Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 16 - Area Of Circle, Sector And Segment

Answer:

Circumference = 39.6 cm
We know:
Circumference of a circle = $2\mathrm{\pi }r$
$$\begin{gathered} \Rightarrow 39.6 =2\times \frac{22}{7}\times r \\ \Rightarrow \frac{39.6\times 7}{2\times 22}=r \\ \Rightarrow r=6.3 \mathrm{cm} \end{gathered}$$

Also,
Area of the circle = $\pi r^2$
                             $$\begin{gathered} =\frac{22}{7}\times 6.3\times 6.3 \\ =124.74 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let the radius of the circle be r.
​Now,
$$\begin{gathered} \mathrm{Area}=98.56 \\ \Rightarrow \mathrm{\pi }{r}^2=98.56 \\ \Rightarrow \frac{22}{7}\times r^2=98.56 \\ \Rightarrow r=5.6 \end{gathered}$$
Now,
Circumference = $2\mathrm{\pi }r=2\times \frac{22}{7}\times 5.6=35.2 \mathrm{cm}$
Hence, the circumference of the circle is 35.2 cm.

Answer:

Let the radius of the circle be r.
Now,
Circumference = Diameter + 45
$$\begin{gathered} \Rightarrow 2\mathrm{\pi }r=2r+45 \\ \Rightarrow 2\mathrm{\pi }r-2r=45 \\ \Rightarrow 2r\left(\frac{22}{7}-1\right)=45 \\ \Rightarrow 2r\times \frac{15}{7}=45 \\ \Rightarrow r=10.5 \mathrm{cm} \end{gathered}$$
∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.

Answer:

Area of the circle = 484 cm²
Area of the square = ${\mathrm{Side}}^2$
$$\begin{gathered} \Rightarrow 484={\mathrm{Side}}^2 \\ \Rightarrow {22}^2={\mathrm{Side}}^2 \\ \Rightarrow \mathrm{Side}=22 \mathrm{cm} \end{gathered}$$
Perimeter of the square = $4\times \mathrm{Side}$
Perimeter of the square = $4\times 22$
                                       = 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
​Thus, we have:
​ $$\begin{gathered} 2\mathrm{\pi r}=88 \\ \Rightarrow 2\times \frac{22}{7}\times \mathrm{r}=88 \\ \Rightarrow \mathrm{r}=14 \end{gathered}$$

Area of the circle = ${\mathrm{\pi r}}^2$
                             $$\begin{gathered} =\frac{22}{7}\times 14\times 14 \\ =616 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the area enclosed by the circle is 616 cm².

Answer:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{a}\mathrm{n} \mathrm{equilateral} \mathrm{t}\mathrm{riangle}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2 \\ \Rightarrow 121\sqrt{3}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2 \\ \Rightarrow 121\times 4={\left(\mathrm{Side}\right)}^2 \\ \Rightarrow \mathrm{Side}=22 \mathrm{cm} \end{gathered}$$

$\mathrm{Perimeter} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangle} = 3\times \mathrm{Side}$
                                                $$\begin{gathered} =3\times 22 \\ =66 \mathrm{cm} \end{gathered}$$

Length of the wire$=66 \mathrm{cm}$
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire
$$\begin{gathered} 2\mathrm{\pi }r\mathit{ }=66 \\ \Rightarrow 2\times \frac{22}{7}\times r=66 \\ \Rightarrow r=\frac{66\times 7}{2\times 22} \\ \Rightarrow r=\frac{21}{2} \\ \Rightarrow r =10.5 \end{gathered}$$
Thus, we have:
Area of the circle =$\mathrm{\pi }{r}^{\mathit{2}}$
                             $$\begin{gathered} =\frac{22}{7}\times 10.5\times 10.5 \\ =346.5 \mathrm{sq}.\mathrm{cm} \end{gathered}$$
Area enclosed by the circle = 346.5 cm²

Answer:

Let the radius of the park be r.
Length of chain = Perimeter of the semicircular park
⇒ 108 = Length of the arc + Diameter
$$\begin{gathered} \Rightarrow 108=\frac{1}{2}\times 2\mathrm{\pi }r+2r \\ \Rightarrow 108=r\left(\frac{22}{7}+2\right) \\ \Rightarrow 108=\frac{36}{7}r \\ \Rightarrow r=21 \mathrm{m} \end{gathered}$$
Now, Area of park$=\frac{1}{2}\mathrm{\pi }{r}^2=\frac{1}{2}\times \frac{22}{7}\times {\left(21\right)}^2=693 {\mathrm{m}}^2$
Hence, the area of the park is 693 m² .

Answer:

Let the radii of the two circles be r₁ cm and r₂ cm.
Now,
Sum of the radii of the two circles = 7 cm
$r_{1 + }{r}_2=7 ...\left(i\right)$

Difference of the circumferences of the two circles = 88 cm
$$\begin{gathered} \Rightarrow 2{\mathrm{\pi r}}_1-2{\mathrm{\pi r}}_2 =8 \\ \Rightarrow 2\mathrm{\pi }({\mathrm{r}}_1-{\mathrm{r}}_2)=8 \\ \Rightarrow ({\mathrm{r}}_1-{\mathrm{r}}_2)=\frac{8}{2\mathrm{\pi }} \\ \Rightarrow {\mathrm{r}}_1-{\mathrm{r}}_2=\frac{8}{2\times {\displaystyle \frac{22}{7}}} \\ \Rightarrow {\mathrm{r}}_{1- }{\mathrm{r}}_2 =\frac{8\times 7}{44} \end{gathered}$$

$$\begin{gathered} r_1-r_2=\frac{56}{44} \\ {r}_1-r_2=\frac{14}{11} ...\left(ii\right) \end{gathered}$$

Adding (i) and (ii), we get:
$$\begin{gathered} 2r_1=\frac{91}{11} \\ {r}_1=\frac{91}{22} \end{gathered}$$

∴ Circumference of the first circle = $2{\mathrm{\pi r}}_1$
                                                    $$\begin{gathered} =2\times \frac{22}{7}\times \frac{91}{22} \\ =26 \mathrm{cm} \end{gathered}$$
Also,
$$\begin{gathered} r_1-r_2=\frac{14}{11} \\ \frac{91}{22}-r_2=\frac{14}{11} \\ \frac{91}{22}-\frac{14}{11}=r_2 \\ {r}_2=\frac{63}{22} \end{gathered}$$

∴ Circumference of the second circle = $2{\mathrm{\pi r}}_2$

                                                             $$\begin{gathered} =2\times \frac{22}{7}\times \frac{63}{22} \\ =18 \mathrm{cm} \end{gathered}$$

Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.

