RD Sharma 2022 Solutions for Class 10 Maths Chapter 13 - Areas Related To Circles

Answer:

We know that the circumference C and area A of a circle of radius r are given by and respectively.

Here,

So substituting the value of r in above formulas,

Circumference of the circle

Area of the circle

Answer:

Let r cm be the radius of the circle. Then

Area of a circle is

We know that the Circumference of circle of radius r is

So substituting the value of r in above formula

Answer:

Let r be the radius of the circle. Then

Circumference of the circle

We know that the area of a circle of radius r is

Substituting the value of r in above formula

Answer:

Width of the road = 7 m
circumference of the circular park = 352 m
$$\begin{gathered} \Rightarrow 2\mathrm{\pi r}=352 \\ \Rightarrow \mathrm{r}=\frac{352}{2\mathrm{\pi }}=56 \mathrm{m} \end{gathered}$$
Area of the road = Area of the circular park including the path − area of the circular park
$$\begin{gathered} =\mathrm{\pi }{\left(\mathrm{r}+7\right)}^2-{\mathrm{\pi r}}^2 \\ =\mathrm{\pi }\left[{\left(56+7\right)}^2-{\left(56\right)}^2\right] \\ =\mathrm{\pi }\left[{63}^2-{56}^2\right] \\ =\mathrm{\pi }\left[7\times 119\right] \\ =2618 {\mathrm{m}}^2 \end{gathered}$$

Answer:

Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.

It is given that,

Now we will calculate the ratio of their areas,

Substituting the value of ,

Hence the ratio of their Areas is.

Answer:

Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.

It is given that the sum of the radii of two circles is 140 cm and difference of their circumferences is 88 cm. So,

……(A)

……(B)

Now, solving (A) and (B)

Thus diameters of circles are,

Answer:

Let the radius of circles be , andrespectively. Then their areas are, and respectively.

It is given that,

We have, and

Substituting the values of ,

Hence, the radius of circle is.

Answer:

Let the radius of circles be, andrespectively. Then their circumferences are, and respectively.

It is given that,

We have, and

Substituting the values of,

Hence the radius of the circle is.

We know that the area A of circle is

Substituting the value of r

Hence the area of the circle is.

Answer:

Let the radius of the inner circle and the race track be R m.
Outer circumference of the race track = 528 m
$$\begin{gathered} \Rightarrow 2\mathrm{\pi }R=528 \\ \Rightarrow R=\frac{528}{2\mathrm{\pi }}=84 \mathrm{m} \end{gathered}$$
Total radius of the outer circle = 84 − 14 = 70 m
Area of the circular track = Area of the outer circle − area of inner circle
$$\begin{gathered} =\mathrm{\pi }{\left(84\right)}^2-\mathrm{\pi }{\left(70\right)}^2 \\ =\mathrm{\pi }\left[\left(84-70\right)\left(84+70\right)\right] \\ =\mathrm{\pi }\left(14\right)\left(154\right) \\ =6776 {\mathrm{m}}^2 \end{gathered}$$
Cost of levelling the track = $0.5\times 6776=\mathrm{Rs} 3388$



 

Answer:

Radius of the park, r = 105 m
Radius of the park with the road, R = 105 + 21 = 126 m
Area of the road = Area of the park with the road − Area of the park
$$\begin{gathered} =\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ =\mathrm{\pi }\left[{\left(126\right)}^2-{\left(105\right)}^2\right] \\ =\mathrm{\pi }\left(21\right)\left(231\right) \\ =15246 {\mathrm{m}}^2 \end{gathered}$$


 

Answer:

Given that, a circle of radius 7.5 cm is inscribed in a square.



Area of circle = $\mathrm{\pi }{r}^2$
                      $$\begin{gathered} =3.14\times 7.5^2 \\ =176.625 {\mathrm{cm}}^2 \end{gathered}$$

Since, the side of square = diameter of circle = 15 cm
Area of square = ${15}^2=225 {\mathrm{cm}}^2$

Thus, the area outside the circle and inside the square is $225-176.625=48.375 {\mathrm{cm}}^2$.

Hence, the required area is 48.375 cm².

Answer:

It is given that the area A of circle inscribed in an equilateral triangle is 154 cm².

We know that the area A of circle inscribed in an equilateral triangle is

Now, we will find the value of r.

