Solution

(b) According to the given condition,

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

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

Solution

(a) According to the given condition,

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

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

Solution

(b) According to the given condition,

Circumference of a circle $=$ Perimeter of square

$2\pi r=4a$

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

$\Rightarrow {\displaystyle \frac{22}{7}}r=2a\Rightarrow 11r=7a$

$\Rightarrow a={\displaystyle \frac{11}{7}}r\Rightarrow r={\displaystyle \frac{7a}{11}}$

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

$\begin{array}{rl}& =\pi {\displaystyle \frac{7a}{1}}{1}^2={\displaystyle \frac{22}{7}}\times {\displaystyle \frac{49a^2}{121}}\\ & ={\displaystyle \frac{14a^2}{11}}\end{array}$

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

From Eqs. (ii) and (iii), $A_1={\displaystyle \frac{14}{11}}{A}_2$

$\therefore A_1>A_2$

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 $B$.

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

So, $\mathrm{△}ABC$ is right angled triangle.

$\therefore$ Area of largest $\mathrm{△}ABC={\displaystyle \frac{1}{2}}\times AB\times CD$

$\begin{array}{rl}& ={\displaystyle \frac{1}{2}}\times 2r\times r\\ & =r^2\text{ sq units }\end{array}$

[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

$$\text{Erroneous nesting of equation structures}$$

Solution

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

$={\pi }^2=\pi {{\displaystyle \frac{16}{2}}}^2=64\pi m^2$

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

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

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

According to the given condition,

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

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

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

Solution

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

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

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

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

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

Solution

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

$\therefore$ Diameter of the circle $=AC=2\times OC=2\times 8=16cm$

which is equal to the diagonal of a square.

Let side of square be $x$.

In right angled $\mathrm{△}ABC,A{C}^2=A{B}^2+B{C}^2$

[by Pythagoras theorem]

$\begin{array}{rl}& \Rightarrow (16{)}^2=x^2+x^2\\ & \Rightarrow 256=2x^2\\ & \Rightarrow x^2=128\\ & \therefore \text{ Area of square }=x^2=128c{m}^2\end{array}$

Alternate Method

Radius of circle $(r)=8cm$

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

Since, square inscribed in circle.

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

Now, $$ Area of square $={\displaystyle \frac{(\text{ Diagonal }{)}^2}{2}}={\displaystyle \frac{(16{)}^2}{2}}={\displaystyle \frac{256}{2}}=128c{m}^2$

Solution

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

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

[given, $d_2=20cm$ ]

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

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

$\Rightarrow D=56cm$

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

$\therefore$ Required radius of circle $={\displaystyle \frac{56}{2}}=28cm$

Solution

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

$\therefore$ Area of first circle $={\pi }_1^2=\pi (24{)}^2=576\pi c{m}^2$

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

According to the given condition,

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

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

Solution

False

Let $ABCD$ be a square of side a.

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

$\therefore$ Radius of circle $={\displaystyle \frac{a}{2}}$

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

Hence, area of the circle is ${\displaystyle \frac{\pi a^2}{4}}c{m}^2$.

Solution

True

Given, radius of circle, $r=acm$

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

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

$=2acm$

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

$=8acm$

Solution

False

Given diameter of circle is $d$.

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

Let side of inner square $EFGH$ be $x$.

$\therefore$ In right angled $\mathrm{△}EFG$,

$\begin{array}{rl}& E{G}^2=E{F}^2+F{G}^2\text{ [by Pythagoras theorem] }\\ & \Rightarrow d^2=x^2+x^2\\ & \therefore d^2=2x^2\Rightarrow x^2={\displaystyle \frac{d^2}{2}}\\ & \text{ But side of the outer square }ABCS=\text{ Diameter of circle }=d\\ & \therefore \text{ Area of outer square }=d^2\\ & \text{ Hence, area of outer square is not equal to four times the area of the inner square. }\end{array}$

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

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 $2r$ with centres $O$ and $O^{\prime }$, respectively.