Answer:

Let r₁ cm and r₂ cm be the radii of the outer and inner boundaries of the ring, respectively.
We have:

$$\begin{gathered} r_1=23 \mathrm{cm} \\ {r}_2=12 \mathrm{cm} \end{gathered}$$

Now,
Area of the outer ring = $\mathrm{\pi }{r_1}^2$

                                    $$\begin{gathered} =\frac{22}{7}\times 23\times 23 \\ =1662.57 {\mathrm{cm}}^2 \end{gathered}$$
 
Area of the inner ring = $\mathrm{\pi }{r_2}^2$

                                    $$\begin{gathered} =\frac{22}{7}\times 12\times 12 \\ =452.57 {\mathrm{cm}}^2 \end{gathered}$$

Area of the ring = Area of the outer ring $-$ Area of the inner ring
                           = 1662.57 $-$ 452.57
                           = 1210 ${\mathrm{cm}}^2$ 

Answer:

(i) The radius (r) of the inner circle is 17 m.
The radius (R) of the outer circle is 25 m.          [Includes path, i.e., (17 + 8)]

Area of the path = $\pi R^2-\pi r^2$
                           $$\begin{gathered} =\mathrm{\pi }\left(R^2-r^2\right) \\ =\frac{22}{7}\left({25}^2-{17}^2\right) \\ =\frac{22}{7}\times (25 - 17) (25 + 17) \\ =\frac{22}{7}\times 8\times 42 \\ =1056 {\mathrm{m}}^2 \end{gathered}$$

∴ Area of the path = 1056 m²                

(ii)


Diameter of the circular park = 7 m
∴ Radius of the circular park, r = $\frac{7}{2}$ = 3.5 m
Width of the path = 0.7 m
∴  Radius of the park including the path, R = 3.5 + 0.7 = 4.2 m
Area of the path
$$\begin{gathered} =\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ =\mathrm{\pi }\left(R^2-r^2\right) \end{gathered}$$
$$\begin{gathered} =\frac{22}{7}\left(4.2^2-3.5^2\right) \\ =\frac{22}{7}\times \left(4.2-3.5\right)\times \left(4.2+3.5\right) \\ =\frac{22}{7}\times 0.7\times 7.7 \\ =16.94 {\mathrm{m}}^2 \end{gathered}$$
Rate of cementing the path = Rs 110/m²      (Given)
∴ Total cost of cementing the path
= 16.94 × 110
= Rs 1863.40
Thus, the expenditure of cementing the path is Rs 1863.40.

Answer:

Let r m and R m be the radii of the inner and outer tracks.
Now,
Circumference of the outer track = $2\mathrm{\pi }R$
$$\begin{gathered} \Rightarrow 396=2\times \frac{22}{7}\times R \\ \Rightarrow R=\frac{396\times 7}{44} \\ \Rightarrow R=63 \end{gathered}$$

Circumference of the inner track = $2\mathrm{\pi }r$
$$\begin{gathered} \Rightarrow 352=2\times \frac{22}{7}\times r \\ \Rightarrow r=\frac{352\times 7}{44} \\ \Rightarrow r=56 \end{gathered}$$

Width of the track = Radius of the outer track $-$ Radius of the inner track
                              $$\begin{gathered} =63-56 \\ =7 \mathrm{m} \end{gathered}$$
   
Area of the outer circle = $\mathrm{\pi }{R}^2$
                                      $$\begin{gathered} =\frac{22}{7}\times 63\times 63 \\ =12474 {\mathrm{m}}^2 \end{gathered}$$

Area of the inner circle = $\mathrm{\pi }{R}^2$
                                      $$\begin{gathered} =\frac{22}{7}\times 56\times 56 \\ =9856 {\mathrm{m}}^2 \end{gathered}$$

Area of the track = 12474 $-$ 9856
                             = 2618 ${\mathrm{m}}^2$

Answer:

Given:
Radius = 2 cm
Angle of sector = ${150}^{\circ }$
Now,
$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{arc}=\frac{2\mathrm{\pi r}\theta }{360} \\ =\frac{2\times {\displaystyle \frac{22}{7}}\times 21\times 150}{360} \\ = 55 \mathrm{cm} \end{gathered}$$

Area of the sector =$\frac{{\mathrm{\pi r}}^2\theta }{360}$

                              $$\begin{gathered} =\frac{22}{7}\times 21\times 21\times \frac{150}{360} \\ =577.5 \mathrm{cm} \end{gathered}$$

Answer:

Radius of the circle, r = 10 cm

Area of sector OPRQ
$$\begin{gathered} =\frac{60°}{360°}\times \mathrm{\pi }{r}^2 \\ =\frac{1}{6}\times 3.14\times {\left(10\right)}^2 \\ =52.33 {\mathrm{cm}}^2 \end{gathered}$$

In ΔOPQ,
∠OPQ = ∠OQP     (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ
$$\begin{gathered} =\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(10\right)}^2 \\ =\frac{100\sqrt{3}}{4}{\mathrm{cm}}^2 \\ =43.30 {\mathrm{cm}}^2 \end{gathered}$$

Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm²

Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
$$\begin{gathered} =\mathrm{\pi }{\left(10\right)}^2-9.03 \\ =314-9.03 \\ =304.97 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Length of the arc = 16.5 cm
$\theta ={54}^{\circ }$
Radius = ?
Circumference=?
We know:
Length of the arc $=\frac{2\mathrm{\pi r\theta }}{360}$
$$\begin{gathered} \Rightarrow 16.5=\frac{2\times {\displaystyle \frac{22}{7}}\times r\times 54}{360} \\ \Rightarrow r=\frac{16.5\times 360\times 7}{44\times 54} \\ \Rightarrow r=17.5 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} c=2\mathrm{\pi r} \\ =2\times \frac{22}{7}\times 17.5 \\ =110 \mathrm{cm} \end{gathered}$$

Circumference = 110 cm

Now,
Area of the circle =${\mathrm{\pi r}}^2$
                             $$\begin{gathered} =\frac{22}{7}\times 17.5\times 17.5 \\ =962.5 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$$\begin{gathered} =\frac{90°}{360°}\mathrm{\pi }{\left(\mathrm{OA}\right)}^2-\frac{1}{2}\times \mathrm{OA}\times \mathrm{OB} \\ =\frac{1}{4}\times \frac{22}{7}\times {\left(7\right)}^2-\frac{1}{2}\times 7\times 7 \\ =\frac{1}{4}\times \frac{22}{7}\times {\left(7\right)}^2-\frac{1}{2}\times 7\times 7 \\ =38.5-24.5 \\ =14 {\mathrm{cm}}^2 \end{gathered}$$
Area of major segment APB = Area of circle − Area of minor segment
$$\begin{gathered} =\mathrm{\pi }{\left(\mathrm{OA}\right)}^2-14 \\ =\frac{22}{7}\times {\left(7\right)}^2-14 \\ =154-14 \\ =140 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of major segment is 140 cm² .

Answer:

Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.
Thus, we have:
$\angle O=\angle A=\angle B=60°$

Length of the arc ACB:
$$\begin{gathered} 2\mathrm{\pi }\times 12\times \frac{60}{360} \\ =4\mathrm{\pi } \\ =12.56 \mathrm{cm} \end{gathered}$$

Length of the arc ADB:

$$\begin{gathered} \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle} - \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{arc} \mathrm{ACB} \\ =2\mathrm{\pi }\times 12-4\mathrm{\pi } \\ =20\mathrm{\pi } \mathrm{cm} \\ = 62.80 \mathrm{cm} \end{gathered}$$

Now,
Area of the minor segment:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{sector} - \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{triangle} \\ =\left[\mathrm{\pi }\times {\left(12\right)}^2\times \frac{60}{360}-\frac{\sqrt{3}}{4}\times {\left(12\right)}^2\right] \\ =13.08 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let O be the centre of the circle and AB be the chord.

Consider $∆$OAB.

$\mathrm{OA}=\mathrm{OB}=5\sqrt{2} \mathrm{cm}$

${\mathrm{OA}}^2+{\mathrm{OB}}^2=50+50=100$

Now,
$\sqrt{100} = 10 \mathrm{cm} = AB$

Thus, $∆$OAB is a right isosceles triangle.