Substituting the value of area,

Let the height of triangle be h. Then

If a is the side of triangle, then

Substituting the value of h,

Hence perimeter of triangle is .

Answer:

Let the radius of wheel be r. Thus, circumference C of the wheel

Since car travels 1 km distance in which wheel makes 450 complete revolutions. Then

We know that,

Hence the radius of wheel is.

Answer:

DISCLAIMER: There is some error in the given question. 
We have solved the question by taking a square inscribed in a circle. Then finding the area inside the circle and outside the square.


Diagonal of the square = 8 cm.
Let the side of the square be a cm.
In triangle BCD, 
$$\begin{gathered} B{C}^2+C{D}^2=B{D}^2 \\ \Rightarrow a^2+a^2=8^2 \\ \Rightarrow a^2=32 .....\left(\mathrm{i}\right) \end{gathered}$$
Radius of the circle will be R = 4 cm
Now area between the circle and the square will be
 $$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{circle}-\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{square} \\ =\mathrm{\pi }{\left(R\right)}^2-a^2 \\ =\mathrm{\pi }{\left(4\right)}^2-32 \\ =\left(16\mathrm{\pi }-32\right) {\mathrm{cm}}^2 \end{gathered}$$
 

Answer:

We have given AB = 4m and circumference of semicircle with radius OA as.

We are asked to find the area between the two semi-circles.

For that we will first find OA.

$\mathrm{\pi }r=163\frac{3}{7}$

Now we will substitute

$$\begin{gathered} \frac{22}{7}\times r=163\frac{3}{7} \\ \Rightarrow r=\frac{1144}{7}\times \frac{7}{22} \\ \Rightarrow r=52 \\ \Rightarrow OA=52 m \end{gathered}$$

Now we will find OB.

$$\begin{gathered} \therefore OB=52+4 \\ \therefore OB=56 \mathrm{m} \end{gathered}$$

Now we will find the area between two semi-circles as given below,
$$\begin{gathered} \therefore \mathrm{Area}=\frac{\mathrm{\pi }\times 56\times 56}{2}-\frac{\mathrm{\pi }\times 52\times 52}{2} \\ =\frac{\mathrm{\pi }}{2}\left(1568-1352\right) {\mathrm{m}}^2 \\ =\frac{\mathrm{\pi }}{2}\times 216 \\ =\frac{22}{7\times 2}\times 216 \\ =339.43 {\mathrm{m}}^2 \end{gathered}$$

Therefore, area of the path is 339.43 m².

Now we will find the cost of gravelling the path.

$$\begin{gathered} \mathrm{Cost}=339.43\times 1.50 \\ =\mathrm{Rs} 509.14 \end{gathered}$$

Therefore, cost of gravelling the path is Rs 509.14.

Now we will find the cost of turfing the plot. For that we will find the area of the plot.

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{plot}=\frac{1}{2}{\mathrm{\pi r}}^2 \\ =\frac{1}{2}\times \frac{22}{7}{\left(52\right)}^2 \\ =\frac{1}{2}\times \frac{22}{7}\times 52\times 52 \\ =4249.14 \\ \mathrm{Cost} \mathrm{of} \mathrm{turfing} \mathrm{the} \mathrm{plot}= 4249.14\times 0.45 \\ =\mathrm{Rs} 1912.11 \end{gathered}$$

Therefore, cost of the turfing the plot is Rs 1912.11.
Disclaimer: Due to some error in the question, we get the different answers.

Answer:

Let the radius of the semicircular plot be r. 
Perimeter of the semi-circular grassy plot = $\mathrm{\pi r}+2\mathrm{r}=72$
$\Rightarrow r=14 \mathrm{cm}$
Given that the width of the plot = 3.5 m
Thus, the outer radius = 3.5 + 14 = 17.5 m
Area of the path = $\frac{{\mathrm{\pi R}}^2}{2}-\frac{{\mathrm{\pi r}}^2}{2}$
$$\begin{gathered} =\frac{\mathrm{\pi }}{2}\left(R^2-r^2\right) \\ =\frac{\mathrm{\pi }}{2}\left({\left(17.5\right)}^2-{\left(14\right)}^2\right) \\ =173.25 {\mathrm{m}}^2 \end{gathered}$$


 

Answer:

