0.49 \\ \therefore & r = 0.7 m \end{matrix} $

So, the radius of the wheel is $0.7 m$.

Distance travelled by a circlular wheel in one revolution $=$ Circumference of circular wheel

$ \begin{aligned} & =2 \pi r \\ & =2 \times \dfrac{22}{7} \times 0.7=\dfrac{22}{5}=4.4 m \quad[\because \text{ circumference of a circle }=2 \pi r] \end{aligned} $

Since, distance travelled by a circular wheel $=176 m$

$\therefore \quad$ Number of revolutions $=\dfrac{\text{ Total } \text{ distance }}{\text{ Distance in one revolution }}=\dfrac{176}{4.4}=40$

Hence, the required number of revolutions made by a circular wheel is 40 .

Solution

Let the radius of the circle be $r$.

$ \begin{aligned} & \therefore \quad O A=O B=r cm \\ & \text{ Given that, length of chord of a circle, } A B=5 cm \\ & \text{ and central angle of the sector } A O B A(\theta)=90^{\circ} \\ & \text{ Now, in } \triangle A O B \quad(A B)^{2}=(O A)^{2}+(O B)^{2} \quad \text{ [by Pythagoras theorem] } \\ & \Rightarrow \quad 2 r^{2}=25 \\ & \therefore \quad r=\dfrac{5}{\sqrt{2}} cm \end{aligned} $

Now, in $\triangle A O B$ we drawn a perpendicular line $O D$, which meets at $D$ on $A B$ and divides chord $A B$ into two equal parts.

So,

$ A D=D B=\dfrac{A B}{2}=\dfrac{5}{2} cm $

[since, the perpendicular drawn from the centre to the chord of a circle divides the chord into two equal parts]

By Pythagoras theorem, in $\triangle A D O$,

$ \begin{aligned} & (O A)^{2}=O D^{2}+A D^{2} \\ & \Rightarrow \quad O D^{2}=O A^{2}-A D^{2} \\ & =\dfrac{5}{\sqrt{2}}^{2}-\dfrac{5}2^{2}=\dfrac{25}{2}-\dfrac{25}{4} \\ & =\dfrac{50-25}{4}=\dfrac{25}{4} \\ & \Rightarrow \quad O D=\dfrac{5}{2} cm \\ & =\dfrac{1}{2} \times 5 \times \dfrac{5}{2}=\dfrac{25}{4} cm^{2} \end{aligned} $

Now, area of sector AOBA $=\dfrac{\pi^{2}}{360^{\circ}} \times \theta=\dfrac{\pi \times \dfrac{5}{\sqrt{2}}^{2}}{360^{\circ}} \times 90^{\circ}$

$ =\dfrac{\pi \times 25}{2 \times 4}=\dfrac{25 \pi}{8} cm^{2} $

$\therefore$ Area of minor segment $=$ Area of sector $A O B A-$ Area of an isosceles $\triangle A O B$

$$ \begin{equation*} =\dfrac{25 \pi}{8}-\dfrac{25}{4} cm^{2} \tag{i} \end{equation*} $$

Now, area of the circle $=\pi^{2}=\pi \quad \dfrac{5}{\sqrt{2}}=\dfrac{25 \pi}{2} cm^{2}$

$\therefore$ Area of major segment $=$ Area of circle - Area of minor segment

$$ \begin{align*} & =\dfrac{25 \pi}{2}-\dfrac{25 \pi}{8}-\dfrac{25}{4} \\ & =\dfrac{25 \pi}{8}(4-1)+\dfrac{25}{4} \\ & =\dfrac{75 \pi}{8}+\dfrac{25}{4} cm^{2} \tag{i} \end{align*} $$

$\therefore$ Difference of the areas of two segments of a circle $=\mid$ Area of major segment - Area of minor segment|

$ \begin{aligned} & =|\dfrac{75 \pi}{8}+\dfrac{25}{4}-\dfrac{25 \pi}{4}-\dfrac{25}{4}| \\ & =|\dfrac{75 \pi}{8}-\dfrac{25 \pi}{8}-\dfrac{25 \pi}{8}+\dfrac{25}{4}| \\ & =|\dfrac{75 \pi-25 \pi}{8}+\dfrac{50}{4}|=|\dfrac{50 \pi}{8}+\dfrac{50}{4}| \\ & =\dfrac{25 \pi}{4}+\dfrac{25}{2} cm^{2} \end{aligned} $

Hence, the required difference of the areas of two segments is $\dfrac{25 \pi}{4}+\dfrac{25}{2} cm^{2}$.

Solution

Given that, radius of the circle $(r)=21 cm$ and central angle of the sector $A O B A(\theta)=120^{\circ}$ So, area of the circle $=\pi r^{2}=\dfrac{22}{7} \times(21)^{2}=\dfrac{22}{7} \times 21 \times 21$

$ =22 \times 3 \times 21=1386 cm^{2} $

Now, area of the minor sector $A O B A$ with central angle $120^{\circ}$

$ \begin{aligned} & =\dfrac{\pi^{2}}{360^{\circ}} \times \theta=\dfrac{22}{7} \times \dfrac{21 \times 21}{360^{\circ}} \times 120 \\ & =\dfrac{22 \times 3 \times 21}{3}=22 \times 21=462 cm^{2} \end{aligned} $

$\therefore$ Area of the major sector $A B O A$

$ \begin{aligned} & =\text{ Area of the circle }- \text{ Area of the sector } A O B A \\ & =1386-462=924 cm^{2} \end{aligned} $

$\therefore$ Difference of the areas of a sector $A O B A$ and its corresponding major sector $A B O A$

$ \begin{aligned} & =\mid \text{ Area of major sector } A B O A-\text{ Area of minor sector } A O B A \mid \\ & =|924-462|=462 cm^{2} \end{aligned} $

Hence, the required difference of two sectors is $462 cm^{2}$.