Thus, we have:
Area of $∆$OAB = $\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}=25 {\mathrm{cm}}^2$

Area of the minor segment = Area of the sector $-$ Area of the triangle
                                            $$\begin{gathered} =\left(\frac{90}{360}\times \mathrm{\pi }\times {\left(5\sqrt{2}\right)}^2\right)-25 \\ =14.25 {\mathrm{cm}}^2 \end{gathered}$$

Area of the major segment = Area of the circle $-$ Area of the minor segment
                                            $$\begin{gathered} =\mathrm{\pi }\times {\left(5\sqrt{2}\right)}^2-14.25 \\ =142.75 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Area}oftheminorsector=\frac{120}{360}\times \mathrm{\pi }\times 42\times 42 \\ =\frac{1}{3}\times \mathrm{\pi }\times 42\times 42 \\ =\mathrm{\pi }\times 14\times 42 \\ =1848 {\mathrm{cm}}^2 \end{gathered}$$

Area of the triangle = $\frac{1}{2}{R}^2\sin \theta$
Here, R is the measure of the equal sides of the isosceles triangle and θ is the angle enclosed by the equal sides.
Thus, we have:
$$\begin{gathered} \frac{1}{2}\times 42\times 42\times \sin \left(120°\right) \\ =762.93 {\mathrm{cm}}^2 \end{gathered}$$

Area of the minor segment = Area of the sector $-$ Area of the triangle
                                            =$1848-762.93=1085.07 {\mathrm{cm}}^2$

Area of the major segment = Area of the circle $-$ Area of the minor segment
                                            $$\begin{gathered} =\left(\mathrm{\pi }\times 42\times 42\right)-1085.07 \\ =5544-1085.07 \\ =4458.93 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let the chord be AB. The ends of the chord are connected to the centre of the circle O to give the triangle OAB.

OAB is an isosceles triangle. The angle at the centre is 60$°$

Area of the triangle = $\frac{1}{2}{\left(30\right)}^2 \sin 60°=450 \times \frac{\sqrt{3}}{2}=389.25 {\mathrm{cm}}^2$

Area of the sector OACBO = $\frac{60}{360}\times \mathrm{\pi }\times 30\times 30=150\mathrm{\pi }=471 {\mathrm{cm}}^2$

Area of the minor segment = Area of the sector $-$ Area of the triangle
                                            =$471-389.25=81.75 {\mathrm{cm}}^2$

Area of the major segment = Area of the circle $-$ Area of the minor segment
                                            $$\begin{gathered} =\left(\mathrm{\pi }\times 30\times 30\right)-81.29 \\ =2744.71 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let the length of the major arc be $x$ cm
Radius of the circle = 10.5 cm
∴ Length of the minor arc = $\frac{x}{5} \mathrm{cm}$

Circumference =  $\left(x+\frac{x}{5}\right)=\frac{6x}{5} \mathrm{cm}$

Using the given data, we get:

$$\begin{gathered} \frac{6x}{5}=2\times \frac{22}{7}\times \frac{21}{2} \\ \Rightarrow \frac{6x}{5}=66 \\ \mathrm{Or}, \\ x = 55 \end{gathered}$$

∴ Area of the sector corresponding to the major arc = $\left(\frac{1}{2}\times 55\times \frac{21}{2}\right)=288.75 {\mathrm{cm}}^2$

Answer:

In 2 days, the short hand will complete 4 rounds.
Length of the short hand = 4 cm
 
Distance covered by the short hand = $4\times 2\mathrm{\pi }\times 4=32\mathrm{\pi } \mathrm{cm}$

In the same 2 days, the long hand will complete 48 rounds.
Length of the long hand = 6 cm
Distance covered by the long hand = $48\times 2\mathrm{\pi }\times 6=576\mathrm{\pi } \mathrm{cm}$

∴ Total distance covered by both the hands = Distance covered by the short hand + Distance covered by the long hand

                                                                       $$\begin{gathered} =32\mathrm{\pi } + 576\mathrm{\pi } \\ = 608\mathrm{\pi } \\ =608 \times 3.14 \\ = 1909.12 \mathrm{cm} \end{gathered}$$

Answer:

Let the radius of the circle be r.
​Now,
$$\begin{gathered} \mathrm{Circumference}=88 \\ \Rightarrow 2\mathrm{\pi }r=88 \\ \Rightarrow r=14 \mathrm{cm} \end{gathered}$$
Now,
Area of quadrant = $\frac{1}{4}\mathrm{\pi }{r}^2=\frac{1}{4}\times \frac{22}{7}\times {\left(14\right)}^2=154 {\mathrm{cm}}^2$
Hence, the area of the quadrant of the circle is 154 cm².

Answer:

r₁ = 16 m
r₂ = 23 m

Amount of additional ground available = Area of the bigger circle $-$ Area of the smaller circle
                                                               $$\begin{gathered} =\mathrm{\pi }\left({r_1}^2-{r_2}^2\right) \\ =\mathrm{\pi }\left({23}^2-{16}^2\right) \\ =\mathrm{\pi }\left(23+16\right)\left(23-16\right) \\ =858 {\mathrm{m}}^2 \end{gathered}$$

Answer:

Radius of the quadrant of the circle = 21 m
The shaded portion shows the part of the field the horse can graze.

Area of the grazed field = Area of the quadrant OPQ
                                       $=\frac{1}{4} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{having} \mathrm{radius} \mathrm{OP}$
                                       $$\begin{gathered} =\frac{1}{4}{\mathrm{\pi r}}^2 \\ =\frac{1}{4}\times \frac{22}{7}\times 21\times 21 \\ =346.5 {\mathrm{m}}^2 \end{gathered}$$

Total area of the field = $70\times 52=3640 {\mathrm{m}}^2$

Area left ungrazed = Area of the field $-$ Area of the grazed field
                               = $3640-346.5=3293.5 {\mathrm{m}}^2$

Answer:

Side of the equilateral triangle = 12 m
Area of the equilateral triangle =$\frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2$
                                                   $$\begin{gathered} =\frac{\sqrt{3}}{4}\times 12\times 12 \\ =62.28 {\mathrm{m}}^2 \end{gathered}$$
Length of the rope = 7 m
Area of the field the horse can graze is the area of the sector of radius 7 m .Also, the angle subtended at the centre is 60$°$

=$\frac{\theta }{360}\times \mathrm{\pi }{r}^{\mathit{2}}$
 
$$\begin{gathered} =\frac{60}{360}\times \frac{22}{7}\times (7{)}^2 \\ =25.67 {\mathrm{m}}^2 \end{gathered}$$

Area of the field the horse cannot graze = Area of the equilateral triangle $-$ Area of the field the horse can graze
                                                                 $=62.28-25.67=36.61 {\mathrm{m}}^2$
                                                       

Answer:

Each cow can graze a region that cannot be accessed by other cows.
∴ Radius of the region grazed by each cow = $\frac{50}{2}=25 \mathrm{m}$

Area that each cow grazes =  $\frac{1}{4}\times \mathrm{\pi }\times r^2$
                                            $$\begin{gathered} =\frac{1}{4}\times 3.14\times 25\times 25 \\ =490.625 {\mathrm{cm}}^2 \end{gathered}$$

Total area grazed = $4\times 490.625=1963.49 {\mathrm{m}}^2$
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{square}=(\mathrm{Side}{)}^2 \\ ={50}^2 \\ =2500 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area left ungrazed = Area of the square $-$ Grazed area
                               = $2500 - 1963.49 = 536.51 \mathrm{m}{ }^2$

Answer:

In a rhombus, all sides are congruent to each other.

Thus, we have:
$OP=PQ=QR=RO$

Now, consider $∆QOP$.

$OQ\mathit{=}OP\mathit{ }(\mathrm{Both} \mathrm{are} \mathrm{radii}.)$

Therefore, $∆QOP$ is equilateral.

Similarly, $∆QOR$ is also equilateral and $∆QOP \cong ∆QOR$.