The area enclosed between the two circles of radii 3.5 cm and 7 cm 
$$\begin{gathered} =\mathrm{\pi }\left(7^2-3.5^2\right) \\ =115.5 {\mathrm{cm}}^2 \end{gathered}$$
Let the radius of the outermost circle be r cm.
Area betweent the circles with radius r and 7 cm=Area between the circles with radius 7 cm and 3.5 cm 
$$\begin{gathered} \mathrm{\pi }\left({\mathrm{r}}^2-7^2\right)=115.5 \\ \Rightarrow \left({\mathrm{r}}^2-7^2\right)=\frac{115.5}{\mathrm{\pi }} \\ \Rightarrow {\mathrm{r}}^2=36.75+49=85.75 {\mathrm{cm}}^2 \\ \Rightarrow \mathrm{r}=9.26 \mathrm{cm} \end{gathered}$$


 

Answer:

Let the three regions be A , B and C.
The diameters are in the ratio 1 : 2 : 3.
Let the diameters be 1x, 2x and 3x
Then the radius will be $\frac{x}{2}, \frac{2x}{2} and \frac{3x}{2}$
Area of region A = ${\mathrm{\pi r}}_{\mathrm{A}}^2=\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2=\frac{{\mathrm{\pi x}}^2}{4}$
Area of region B = ${\mathrm{\pi r}}_{\mathrm{B}}^2-{\mathrm{\pi r}}_A^2=\mathrm{\pi }{\left(\mathrm{x}\right)}^2-\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2=\frac{3\mathrm{\pi }{\left(\mathrm{x}\right)}^2}{4}$
Area of region C = ${\mathrm{\pi r}}_C^2-{\mathrm{\pi r}}_{\mathrm{B}}^2-{\mathrm{\pi r}}_A^2=\mathrm{\pi }{\left(\frac{3\mathrm{x}}{2}\right)}^2-\mathrm{\pi }{\left(\mathrm{x}\right)}^2-\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2=\mathrm{\pi }{\left(\frac{3\mathrm{x}}{2}\right)}^2-\frac{3{\mathrm{\pi x}}^2}{4}=\frac{5{\mathrm{\pi x}}^2}{4}$
Thus, ratio of the areas of regions A, B and C  will be 
$\frac{{\mathrm{\pi x}}^2}{4}$:$\frac{3\mathrm{\pi }{\left(\mathrm{x}\right)}^2}{4}$:$\frac{5{\mathrm{\pi x}}^2}{4}$
⇒1:3:5

Answer:

The arc length l of a sector of an angle θ in a circle of radius r is given by

It is given that and. Substituting the value of r and θ in above equation,

Answer:

We know that the arc length l of a sector of an angle θ in a circle of radius r is

It is given that and length . Substituting these value in above equation,

Hence, the angle subtended at the centre of circle is.

Answer:

We know that the arc length l of a sector of an angle θ in a circle of radius r is

It is given and angle.

Now we substitute the value of l and θ in above formula to find the value of radius r of circle.

Answer:

We know that the arc length l of a sector of an angle θ in a circle of radius r is

It is given that and angle.

Now we substitute the value of l and θ in above formula to find the value of radius r of circle.

Answer:

We know that the area A of a sector of an angle θ in the circle of radius r is given by

It is given that and angle.

Now we substitute the value of r and θ in above formula,

Answer:

We know, $A=\frac{\theta }{360}\times \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{circle}$              .....(i)
Let the area of the circle be Ar. 
Thus area of the sector = $\frac{1}{12}Ar$                       .....(ii)
From (i) and (ii) we have
$$\begin{gathered} \frac{1}{12}Ar=\frac{\theta }{360}\times Ar \\ \Rightarrow \frac{360}{12}=\theta \\ \Rightarrow \theta =30° \end{gathered}$$

 

Answer:

We know that the arc length l and area A of a sector of an angle θ in the circle of radius r is given by and respectively.

It is given that, and.

We will calculate the arc length using the value of r and θ,

Now, we will find the value of area A of the sector

Answer:

Angle make by the minute hand in 1 minute = 6^∘
Angle make by the minute hand in 5 minute = 5 ⨯ 6^∘ = 30^∘
Area of the sector having central angle is given by
$$\begin{gathered} \frac{30°}{360°}\mathrm{\pi }{\left(14\right)}^2 \\ =\frac{1}{12}\times \frac{22}{7}{\left(14\right)}^2 \\ =51.33 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area swept by minute hand in 5 minutes is 51.33 cm²

Answer:

We know that the arc length l and area A of a sector of circle at an angle θ of radius r is given by and angle.