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

Now, let ${\theta }_1$ be the angle subtended by arc $\widehat{AB}$ of ${\theta }_2$ be the angle subtended by arc $\widehat{CD}$ at the centre.

$\begin{array}{rl}& \widehat{AB}=l={\displaystyle \frac{Q_1}{360}}\times 2\pi r\\ & \widehat{CD}=l={\displaystyle \frac{{\theta }_2}{360}}\times 2\pi (2r)={\displaystyle \frac{{\theta }_2}{360}}\times 4\pi r\end{array}$

From Eqs. (i) and (ii),

$\begin{array}{r}\Rightarrow {\displaystyle \frac{{\theta }_1}{360}}\times 2\pi r={\displaystyle \frac{{\theta }_2}{360}}\times 4\pi r\\ {\theta }_1=2{\theta }_2\end{array}$

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 $\pi {{\displaystyle \frac{b}{2}}}^{2}cm$, where ${\displaystyle \frac{b}{2}}$ 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 $r$.

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

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

i.e. $r_1=15cm\text{ and }{r}_2=18cm$

Now, by given condition,

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

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

$\Rightarrow r=r_1+r_2$

$\Rightarrow r=15+18$

$\therefore r=33cm$

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

Solution

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

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

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

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

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

$\Rightarrow$ Diameter of circle $=8$

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

$\Rightarrow r={\displaystyle \frac{8}{2}}=4cm$

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

$=16\pi \times c{m}^2$

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

$=32c{m}^2$

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

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

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

Solution

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

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

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

$\begin{array}{rl}& ={\displaystyle \frac{22}{7}}\times {\displaystyle \frac{(28{)}^2}{360}}\times {45}^{\circ }\\ & ={\displaystyle \frac{22\times 28\times 28}{7}}\times {\displaystyle \frac{{45}^{\circ }}{{360}^{\circ }}}\\ & =22\times 4\times 28\times {\displaystyle \frac{1}{8}}\\ & =22\times 14\\ & =308c{m}^2\end{array}$

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

Solution

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

Circumference of the wheel $=2\pi$

$\begin{array}{rl}& =2\times {\displaystyle \frac{22}{7}}\times 35\\ & =220cm\end{array}$

But speed of the wheel $=66km{h}^{-1}={\displaystyle \frac{66\times 1000}{60}}m/min$

$\begin{array}{rl}& =1100\times 100cmmi{n}^{-1}\\ & =110000cmmi{n}^{-1}\end{array}$

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

Hence, required number of revolutions per minute is 500 .

Solution

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

$\therefore$ Area of the field in which the cow graze $=$ Area of sector $AFEG={\displaystyle \frac{\theta }{{360}^{\circ }}}\times \pi r^2$

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

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

$\begin{array}{rl}& ={\displaystyle \frac{1}{4}}\times {\displaystyle \frac{22}{7}}\times 196\\ & =154m^2\end{array}$

Solution

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

$\therefore$ Area of rectangle $ACDF=$ Length $\times$ Breadth $=38\times 10=380c{m}^2$

Both ends of flower bed are semi-circles.

$\therefore$ Radius of semi-circle $={\displaystyle \frac{DF}{2}}={\displaystyle \frac{10}{2}}=5cm$

$\therefore$ Area of one semi-circles $={\displaystyle \frac{\pi r^2}{2}}={\displaystyle \frac{\pi }{2}}(5{)}^2={\displaystyle \frac{25\pi }{2}}c{m}^2$

$\therefore$ Area of two semi-circles $=2\times {\displaystyle \frac{25}{2}}\pi =25\pi c{m}^2$