$\mathrm{Ar}.\left(QROP\right) = \mathrm{Ar}.\left(∆QOP\right)+A\left(∆QOR\right)=2\mathrm{Ar}.\left[∆QOP\right]$

$$\begin{gathered} \mathrm{Ar}.\left(∆\mathrm{QOP}\right)=\frac{1}{2}\times 32\sqrt{3}=16\sqrt{3} \\ Or, \\ 16\sqrt{3}=\frac{\sqrt{3}}{4}{s}^2 (\mathrm{where} s \mathrm{is} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{rhombus}) \\ Or, \\ {s}^2=16\times 4=64 \\ \Rightarrow s=8 \mathrm{cm} \end{gathered}$$

∴ OQ = 8 cm

Hence, the radius of the circle is 8 cm.

Answer:

(​i)​ If a circle is inscribed in a square, then the side of the square is equal to the diameter of the circle.
Side of the square = 10 cm
Side = Diameter = 10
∴ Radius = 5 cm
Area of the inscribed circle = ${\mathrm{\pi r}}^2$
                                            $$\begin{gathered} =3.14\times 5\times 5 \\ =78.5 {\mathrm{cm}}^2 \end{gathered}$$

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.
$$\begin{gathered} \mathrm{Diagonal} \mathrm{of} \mathrm{the} \mathrm{square} = \sqrt{2}\times \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{square} \\ =\sqrt{2}\times 10 \\ =10\sqrt{2} \mathrm{cm} \end{gathered}$$

Diagonal = Diameter = $10\sqrt{2} \mathrm{cm}$
∴ $r=5\sqrt{2}$ cm

Now,
Area of the circumscribed circle = ${\mathrm{\pi r}}^2$
                                                     $$\begin{gathered} =3.14\times {\left(5\sqrt{2}\right)}^2 \\ =3.14\times 50 \\ =157 {\mathrm{cm}}^2 \end{gathered}$$
                          

Answer:

If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.
Let the diagonal of the square be d cm.
Thus, we have:
Radius, r = $\frac{d}{2} \mathrm{cm}$

$\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{circle}=\mathrm{\pi }{r}^2$
                       = $\mathrm{\pi }\frac{d^2}{4} {\mathrm{cm}}^2$

$$\begin{gathered} \mathrm{We} \mathrm{know}: \\ d=\sqrt{2}\times \mathrm{Side} \\ \Rightarrow \mathrm{Side}=\frac{d}{\sqrt{2}} \mathrm{cm} \end{gathered}$$

 $$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{square}=(\mathrm{Side}{)}^2 \\ ={\left(\frac{d}{\sqrt{2}}\right)}^2 \\ =\frac{d^{\mathit{2}}}{2} {\mathrm{cm}}^2 \end{gathered}$$

Ratio of the area of the circle to that of the square:
$$\begin{gathered} =\frac{\pi {\displaystyle \frac{d^2}{4}}}{{\displaystyle \frac{d^2}{2}}} \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$
Thus, the ratio of the area of the circle to that of the square is $\mathrm{\pi }:2$.

Answer:

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 ${\mathrm{cm}}^2$
We know:
Area of the circle =$\pi r^2$
$$\begin{gathered} \Rightarrow 154=\frac{22}{7}{r}^2 \\ \Rightarrow \frac{154\times 7}{22}=r^2 \\ \Rightarrow r^2=49 \\ \Rightarrow r=7 \end{gathered}$$
In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO:OD = 2:1  

Now,
Let the altitude be h cm.
We have:
$$\begin{gathered} \angle ADB ={90}^{\circ } \\ OD = \frac{1}{3}AD \\ OD=\frac{h}{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow h=3r \\ \Rightarrow h=21 \end{gathered}$$

Let each side of the triangle be a cm.
$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{right}-\mathrm{angled} ∆\mathrm{ADB}, \mathrm{we} \mathrm{have}: \\ A{B}^2=A{D}^2+D{B}^2 \\ {a}^2=h^2+{\left(\frac{a}{2}\right)}^2 \\ 4a^2=4h^2+a^2 \\ 3a^2=4h^2 \\ {a}^2=\frac{4h^2}{3} \\ a=\frac{2h}{\sqrt{3}} \\ a=\frac{42}{\sqrt{3}} \end{gathered}$$
∴ Perimeter of the triangle = 3a
​                                            $$\begin{gathered} =3\times \frac{42}{\sqrt{3}} \\ =\sqrt{3}\times 42 \\ =72.66 \mathrm{cm} \end{gathered}$$

Answer:

Radius of the wheel = 42 cm
​Circumference of the wheel = $2\mathrm{\pi r}$
                                             $$\begin{gathered} =2\times \frac{22}{7}\times 42 \\ =264 \mathrm{cm} \\ =\frac{264}{100} \mathrm{m} \\ =2.64 \mathrm{m} \end{gathered}$$

Distance covered by the wheel in 1 revolution = 2.64 m

Total distance = 19.8 km or 19800 m

∴ Number of revolutions taken by the wheel = $\frac{19800}{2.64}=7500$

Answer:

Radius of the wheel = 2.1 m
Circumference of the wheel = $2\mathrm{\pi r}$
                                             $$\begin{gathered} =2\times \frac{22}{7}\times 2.1 \\ =13.2 \mathrm{m} \end{gathered}$$
Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions = $(13.2\times 75)=\left(990\times \frac{1}{100}\right) \mathrm{m}$
                                                                                                $=\left(990\times \frac{1}{1000}\right) \mathrm{km}$
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 75 revolutions =$\frac{990}{1000}$ km

∴ Distance covered by the wheel in 1 hour = $\frac{990}{1000}\times 60$
                                                                      = $59.4 \mathrm{km}/\mathrm{h}$

                                                                                              

Answer:

Distance = 4.95 km = $4.95\times 1000\times 100 \mathrm{cm}$
∴ Distance covered by the wheel in 1 revolution $=\frac{\mathrm{Total} \mathrm{distance} \mathrm{covered} }{\mathrm{Number} \mathrm{of} \mathrm{revolutions}}$

                                                                               $$\begin{gathered} =\frac{4.95\times 1000\times 100}{2500} \\ =198 \mathrm{cm} \end{gathered}$$

Now,
Circumference of the wheel = 198 cm
$$\begin{gathered} \Rightarrow 2\pi r=198 \\ \Rightarrow 2\times \frac{22}{7}\times r=198 \\ \Rightarrow r=\frac{198\times 7}{44} \\ \Rightarrow r=31.5\mathrm{cm} \end{gathered}$$

∴ Diameter of the wheel = 2r
                                         = 2(31.5)
                                         = 63 cm

Answer:

Diameter of the wheel = 60 cm
​∴ Radius of the wheel = 30 cm
​Circumference of the wheel = $2\mathrm{\pi r}$
                                             $$\begin{gathered} =2\times \frac{22}{7}\times 30 \\ =\frac{1320}{7} \mathrm{cm} \end{gathered}$$
Distance covered by the wheel in 1 revolution = $\frac{1320}{7} \mathrm{cm}$
∴ Distance covered by the wheel in 140 revolutions = $\left(\frac{1320}{7}\times 140\times \frac{1}{100}\right) \mathrm{m}$
                                                                                    $=\left({\displaystyle \frac{1320\times 140}{7\times 100}}\times {\displaystyle \frac{1}{1000}}\right) \mathrm{km}=\frac{264}{1000} \mathrm{km}$

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions = $\frac{264}{1000} \mathrm{km}$

∴ Distance covered by the wheel in 1 hour = $\frac{264}{1000}\times 60=15.84 \mathrm{km}/\mathrm{h}$

Hence, the speed at which the boy is cycling is 15.84 km/h.