Let OAB is the given sector.

It is given that and angle.

Now using the value of r and θ, we will find the value of l and A,

Arc length,

Area of sector,

Answer:

Here, we have θ = 60° and r = 21 cm

(i) The length of the arc is given by
$$\begin{gathered} \frac{60°}{360°}\times 2\mathrm{\pi }\left(21\right) \\ =\frac{1}{6}\times 2\times \frac{22}{7}\times 21 \\ =22 \mathrm{cm} \end{gathered}$$


(ii) Area of the sector formed by the arc is given by
$$\begin{gathered} \frac{60°}{360°}\mathrm{\pi }{\left(21\right)}^2 \\ =\frac{1}{6}\times \frac{22}{7}{\left(21\right)}^2 \\ =231 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Minute hand of the clock describes a circle of radius equal to its length i.e 5 cm.
The minute hand rotates through 6^o in one minute. 
So, the minute hand will sweep 210^o in one minute.
Area swept by the minute hand in one minute is the area of a sector of angle 6^o in a circle of radius 5 cm. 
Area swept by the minute hand in 35 minutes is the area of a sector of angle 210^o in a circle of radius 5 cm
$\mathrm{Area} \mathrm{of} \mathrm{sector}=\frac{\theta }{360}\times \mathrm{\pi }{r}^2$
$$\begin{gathered} =\frac{210}{360}\times \mathrm{\pi }\times {\left(5\right)}^2 \\ =45\frac{5}{6} {\mathrm{cm}}^2 \end{gathered}$$
 

Answer:

We know that the area A of a sector of circle of radius r and arc length l is given by

Let OAB is the given sector. Then,

arc AB = 15.8 m

So, l = 15.8 m

Now substituting the value of r and l in above formula,

$$\begin{gathered} A=\frac{1}{2}\times 15.8\times 5.7 \\ =45.03 m^2 \end{gathered}$$

Answer:

We have to find the area of the sector AOB formed by the chord AB.

We have and. So,

Let. Then,

In, we have

Hence,

Now, using the value of and r we will find the area of sector AOB,

Answer:

It is given that the radius of circle, and angle at the centre of circle.

(i) We know that the Circumference C of circle of radius r is,

$\begin{array}{rcl}C& =& 2\mathrm{\pi r}\\ & =& 2\times \frac{22}{7}\times 6\\ & =& \frac{264}{7}\\ & =& 37.71 \mathrm{cm}\end{array}$

(ii) We know that the Area A of circle of radius r is,

(iii) We know that the arc length l of a sector of an angle θ in a circle of radius r is

(iv) We know that the area A of a sector of an angle θ in the circle of radius r is given by

Answer:

It is given that the radius of circle is r and the angle.

In,

It is given that.

(i) We know that the arc length l of a sector of an angle θ in a circle of radius r is

Now we substitute the value of OC, OA and l to find the perimeter of sector AOC,

Hence,

(ii) We know that area A of the sector at an angle θ in the circle of radius r is

. 

Thus

Hence,

Answer:

It is given that the radius of circle is r cm and angle.

(i) We know that the arc length l of a sector of an angle θ in a circle of radius r is

Now we substitute the value of OB, OA and l to find the perimeter of sector AOB,

(ii) We know that area A of the sector at an angle θ in the circle of radius r is

. Thus

Substituting the value of θ,

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that the chord PQ divides the circle in two segments.

We have and. So,

Since,

In, we have

Thus the radius of circle is .

Now using the value of radius r and angle θ we will find the area of minor segment

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that,

Substituting these values in above formula

Answer:

We know area of minor segment of the circle is $A=\left\{\frac{\mathrm{\pi \theta }}{360}-\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right\}{r}^2$
$$\begin{gathered} \Rightarrow A=\left\{\frac{\mathrm{\pi }\times 90°}{360}-\sin \frac{90}{2}\cos \frac{90}{2}\right\}{\left(20\right)}^2 \\ \Rightarrow A=\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right)\left(400\right) \end{gathered}$$
Area of the major segment = Area of the circle − area of the minor segment
$$\begin{gathered} =\mathrm{\pi }{\left(20\right)}^2-\left(400\right)\left[\frac{\mathrm{\pi }}{2}-\frac{1}{2}\right] \\ =\left(400\right)\left[\mathrm{\pi }-\frac{\mathrm{\pi }}{2}+\frac{1}{2}\right] \\ =1142 {\mathrm{cm}}^2 \end{gathered}$$

DISCLAIMER: The answer given in the book is incorrect. If we take the radius 10 instead of 20, then the answer will match.
 