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

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

Solution

Given,

$AC=6cm$ and $BC=8cm$

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

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

In right angled $\mathrm{△}ACB$, use Pythagoras theorem,

$\begin{array}{ccc}\therefore & & A{B}^2=A{C}^2+C{B}^2\\ \Rightarrow & & A{B}^2=6^2+8^2=36+64\\ \Rightarrow & & A{B}^2=100\\ \Rightarrow & & AB=10cm\text{ [since, side cannot be negative] }\\ \therefore & & \text{ Area of }\mathrm{△}ABC={\displaystyle \frac{1}{2}}\times BC\times AC={\displaystyle \frac{1}{2}}\times 8\times 6=24c{m}^2\end{array}$

Here, diameter of circle, $AB=10cm$

$\therefore$ Radius of circle, $r={\displaystyle \frac{10}{2}}=5cm$

$\begin{array}{r}\text{ Area of circle }={\pi }^2=3.14\times (5{)}^2\\ & =3.14\times 25=78.5c{m}^2\end{array}$

$\therefore$ Area of the shaded region $=$ Area of circle - Area of $\mathrm{△}ABC$

$=78.5-24=54.5c{m}^2$

Solution

In a figure, join $ED$

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

Now, $$ area of rectangle $ABCD=BC\times AB=8\times 4=32m^2$

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

$\therefore$ Area of shaded region $=$ Area of rectangle $ABCD+$ Area of semi-circle DFE $=(32+2\pi )m^2$

Solution

Join $GH$ and $FE$

Here, $$ breadth of the rectangle $BC=12m$

$\therefore$ Breadth of the inner rectangle $EFGH=12-(4+4)=4cm$

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

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

$\therefore$ Length of inner rectangle $EFGH=26-(5+5)=16m$

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

Now, area of inner rectangle $EFGH=EH\times FG=16\times 4=64m^2$

and $$ area of outer rectangle $ABCD=26\times 12=312m^2$

$\therefore$ 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 $(r)=14cm$

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

Since, in $\mathrm{△}AOB,OA=OB=$ Radius of circle i.e., $\mathrm{△}AOB$ is isosceles.

$\Rightarrow \mathrm{\angle }OAB=\mathrm{\angle }OBA=\theta$

Now, in $\mathrm{△}OAB\mathrm{\angle }AOB+\mathrm{\angle }OAB+\mathrm{\angle }OBA={180}^{\circ }$

[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. $\mathrm{\angle }OAB=\mathrm{\angle }OBA={60}^{\circ }=\mathrm{\angle }AOB$

Since, all angles of $\mathrm{△}AOB$ are equal to ${60}^{\circ }$ i.e., $\mathrm{△}AOB$ is an equilateral triangle.

Also,

$OA=OB=AB=14cm$

So, $$ Area of $\mathrm{△}OAB={\displaystyle \frac{\sqrt{3}}{4}}(\text{ side }{)}^2$

$\begin{array}{rl}& ={\displaystyle \frac{\sqrt{3}}{4}}\times (14{)}^2[\because \text{ area of an equilateral triangle }={\displaystyle \frac{\sqrt{3}}{4}}(\text{ side }{)}^2]\\ & ={\displaystyle \frac{\sqrt{3}}{4}}\times 196=49\sqrt{3}c{m}^2\end{array}$

and $$ area of sector $OBAO={\displaystyle \frac{{\pi }^2}{{360}^2}}\times \theta$

$\begin{array}{rl}& ={\displaystyle \frac{22}{7}}\times {\displaystyle \frac{14\times 14}{360}}\times {60}^{\circ }\\ & ={\displaystyle \frac{22\times 2\times 14}{6}}={\displaystyle \frac{22\times 14}{3}}={\displaystyle \frac{308}{3}}c{m}^2\end{array}$

$\therefore$ Area of minor segment $=$ Area of sector OBAO - Area of $\mathrm{△}OAB$

$={\displaystyle \frac{308}{3}}-49\sqrt{3}c{m}^2$

Hence, the required area of the minor segment is ${\displaystyle \frac{308}{3}}-49\sqrt{3}c{m}^2$.

Solution

Given, side of a square $BC=12cm$

Since, $Q$ is a mid-point of $BC$.