Answer:

The radius of wheel of a motorcycle = 35 cm = 0.35 m.
So, the distance covered by this wheel in 1 revolution will be equal to perimeter of wheel i.e. $2\mathrm{\pi }r=2\times \frac{22}{7}\times 0.35=2.2 \mathrm{m}.$
Since speed is given to be $66 \mathrm{km}/\mathrm{hr}=66\times \frac{1000\mathrm{m}}{60\min }=1100 \mathrm{m}/\min .$
As we know $\mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}}$
$$\begin{gathered} \Rightarrow 1100=\frac{\mathrm{number} \mathrm{of} \mathrm{revolutions}\times \mathrm{perimeter} \mathrm{of} \mathrm{wheel}}{\mathrm{time}} \\ \Rightarrow 1100=\frac{\mathrm{number} \mathrm{of} \mathrm{revolutions}\times 2.2}{1} \\ \Rightarrow \mathrm{number} \mathrm{of} \mathrm{revolutions}=\frac{1100}{2.2}=500 \mathrm{revolutions}. \end{gathered}$$

Answer:

Radius of the front wheel = $40 \mathrm{cm} = \frac{2}{5} \mathrm{m}$

Circumference of the front wheel = $\left(2\mathrm{\pi }\times \frac{2}{5}\right) \mathrm{m}=\frac{4\mathrm{\pi }}{5} \mathrm{m}$

Distance covered by the front wheel in 800 revolutions = $\left(\frac{4\mathrm{\pi }}{5}\times 800\right) \mathrm{m}=\left(640\mathrm{\pi }\right) \mathrm{m}$
Radius of the rear wheel = 1 m
Circumference of the rear wheel = $\left(2\mathrm{\pi }\times 1\right)=2\mathrm{\pi } \mathrm{m}$

∴ Required number of revolutions = $\frac{\mathrm{Distance} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{front} \mathrm{wheel} \mathrm{in} 800 \mathrm{revolutions}}{\mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{rear} \mathrm{wheel}}$
                                                        $$\begin{gathered} =\frac{640\mathrm{\pi }}{2\mathrm{\pi }} \\ =320 \end{gathered}$$

Answer:

Side of the square = 14 cm
Radius of the circle $=\frac{14}{2}$= 7 cm
Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^2$
                                                        $$\begin{gathered} =\frac{1}{4}\times \frac{22}{7}\times 7\times 7 \\ =38.5 {\mathrm{cm}}^2 \end{gathered}$$
Area of the quadrants of four circles = $38.5\times 4=154 {\mathrm{cm}}^2$

Now,
Area of the square = ${\left(\mathrm{Side}\right)}^2$
                               $$\begin{gathered} ={14}^2 \\ =196 {\mathrm{cm}}^2 \end{gathered}$$

Area of the shaded region = Area of the square $-$ Area of the quadrants of four circles
                                           = 196 $-$ 154
                                           = 42 cm²

Answer:

Radius = 5 cm 
AB = BC = CD = AD = 10 cm
All sides are equal, so it is a square.
Area of a square = ${\mathrm{Side}}^2$
Area of the square = ${10}^2=100 {\mathrm{cm}}^2$
Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^2$
  
                                                        $$\begin{gathered} =\frac{1}{4}\times \frac{22}{7}\times 5\times 5 \\ =19.64 {\mathrm{cm}}^2 \end{gathered}$$
Area of the quadrants of four circles = $19.64\times 4=78.57 {\mathrm{cm}}^2$
Area of the shaded portion = Area of the square $-$ Area of the quadrants of four circles
                                            $$\begin{gathered} =100-78.57 \\ =21.43 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

When four circles touch each other, their centres form the vertices of a square. The sides of the square are 2a units.

Area of the square = ${\left(2a\right)}^2=4a^2 \mathrm{sq}. \mathrm{units}$

Area occupied by the four sectors
$$\begin{gathered} =4\times \frac{90}{360}\times \mathrm{\pi }\times a^2 \\ =\pi a^2 \mathrm{sq}. \mathrm{units} \end{gathered}$$

Area between the circles = Area of the square $-$ Area of the four sectors
$$\begin{gathered} =\left(4-\frac{22}{7}\right)a^2 \\ =\frac{6}{7}{a}^2 \mathrm{sq}. \mathrm{units} \end{gathered}$$

Answer:

Join ABC. All sides are equal, so it is an equilateral triangle.
Now,
Area of the equilateral triangle =$\frac{\sqrt{3}}{4}\times {\mathrm{Side}}^2$

                                                   $$\begin{gathered} =\frac{1.73}{4}\times 12\times 12 \\ =62.28 {\mathrm{cm}}^2 \end{gathered}$$


$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{arc} \mathrm{of} \mathrm{the} \mathrm{circle}=\frac{60}{360}{\mathrm{\pi r}}^2 \\ =\frac{1}{6}{\mathrm{\pi r}}^2 \\ =\frac{1}{6}\times \frac{22}{7}\times 6\times 6 \\ =18.86 {\mathrm{cm}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{sectors}=3\times 18.86=56.57 {\mathrm{cm}}^2 \end{gathered}$$

                                      
Area of the shaded portion = Area of the triangle $-$ Area of the three quadrants

$$\begin{gathered} =62.28-56.57 \\ = 5.71 {\mathrm{cm}}^2 \end{gathered}$$
                                

Answer:

When three circles touch each other, their centres form an equilateral triangle, with each side being 2a.

Area of the triangle = $\frac{\sqrt{3}}{4}\times 2a\times 2a=\sqrt{3}{a}^2$

Total area of the three sectors of circles = $3\times \frac{60}{360}\times \frac{22}{7}\times a^2=\frac{1}{2}\times \frac{22}{7}\times a^2=\frac{11}{7}{a}^2$

Area of the region between the circles = $\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{triangle} - \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{sectors}$
 $$\begin{gathered} =\left(\sqrt{3}-\frac{11}{7}\right)a^2 \\ =\left(1.73-1.57\right)a^2 \\ =0.16a^2 \\ =\frac{4}{25}{a}^2 \end{gathered}$$

Answer:

Area of trapezium = $\frac{1}{2}\left(\mathrm{AD}+\mathrm{BC}\right)\times \mathrm{AB}$
$$\begin{gathered} \Rightarrow 24.5=\frac{1}{2}\left(10+4\right)\times \mathrm{AB} \\ \Rightarrow \mathrm{AB}=3.5 \mathrm{cm} \end{gathered}$$
Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE
$$\begin{gathered} =24.5-\frac{1}{4}\mathrm{\pi }{\left(\mathrm{AB}\right)}^2 \\ =24.5-\frac{1}{4}\times \frac{22}{7}{\left(3.5\right)}^2 \\ =24.5-9.625 \\ =14.875 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 14.875 cm²

Answer:

(i) Area of fours sector = Area of sector having central angle 60° + Area of sector having central angle 90° + Area of sector having central angle 90° + Area of sector having central angle 120°
$$\begin{gathered} =\frac{60°}{360°}\mathrm{\pi }{\left(14\right)}^2+\frac{90°}{360°}\mathrm{\pi }{\left(14\right)}^2+\frac{90°}{360°}\mathrm{\pi }{\left(14\right)}^2+\frac{120°}{360°}\mathrm{\pi }{\left(14\right)}^2 \\ =\left(\frac{60°}{360°}+\frac{90°}{360°}+\frac{90°}{360°}+\frac{120°}{360°}\right)\mathrm{\pi }{\left(14\right)}^2 \\ =\left(\frac{360°}{360°}\right)\mathrm{\pi }{\left(14\right)}^2 \\ =616 {\mathrm{m}}^2 \end{gathered}$$

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants
$$\begin{gathered} =\frac{1}{2}\left(\mathrm{AD}+\mathrm{BC}\right)\times \mathrm{AB}-616 \\ =\frac{1}{2}\left(55+45\right)\times 30-616 \\ =1500-616 \\ =884 {\mathrm{m}}^2 \end{gathered}$$