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that the chord AB divides the circle in two segments.

We have and. So,

Let. Then,

In, we have

Hence,

Now using the value of r and θ, we will find the area of minor segment

Answer:

We know area of minor segment of the circle is $A=\left\{\frac{\mathrm{\pi \theta }}{360}-\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right\}{r}^2$
$$\begin{gathered} A=\left\{\frac{\mathrm{\pi }90}{360}-\sin \frac{90}{2}\cos \frac{90}{2}\right\}{\left(5\right)}^2 \\ A=\left\{\frac{\mathrm{\pi }}{4}-\sin 45\times \cos 45\right\}{\left(5\right)}^2 \\ A=\left\{\frac{\mathrm{\pi }}{4}-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\right\}25 \\ A=\left\{\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right\}25=7.125 {\mathrm{cm}}^2 \end{gathered}$$
Area of the major segment = Area of the sector − area of minor segment
$$\begin{gathered} =\mathrm{\pi }{\left(5\right)}^2-7.125 \\ =71.375 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

.

It is given that,

So,

Area, A of minor segment cutoff by AC at angle

Now, since and

We know that the area of sector of a circle of radius r at an angle θ is

So, the area of sector BOC,

It is given that,

Answer:

We know that the area of circle and area of minor segment of angle θ in a circle of radius r is given by, andrespectively.

It is given that,

Answer:

It is given that a plot is in form of rectangle ABCD having a semicircle on BC.

Since BC is the diameter of semicircle. Then, radius of semicircle is

                               =308 m²

                                          = 1680 m²

Now,

Area of plot = Area of rectangle + Area of semicircle
                   = 1680 + 308 m²
                   = 1988 m²
 

Answer:

Area of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle.

We know that area of the four sectors with radius is equal to area of one circle.

Therefore, area of the shaded region is

Answer:

It is given that four cows are tethered at four corner of square ABCD. We have to find the area of plot that will left ungrazed.

Let the side of square is a.

a = 25 + 25 m = 50 m

Area of square = a²
                        $$\begin{gathered} =50\times 50 \\ =2500 m^2 \end{gathered}$$

Answer:

Shaded portion shows the area that is grazed by the cow. It is in the form of a quadrant of a circle with radius 14 m.
$$\begin{gathered} \mathrm{Area} \mathrm{grazed} \mathrm{by} \mathrm{the} \mathrm{cow}=\frac{1}{4}\mathrm{\pi }{r}^2 \\ =\frac{1}{4}\mathrm{\pi }{\left(14\right)}^2 \\ =154 {\mathrm{m}}^2 \end{gathered}$$

Answer:

The area grazed by the calf is in the form of a quadrant of a circle with radius 6 m.
$$\begin{gathered} \mathrm{Area} \mathrm{grazed} \mathrm{by} \mathrm{the} \mathrm{calf} \mathrm{with} \mathrm{rope} 6 \mathrm{m}=\frac{1}{4}\mathrm{\pi }{\left(6\right)}^2 \\ =28.28 {\mathrm{m}}^2 \end{gathered}$$
When the rope length is increased then total rope length = 6 + 5.5 m = 11.5 m
$$\begin{gathered} \mathrm{Area} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{calf} \mathrm{for} \mathrm{grazing} \mathrm{with} \mathrm{rope} 11.5 \mathrm{m}=\frac{1}{4}\mathrm{\pi }{\left(11.5\right)}^2 \\ =103.91 {\mathrm{m}}^2 \end{gathered}$$
Hence, increase in area of grassy lawn that is grazed = 103.91 − 28.28 = 75.63 m²

Answer:

ABCD is a square.