$\therefore$ Radius $=BQ={\displaystyle \frac{12}{2}}=6cm$

Now,

Now, area of quadrant $BPQ={\displaystyle \frac{{\pi }^2}{4}}={\displaystyle \frac{3.14\times (6{)}^2}{4}}={\displaystyle \frac{113.04}{4}}c{m}^2$ Area of four quadrants $={\displaystyle \frac{4\times 113.04}{4}}=1123.04c{m}^2$

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

$=144-113.04=30.96c{m}^2$

Solution

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

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

$\begin{array}{r}\therefore AE=EC=CD=BD=BF=FA=5cm\\ \text{ Now, }\text{ area of sector }CDE={\displaystyle \frac{\theta \pi r^2}{360}}={\displaystyle \frac{60\times 3.14}{360}}(5{)}^2\\ ={\displaystyle \frac{3.14\times 25}{6}}={\displaystyle \frac{78.5}{6}}=13.0833c{m}^2\\ \therefore \text{ Area of shaded region }=3(\text{ Area of sector }CDE)\\ =3\times 13.0833\\ =39.25c{m}^2\end{array}$

Solution

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

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

$={\displaystyle \frac{\mathrm{\angle }P}{{360}^{\circ }}}\times \pi \times (14{)}^{2}c{m}^2$

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

Area of the sector with central angle $={\displaystyle \frac{\mathrm{\angle }Q}{{360}^{\circ }}}\times \pi r^2={\displaystyle \frac{\mathrm{\angle }Q}{{360}^{\circ }}}\times \pi \times (14{)}^{2}c{m}^2$

and area of the sector with central angle $R={\displaystyle \frac{\mathrm{\angle }R}{{360}^{\circ }}}\times {\pi }^2={\displaystyle \frac{\mathrm{\angle }R}{{360}^{\circ }}}\times \pi \times (14{)}^{2}c{m}^2$

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

$\begin{array}{rl}& ={\displaystyle \frac{\mathrm{\angle }P}{{360}^{\circ }}}\times \pi \times (14{)}^2+{\displaystyle \frac{\mathrm{\angle }\theta }{{360}^{\circ }}}\times \pi \times (14{)}^2+{\displaystyle \frac{\mathrm{\angle }R}{{360}^{\circ }}}\times \pi \times (14{)}^2\\ & ={\displaystyle \frac{\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R}{360}}\times 196\times \pi ={\displaystyle \frac{{180}^{\circ }}{{360}^{\circ }}}\times 196\pi c{m}^2\end{array}$

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

$=98\pi c{m}^2=98\times {\displaystyle \frac{22}{7}}$

$=14\times 22=308c{m}^2$

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

Solution

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

Width of the road $=21m$

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

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

$r_e=105+21=126m$

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

$\begin{array}{rl}\text{-Area of circular park }& ⟶\\ & =\pi r_e^2-{\pi }_i^2\\ & =\pi (r_e^2-r_i^2)\\ & =\pi ({126}^2-(105{)}^2.\\ & ={\displaystyle \frac{22}{7}}\times (126+105)(126-105)\\ & ={\displaystyle \frac{22}{7}}\times 231\times 21[\because \text{ area of circle }=\pi r^2]\\ & =66\times 231\\ & =15246c{m}^2\end{array}$

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

Solution

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

Area of sector with $\mathrm{\angle }A={\displaystyle \frac{\mathrm{\angle }A}{{360}^{\circ }}}\times \pi r^2={\displaystyle \frac{\mathrm{\angle }A}{{360}^{\circ }}}\times \pi \times (21{)}^{2}c{m}^2$

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

Area of sector with $\mathrm{\angle }B={\displaystyle \frac{\mathrm{\angle }B}{{360}^{\circ }}}\times \pi r^2={\displaystyle \frac{\mathrm{\angle }B}{{360}^{\circ }}}\times \pi \times (21{)}^{2}c{m}^2$