Answer:

In equilateral traingle all the angles are of  60°
∴ ∠ABO = ∠AOB = 60°
 Area of the shaded region = (Area of triangle  AOB − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)
$$\begin{gathered} =\frac{\sqrt{3}}{4}{\left(\mathrm{AB}\right)}^2-\frac{60°}{360°}\mathrm{\pi }{\left(6\right)}^2+\frac{300°}{360°}\mathrm{\pi }{\left(6\right)}^2 \\ =\frac{1.73}{4}{\left(12\right)}^2-\frac{1}{6}\times 3.14{\left(6\right)}^2+\frac{5}{6}\times 3.14{\left(6\right)}^2 \\ =62.28-18.84+94.2 \\ =137.64 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 137.64 cm²

Answer:

We know that the opposite sides of a rectangle are equal
AD = BC =  70 cm
In right triangle AED
AE² = AD² − DE²  
= (70)² − (42)² 
= 4900 − 1764
= 3136
∴ AE² = 3136
⇒ AE = 56
= Area of the shaded region = Area of rectangle − (Area of triangle  AED + Area of semicircle)
$$\begin{gathered} =\mathrm{AB}\times \mathrm{BC}-\left[\frac{1}{2}\times \mathrm{AE}\times \mathrm{DE}+\frac{1}{2}\mathrm{\pi }{\left(\frac{\mathrm{BC}}{2}\right)}^2\right] \\ =80\times 70-\left[\frac{1}{2}\times 56\times 42+\frac{1}{2}\times \frac{22}{7}{\left(\frac{70}{2}\right)}^2\right] \\ =5600-3101 \\ =2499 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 2499 cm²

Answer:

In right triangle AED
AD² = AE² + DE²  
= (9)² + (12)² 
= 81 + 144
= 225
∴ AD² = 225
⇒ AD = 15 cm
We know that the opposite sides of a rectangle are equal
AD = BC =  15 cm
= Area of the shaded region = Area of rectangle − Area of triangle  AED + Area of semicircle
$$\begin{gathered} =\mathrm{AB}\times \mathrm{BC}-\frac{1}{2}\times \mathrm{AE}\times \mathrm{DE}+\frac{1}{2}\mathrm{\pi }{\left(\frac{\mathrm{BC}}{2}\right)}^2 \\ =20\times 15-\frac{1}{2}\times 9\times 12+\frac{1}{2}\times 3.14{\left(\frac{15}{2}\right)}^2 \\ =300-54+88.31 \\ =334.31 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 334.31 cm²

Answer:

In right triangle ABC
BC² = AB² + AC²  
= (7)² + (24)² 
= 49 + 576
= 625
∴ BC² = 625
⇒ BC = 25
Now, ∠COD + ∠BOD = 180°            (Linear pair angles)
⇒∠COD = 180° − 90° = 90°
 Now, Area of the shaded region = Area of sector having central angle (360° − 90°) −  Area of triangle  ABC
$$\begin{gathered} =\frac{270°}{360°}\mathrm{\pi }{\left(\frac{\mathrm{BC}}{2}\right)}^2-\frac{1}{2}\mathrm{AB}\times \mathrm{AC} \\ =\frac{3}{4}\times 3.14{\left(\frac{25}{2}\right)}^2-\frac{1}{2}\times 7\times 24 \\ =367.97-84 \\ =283.97 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 283.97 cm²

Answer:

We can find the radius of the incircle by using the formula
$r=\frac{2\times \mathrm{Area} \mathrm{of} \mathrm{triangle}}{\mathrm{Perimeter} \mathrm{of} \mathrm{triangle}}=\frac{2\times {\displaystyle \frac{\sqrt{3}}{4}}\times {\left(12\right)}^2}{3\times 12}=2\sqrt{3} \mathrm{cm}$
Now, area of shaded region = Area of triangle − Area of circle
$$\begin{gathered} =\frac{\sqrt{3}}{4}\times {\left(12\right)}^2-3.14\times {\left(2\sqrt{3}\right)}^2 \\ =62.28-37.68 \\ =24.6 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 24.6 cm²

Answer:

Construction:  Join AO and extend it to D on BC.


Radius of the circle, r = 42 cm
∠OCD= 30°

$$\begin{gathered} \cos 30°=\frac{\mathrm{DC}}{\mathrm{OC}} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{DC}}{42} \\ \Rightarrow \mathrm{DC}=21\sqrt{3} \\ \Rightarrow \mathrm{BC}=2\times \mathrm{DC}=42\sqrt{3} =72.66 \mathrm{cm} \\ \sin 30°=\frac{\mathrm{OD}}{\mathrm{OC}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{OD}}{42} \\ \Rightarrow \mathrm{OD}=21 \mathrm{cm} \\ \mathrm{Now}, \mathrm{AD}=\mathrm{AO}+\mathrm{OD}=42+21=63 \mathrm{cm} \end{gathered}$$

Area of shaded region = Area of circle − Area of triangle ABC
$$\begin{gathered} =\mathrm{\pi }{\left(\mathrm{OA}\right)}^2-\frac{1}{2}\times \mathrm{AD}\times \mathrm{AB} \\ =\frac{22}{7}{\left(42\right)}^2-\frac{1}{2}\times 63\times 72.66 \\ =5544-2288.79 \\ =3255.21 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let the radius of the circle be r
Now, Perimeter of quadrant = $\frac{1}{4}\left(2\mathrm{\pi }r\right)+2r$
$$\begin{gathered} \Rightarrow 25=\frac{1}{2}\times \frac{22}{7}\times r+2r \\ \Rightarrow 25=\frac{25r}{7} \\ \Rightarrow r=7 \mathrm{cm} \end{gathered}$$
Area of quadrant = $\frac{1}{4}\mathrm{\pi }{r}^2=\frac{1}{4}\times \frac{22}{7}\times 7\times 7=38.5 {\mathrm{cm}}^2$
Hence, the area of quadrant is 38.5 cm²

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$$\begin{gathered} =\frac{\mathrm{\theta }}{360°}\mathrm{\pi }{\left(\mathrm{OA}\right)}^2-\frac{1}{2}\times \mathrm{OA}\times \mathrm{OB} \\ =\frac{90°}{360°}\times 3.14{\left(10\right)}^2-\frac{1}{2}\times 10\times 10 \\ =78.5-50 \\ =28.5 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of minor segment is 28.5 cm²

Answer:

Area of the road = Area of outer circle − Area of inner circle
$$\begin{gathered} =\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ =\mathrm{\pi }\left[{\left(110\right)}^2-{\left(100\right)}^2\right] \\ =3.14\times 2100 \\ =6594 {\mathrm{m}}^2 \end{gathered}$$
Cost of levelling the road = Area of the road ⨯ Rate
= 6594 ⨯ 20
= Rs 131880

Answer:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle}=49\sqrt{3} \\ \Rightarrow \frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2=49\sqrt{3} \\ \Rightarrow {\left(\mathrm{Side}\right)}^2=7^2\times 2^2 \\ \Rightarrow \mathrm{Side}=14 \mathrm{cm} \end{gathered}$$

Radius of the circle = Half of the side of the triangle = 7 cm
Area of triangle not included in the circle = Area of triangle − Area of 3 sectors having central angle 60^∘
$$\begin{gathered} =49\sqrt{3}-3\times \frac{60°}{360°}\times \frac{22}{7}\times 7\times 7 \\ =84.77-77 \\ =7.77 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the required area is 7.77 cm²

Answer:

CD = 8 cm
BP = HQ = 4 cm
DE = EF = 5 cm
Area of the parallelogram ABCD = $B\times H$
                                                     $$\begin{gathered} =8\times 4 \\ =32 \mathrm{sq}.\mathrm{cm} \end{gathered}$$