Side of the square = AB = $2\sqrt{2} \mathrm{cm}$

We know

Length of the diagonal of square = $\sqrt{2}$ × Side of the square

∴ BD = Diameter of the circle = Length of the diagonal of square = $\sqrt{2}\times 2\sqrt{2}$ = 4 cm

⇒ Radius of the circle = $\frac{\mathrm{BD}}{2}$ = 2 cm

Now,

Area of shaded region

= Area of the circle − Area of the square ABCD

$$\begin{gathered} =\mathrm{\pi }{\left(\frac{\mathrm{BD}}{2}\right)}^2-{\left(\mathrm{AB}\right)}^2 \\ =3.14\times {\left(2\right)}^2-{\left(2\sqrt{2}\right)}^2 \\ =3.14\times 4-8 \end{gathered}$$
$$\begin{gathered} =12.56-8 \\ =4.56 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the area of the shaded region is 4.56 cm².

Answer:

OABC is a square.

Side of the square = OA = 15 cm

We know

Length of the diagonal of square = $\sqrt{2}$ × Side of the square

∴ OB = Radius of the quadrant of the circle = Length of the diagonal of square = $15\sqrt{2} \mathrm{cm}$

Now,

Area of shaded region

= Area of quadrant OPQO − Area of the square OABC

$$\begin{gathered} =\frac{1}{4}\times \mathrm{\pi }{\left(\mathrm{OB}\right)}^2-{\left(\mathrm{OA}\right)}^2 \\ =\frac{1}{4}\times 3.14\times {\left(15\sqrt{2}\right)}^2-{\left(15\right)}^2 \\ =\frac{1}{4}\times 3.14\times 450-225 \end{gathered}$$
$$\begin{gathered} =353.25-225 \\ =128.25 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the area of the shaded region is 128.25 cm².

Answer:

Answer:

It is given that a circle of radius 4.2 cm and two quadrants of radius 1.4 cm are cut from a square of side 8 cm.

Let the side of square be a. Then,

Since the radius of circle is 4.2 cm. So, 

$$\begin{gathered} Area of circle={\mathrm{\pi r}}^2 \\ =\frac{22}{7}\times 4.2\times 4.2 \\ =55.44 {\mathrm{cm}}^2 \end{gathered}$$

Now area of quadrant of circle of radius 1.4 cm is,

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{shaded} \mathrm{region} = \mathrm{Area} \mathrm{of} \mathrm{square}-\mathrm{Area} \mathrm{of} \mathrm{circle}-2\times \mathrm{Area} \mathrm{of} \mathrm{quadrant} \\ =64-55.44-2\times 1.54 \\ =5.48 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of rectangle − area of the semi-circle with diameter DC triangle + 2 area of two semicircles with diameters AD and BC

Substituting we get,

Therefore, area of the shaded region is.

Answer:

(i) We have given two semi-circles and a rectangle.

Area of the shaded region = Area of the rectangle − Area of the two semicircles ……..(1)

Substituting we get,

Therefore, area of shaded region is.

(ii) Now we will find length of the boundary of the shaded region.

Therefore, length of the boundary of the shaded region is.

Answer:

It is given that OACB is a quadrant of circle with centre at O and radius 3.5 cm.

(i) We know that the area of quadrant of circle of radius r is,

Substituting the value of radius,

Hence, the area of OACB is.

(ii) It is given that radius of quadrant of small circle is 2 cm.

Let the area of quadrant of small circle be.

It is clear from the above figure that area of shaded region is the difference of larger quadrant and the smaller one. Hence,

Answer:

Construction: Join OB

In right triangle AOB
OB² = OA² + AB²
= 21² + 21²
= 441 + 441
= 882
∴ OB² = 882
Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC
$$\begin{gathered} =\frac{1}{4}\mathrm{\pi }{\left(\mathrm{OB}\right)}^2-{\left(\mathrm{OA}\right)}^2 \\ =\frac{1}{4}\times \frac{22}{7}\times 882-441 \\ =693-441 \\ =252 {\mathrm{cm}}^2 \end{gathered}$$
14π(OB)2−(OA)2=14×3.14×800−400=628−400=228 cm2
Hence, the area of the shaded region is 252 cm².

Answer:

(i) Area of shaded region = Area of square OABC − Area of quadrant OAPC
$$\begin{gathered} ={\left(\mathrm{Side}\right)}^2-\frac{1}{4}\mathrm{\pi }{r}^2 \\ ={\left(7\right)}^2-\frac{1}{4}\times \frac{22}{7}\times 7\times 7 \\ =49-38.5 \\ =10.5 {\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of the shaded region is 10.5 cm²
 

(ii) It is given a triangle ABC is cut from a circle.