Area of sector with $\mathrm{\angle }C={\displaystyle \frac{\mathrm{\angle }C}{{360}^{\circ }}}\times \pi r^2={\displaystyle \frac{\mathrm{\angle }C}{{360}^{\circ }}}\times \pi \times (21{)}^{2}c{m}^2$

and area of sector with $\mathrm{\angle }D={\displaystyle \frac{\mathrm{\angle }D}{{360}^{\circ }}}\times {\pi }^2={\displaystyle \frac{\mathrm{\angle }D}{{360}^{\circ }}}\times \pi \times (21{)}^{2}c{m}^2$

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

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

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

Solution

Length of arc of circle $=20cm$

Here,

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

$\therefore$ $\text{Length of arc }={\displaystyle \frac{\theta }{{360}^{\circ }}}\times 2\pi r$

$\Rightarrow$ $20={\displaystyle \frac{{60}^{\circ }}{{360}^{\circ }}}\times 2\pi r\Rightarrow {\displaystyle \frac{20\times 6}{2\pi }}=r$

$\therefore$ $r={\displaystyle \frac{60}{\pi }}cm$

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

Solution

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

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

Solution

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

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

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

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

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

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

$=352000cm$

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

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

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 $(r)=7m$

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

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

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

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

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

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

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

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

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

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

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 $(r)=12cm$

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

$\therefore$ Area of sector $OBCA={\displaystyle \frac{\pi r^2}{360}}\times \theta$ [here, $OBCA=$ sector and $ABCA=$ segment]

$\begin{array}{rl}& ={\displaystyle \frac{3.14\times 12\times 12}{{360}^{\circ }}}\times {60}^{\circ }\\ & =3.14\times 2\times 12\\ & =3.14\times 24=75.36c{m}^2\end{array}$

Since, $\mathrm{△}OAB$ is an isosceles triangle.

$\begin{array}{cc}\text{ Let }& \mathrm{\angle }OAB=\mathrm{\angle }OBA={\theta }_1\\ \text{ and }& OA=OB=12cm\\ \therefore & \mathrm{\angle }OAB+\mathrm{\angle }OBA+\mathrm{\angle }AOB={180}^{\circ }[\because \text{ sum of all interior angles of a triangle is }{180}^{\circ }]\\ \Rightarrow & {\theta }_1+{\theta }_1+{60}^{\circ }={180}^{\circ }\\ \Rightarrow & 2{\theta }_1={120}^{\circ }\\ \Rightarrow & {\theta }_1={60}^{\circ }\\ \therefore & {\theta }_1=\theta ={60}^{\circ }\end{array}$

So, the required $\mathrm{△}AOB$ is an equilateral triangle.

Now, $$ area of $\mathrm{△}AOB={\displaystyle \frac{\sqrt{3}}{4}}(\text{ side }{)}^2[\because .$ area of an equilateral triangle $.={\displaystyle \frac{\sqrt{3}}{4}}(\text{ side }{)}^2]$

$\begin{array}{rl}& ={\displaystyle \frac{\sqrt{3}}{4}}(12{)}^2\\ & ={\displaystyle \frac{\sqrt{3}}{4}}\times 12\times 12=36\sqrt{3}c{m}^2\end{array}$

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

$ABCA=$ Area of sector $OBCA-$ Area of $\mathrm{△}AOB$

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

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

Solution

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

The diameter of circular pond $=17.5m$

$\begin{array}{c}\therefore \text{ Radius of circular pond }(r_i)={\displaystyle \frac{\text{ Diameter }}{2}}\\ \text{ i.e., }& OA=r_i={\displaystyle \frac{17.5}{2}}=8.75m\end{array}$

and the width of the path $=2m$

i.e., $AB=2m$

Now, length of $OB=OA+AB=r_i+AB$

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

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

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

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

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

$=\text{₹}3061.50$

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

Solution

Given-

$ABCD$ is a trapezium. $AB|DC$ and $AB=18\mathrm{cm},DC=32\mathrm{cm}$.