Area of parallelogram FGHI = $B\mathit{\times }H$
                                               $$\begin{gathered} =8\times 4 \\ =32 \mathrm{sq}.\mathrm{cm} \end{gathered}$$
Area of the square = ${\mathrm{Side}}^2$
                               =$8^2 =64 \mathrm{sq}.\mathrm{cm}$
    
In $∆$ELF, we have:
$$\begin{gathered} {\mathrm{EL}}^2=5^2-4^2 \\ {\mathrm{EL}}^2=9 \\ \mathrm{EL}=3 \mathrm{cm} \end{gathered}$$

Area of $△$DEF = $\frac{1}{2}\times B\times H$
                        $$\begin{gathered} =\frac{1}{2}\times 8\times 3 \\ =12 \mathrm{sq}.\mathrm{cm} \end{gathered}$$

Area of the semicircle =$\frac{1}{2}{\mathrm{\pi r}}^2$

                                    $$\begin{gathered} =\frac{1}{2}\times \frac{22}{7}\times 16 \\ =25.12 \mathrm{sq}.\mathrm{cm} \end{gathered}$$
∴ Total Area =  Area of the parallelogram ABCD + Area of the parallelogram FGHI + Area of the triangle DEF + Area of the semicircle CKI + Area of the square
Total Area = 165.12 cm²

Answer:

Area of sector having central angle 150° =  $=\frac{150°}{360°}\mathrm{\pi }{\left(6\right)}^2=\frac{5}{12}\times \mathrm{Area} \mathrm{of} \mathrm{circular} \mathrm{disc}$

Now,  Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
$$\begin{gathered} =\frac{90°}{360°}\mathrm{\pi }{\left(6\right)}^2:\frac{120°}{360°}\mathrm{\pi }{\left(6\right)}^2:\frac{150°}{360°}\mathrm{\pi }{\left(6\right)}^2 \\ =\frac{1}{4}:\frac{1}{3}:\frac{5}{12} \\ =3:4:5 \end{gathered}$$

Answer:

Join each vertex of the hexagon to the centre of the circle.

The hexagon is made up of six triangles.

$$\begin{gathered} \mathrm{Total} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{design}=\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{circle}-\mathrm{Area} \mathrm{of} \mathrm{six} \mathrm{triangles} \\ = \left[\frac{22}{7}\times 35\times 35\right]-\left[6\times \left(\frac{\sqrt{3}}{4}\times 35\times 35\right)\right] \\ =\left(35\times 35\right)\left[\frac{22}{7}-\left(\frac{6\times 1.732}{4}\right)\right] \\ =1225\left(\frac{22}{7}-2.598\right) \\ =1225\times 0.542 \\ =663.95 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

In the right $∆$RPQ, we have:
$$\begin{gathered} RQ=\sqrt{R{P}^2+P{Q}^2} \\ =\sqrt{7^2+{24}^2} \\ =\sqrt{49 + 576} \\ = 25 \mathrm{cm} \end{gathered}$$
​
OR = OQ = 12.5 cm
Now,
Area of the circle = ${\mathrm{\pi r}}^2$
                             $$\begin{gathered} =3.14\times 12.5\times 12.5 \\ =490.625 \mathrm{sq}.\mathrm{cm} \end{gathered}$$

Area of the semicircle = $\frac{490.625}{2}=245.31 \mathrm{sq}.\mathrm{cm}$
Area of the triangle $=\frac{1}{2}\times b\times h=\frac{1}{2}\times 7\times 24=84 \mathrm{sq}.\mathrm{cm}$
Thus, we have:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{shaded} \mathrm{part} =\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{semicircle}- \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{triangle} \\ = 245.31 - 84 \\ = 161.31 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Using Pythagoras' theorem for triangle ABC, we have:

$C{A}^2+A{B}^2=B{C}^2$

 $$\begin{gathered} CA=\sqrt{B{C}^2-A{B}^2} \\ =\sqrt{100-36} \\ =\sqrt{64} \\ = 8 \mathrm{cm} \end{gathered}$$

Now, we must find the radius of the incircle. Draw OE, OD and OF perpendicular to AC, AB and BC, respectively.



Consider quadrilateral AEOD.
Here,
$EO=OD (\mathrm{Both} \mathrm{are} \mathrm{radii}.)$

Because the circle is an incircle, AE and AD are tangents to the circle.
$\angle AEO=\angle ADO =90°$

Also,
$\angle A=90°$
Therefore, AEOD is a square.
Thus, we can say that $AE=EO=OD=AD=r$.

$$\begin{gathered} CE=CF=8-r \\ BF=BD=6-r \\ CF+BF=10 \\ \Rightarrow \left(8-r\right)+\left(6-r\right)=10 \\ \Rightarrow 14-2r=10 \\ \Rightarrow r=2 \mathrm{cm} \end{gathered}$$

Area of the shaded part = Area of the triangle $-$ Area of the circle
                                       $$\begin{gathered} =\left\{\frac{1}{2}\times 6\times 8\right\}-\left\{\mathrm{\pi }\times 2\times 2\right\} \\ =24-12.56 \\ =11.44 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Perimeter (circumference of the circle) = $2\mathrm{\pi r}$
We know:
Perimeter of a semicircular arc = $\mathrm{\pi r}$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS = $\mathrm{\pi r}=6\mathrm{\pi } \mathrm{cm}$

For the arc QES, radius is 4 cm.
​∴ Circumference of the semicircle QES = $\mathrm{\pi r}=4\mathrm{\pi } \mathrm{cm}$

For the arc PBQ, radius is 2 cm.
∴ Circumference of the semicircle PBQ = $\mathrm{\pi r}=2\mathrm{\pi } \mathrm{cm}$

Now,
Perimeter of the shaded region = $6\mathrm{\pi }+4\mathrm{\pi }+2\mathrm{\pi }$
                                                   $=12\mathrm{\pi cm}$
                                                   $$\begin{gathered} =12\times 3.14 \\ =37.68 \mathrm{cm} \end{gathered}$$

Area of the semicircle PBQ = $\frac{1}{2}{\mathrm{\pi r}}^2$

                                            $$\begin{gathered} =\frac{1}{2}\times 3.14\times 2\times 2 \\ =6.28 {\mathrm{cm}}^2 \end{gathered}$$

Area of the semicircle PTS = $\frac{1}{2}{\mathrm{\pi r}}^2$

                                            $$\begin{gathered} =\frac{1}{2}\times 3.14\times 6\times 6 \\ =56.52 {\mathrm{cm}}^2 \end{gathered}$$

Area of the semicircle QES = $\frac{1}{2}{\mathrm{\pi r}}^2$
  
                                            $$\begin{gathered} =\frac{1}{2}\times 3.14\times 4\times 4 \\ =25.12 {\mathrm{cm}}^2 \end{gathered}$$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES
                                           $=6.28 + 56.52 - 25.12 = 37.68 {\mathrm{cm}}^2$

Answer:

Length of the inner curved portion $(400-2\times 90)=220 \mathrm{m}$
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
​Thus, we have:
$$\begin{gathered} \Rightarrow \mathrm{\pi r}=110 \\ \Rightarrow \frac{22}{7}\mathrm{r}=110 \\ \Rightarrow r=\frac{110\times 7}{22} \\ \Rightarrow r=35 \mathrm{m} \end{gathered}$$

Inner radius = 35 m
Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each $(90\times 14) \mathrm{m}$] + Area of the circular ring with R = 49 m and r = 35 m)}
$$\begin{gathered} =(2\times 90\times 14)+\frac{22}{7}\times \left[{\left(49\right)}^2-{\left(35\right)}^2\right] \\ =2520+\frac{22}{7}\times \left(2401-1225\right) \\ =2520+\frac{22}{7}\times 1176 \\ =2520+3696 \\ =6216 {\mathrm{m}}^2 \end{gathered}$$

​Length of the outer boundary of the track
$$\begin{gathered} =\left(2\times 90+2\times \frac{22}{7}\times 49\right) \\ =488 \mathrm{m} \end{gathered}$$

Therefore, the length of the outer boundary of the track is 488 m and the area of the track is 6216 sq. m.

Answer:

Area of the shaded region

= Area of the rectangle − Area of the semicircle 

$$\begin{gathered} =21\times 14-\left\{\frac{1}{2}\times \mathrm{\pi }\times {\left(\frac{14}{2}\right)}^2\right\} \\ =294-\left\{\frac{1}{2}\times \frac{22}{7}\times 7\times 7\right\} \\ =294-77 \\ =217 {\text{cm}}^2 \end{gathered}$$

Therefore, area of shaded region is 217  cm².

Length of the boundary (or perimeter) of the shaded region

$$\begin{gathered} =\mathrm{AB}+\mathrm{AD}+\mathrm{DC}+\mathrm{arc} \mathrm{BC} \\ =\mathrm{AB}+\mathrm{AD}+\mathrm{DC}+\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{14}{2}\right) \\ =21+14+21+\frac{22}{7}\times 7 \\ =56+22 \\ =78 \mathrm{cm} \end{gathered}$$

Therefore, the perimeter of the shaded region is 78 cm.

Answer:

Given: Radius of the inner circle with radius OC, r = 21 cm
Radius of the inner circle with radius OA, R = 42 cm
∠AOB = 60°

Area of the circular ring

$$\begin{gathered} =\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ =\mathrm{\pi }\left[R^2-r^2\right] \\ =\mathrm{\pi }\left[{42}^2-{21}^2\right] {\mathrm{cm}}^2 \end{gathered}$$

Area of ACDB = area of sector AOB − area of COD
$$\begin{gathered} =\frac{60}{360}\times \mathrm{\pi }\times R^2-\frac{60}{360}\times \mathrm{\pi }\times r^2 \\ =\frac{60}{360}\times \mathrm{\pi }\left[R^2-r^2\right] \\ =\frac{60}{360}\times \mathrm{\pi }\left[{42}^2-{21}^2\right] \end{gathered}$$

Area of shaded region = area of circular ring − area of ACDB
$$\begin{gathered} =\mathrm{\pi }\left[{42}^2-{21}^2\right]-\frac{60}{360}\mathrm{\pi }\left[{42}^2-{21}^2\right] \\ =\mathrm{\pi }\left[{42}^2-{21}^2\right]\left[1-\frac{60}{360}\right] \\ =\frac{22}{7}\left(42-21\right)\left(42+21\right)\times \frac{300}{360} \\ =3465 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Area of the shaded region

= Area of the semi-circle with diameter of 9 cm − Areas of two semi-circles with diameter 3 cm − Area of the circle with diameter 4.5 cm + Area of semi-circle with diameter 3 cm

= Area of the semi-circle with radius of 4.5 cm − 2 × Area of semi-circle with radius 1.5 cm − Area of the circle with radius 2.25 cm + Area of semi-circle with radius 1.5 cm



= $3.9375\times \frac{22}{7}$

= 12.375 cm²

Thus, the area of the shaded region is 12.375 cm².

Answer:

We have,
Side of square = 28 cm and radius of each circle = $\frac{28}{2}$ cm


Area of the shaded region

= Area of the square + Area of the two circles − Area of the two quadrants

$$\begin{gathered} ={\left(28\right)}^2 + 2\times \mathrm{\pi }\times {\left(\frac{28}{2}\right)}^2 - 2\times \frac{1}{4}\times \mathrm{\pi }\times {\left(\frac{28}{2}\right)}^2 \\ ={\left(28\right)}^2 + \frac{3}{2}\times \mathrm{\pi }\times {\left(\frac{28}{2}\right)}^2 \\ ={\left(28\right)}^2\left(1+\frac{3}{2}\times \frac{22}{7}\times \frac{1}{2}\times \frac{1}{2}\right) \\ ={\left(28\right)}^2\left(1+\frac{33}{28}\right) \\ ={\left(28\right)}^2\times \frac{61}{28} \\ =28\times 61 \\ =1708 {\mathrm{cm}}^2 \end{gathered}$$

Therefore, the area of the shaded region is 1708 cm².

Answer:

Given that ABC is a triangle with sides AB = 14 cm, BC = 48 cm, CA = 50 cm.
Clearly it is a right angled triangle.
Area of $△\mathrm{ABC}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{2}\times 48\times 14=336 {\mathrm{cm}}^2.$

Now we need to remove the area of 3 arcs each of radius 5 cm from this area of triangle.
Here, $\angle \mathrm{B}=90°, \angle \mathrm{A}+\angle \mathrm{C}=90°.$
The area of these 3 arcs would exactly be equal to a sector of circle with radius 5 cm and angle = $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$ i.e. a semi circle.
Since we know that area of semi circle is $\frac{1}{2}\mathrm{\pi }{r}^2=\frac{1}{2}\times 3.14\times 5\times 5=\frac{1}{2}\times \frac{314}{100}\times 25=\frac{314}{8}=\frac{157}{4} {\mathrm{cm}}^2.$
Hence, the area of shaded region is = $$\begin{gathered} \mathrm{area} \mathrm{of} △\mathrm{ABC}-\mathrm{combined} \mathrm{area} \mathrm{of} 3 \mathrm{arcs}=336-\frac{157}{4} \\ =336-39.25=296.75 {\mathrm{cm}}^2. \end{gathered}$$

Answer:

The diameters of concentric circles are in ratio 1 : 2 : 3.
let the diameters be d₁ = 1x, d₂ = 2x, d₃ = 3x.
So, the radii are $r_1=\frac{x}{2}, r_2=x, r_3=\frac{3x}{2}.$
The areas of three regions are given be 
$$\begin{gathered} A_1=\mathrm{\pi }{r_1}^2=\mathrm{\pi }{\left(\frac{x}{2}\right)}^2=\frac{\mathrm{\pi }{x}^2}{4}, \\ {\mathrm{A}}_2=\mathrm{\pi }{r_2}^2-\mathrm{\pi }{r_1}^2=\mathrm{\pi }{x}^{\mathit{2}}-\frac{\mathrm{\pi }{x}^2}{4}=\frac{3\mathrm{\pi }{x}^2}{4}, \\ {\mathrm{A}}_3=\mathrm{\pi }{r_3}^2-\mathrm{\pi }{r_2}^2=\mathrm{\pi }{\left(\frac{3x}{2}\right)}^2-\mathrm{\pi }{x}^2=\frac{5\mathrm{\pi }{x}^2}{4} \end{gathered}$$
So, their ratio is $A_1:A_2:A_3=\frac{\mathrm{\pi }{x}^{\mathit{2}}}{4}:\frac{3\mathrm{\pi }{x}^2}{4}:\frac{5\mathrm{\pi }{x}^2}{4}=1:3:5.$

Answer:

Let the radius of the circle be r and circumference C.
​
Now,
$$\begin{gathered} C-r=37 \\ \Rightarrow 2\mathrm{\pi }r-r=37 \\ \Rightarrow r\left(2\times \frac{22}{7}-1\right)=37 \\ \Rightarrow r=\frac{37\times 7}{37}=7 \mathrm{cm} \end{gathered}$$
Now, $C=2\mathrm{\pi }r=2\times \frac{22}{7}\times 7=44 \mathrm{cm}$
Hence, the circumference of the circle is 44 cm.