In,

, Since any angle inscribed in semicircle is always right angle.

By applying Pythagoras theorem,

We know that the area A of circle of radius r is

Substituting the value of radius r,

Answer:

We know that a regular hexagon is made up of 6 equilateral triangles.

We have given area of the one of the triangles.

We know that if a regular hexagon is inscribed in the circle, then the radius of the circle is same as the side of the regular hexagon.

We also know that a regular hexagon is made up of 6 equilateral triangles and we have area of one of the equilateral triangle.

Substituting the value of the given equilateral triangle we get,

Now we will find the area of the circle.

Substituting the values we get,

Now we will substitute we get,

Therefore, area of the hexagon and area of the circle are and respectively.

Answer:

Perimeter of the shaded portion = 4 ⨯ Length of the arc having central angle 90^∘
$$\begin{gathered} =4\times \frac{90°}{360°}\times 2\times \frac{22}{7}\times \frac{14}{2} \\ =4\times \frac{1}{4}\times 2\times \frac{22}{7}\times 7 \\ =44 \mathrm{cm} \end{gathered}$$
Area of shaded portion = Area of square ABCD − 4 ⨯ Area of the arc having central angle 90^∘
$$\begin{gathered} ={\left(14\right)}^2-4\times \frac{90°}{360°}\times \frac{22}{7}\times {\left(\frac{14}{2}\right)}^2 \\ =196-4\times \frac{1}{4}\times \frac{22}{7}\times {\left(7\right)}^2 \\ =196-154 \\ =42 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

It is given that AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of small circle.

It is given that,

So, radius r of small circle is

We know that the area A of circle of radius r is.

Substituting the value of r in above formula,

Now, let the area of large circle be.

Using the value radius OA,

Hence,

Answer:

Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ

$$\begin{gathered} =\frac{1}{2}\times 2\mathrm{\pi }{r}_1+\frac{1}{2}\times 2\mathrm{\pi }{r}_2+\frac{1}{2}\times 2\mathrm{\pi }{r}_3 \\ =\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{7}{2}\right)+\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{10}{2}\right)+\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{3}{2}\right) \\ =\frac{7}{2}\mathrm{\pi }+5\mathrm{\pi }+\frac{3}{2}\mathrm{\pi } \\ =10\mathrm{\pi } \\ =31.4 \mathrm{cm} \end{gathered}$$
=12×2πr1+12×2πr2+12×2πr3=12×2π(72)+12×2π(102)+12×2π(32)=72π+5π+32π=72π+32π+5π=5π+5π=10π=31.4 cm
Hence, the perimeter of shaded region is 31.4 cm.

Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of circle with radius AC − area of circle with radius BC

We have given radius of the outer circle that is 8 cm but we don’t know the radius of the inner circle.

We can calculate the radius of the inner circle as shown below,

Substituting we get,

Therefore, area of the shaded region is.

Answer:

(i) Let us find the perimeter of the shaded region.

Therefore, perimeter of the shaded region is .

Now we will find the area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of the semi-circle − area of the right angled triangle

First, we will find the length of AB as shown below,

Substituting we get,

Therefore, area of the shaded region is.

Answer:

We will find the area of the shaded region as shown below,

Area of the shaded region = area of quadrant + area of isosceles triangle ……..(1)

Substituting we get,

Therefore, area of the shaded region is.

Answer:

Answer:

We have given a trapezium. We are asked to find the area of the shaded region.

We can find the area of the remaining part that is area of the shaded region as shown below.

……..(1)

Now we find the value of CD.

………(Since, CE is radius of the sector, therefore, CE = 3.5)

Substituting the values of CD and in equation (1),

Therefore, area of the remaining part is.

Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of equilateral triangle − 3 area of circular arc

Substituting and we get,

Therefore, area of the shaded region is.

Answer:

It is given that, and

We know that the circumference C of semicircle of radius be r is

$C=\mathrm{\pi }r$

The inside perimeter of running track is the sum of twice the length of straight portion and circumferences of semicircles. So,
inside perimeter of running track = 400 m
$2l+2\mathrm{\pi }r\mathit{=}400 m$
$\Rightarrow 2\times 90+2\times \frac{22}{7}\times r=400 m$
$\Rightarrow r=\frac{220\times 7}{2\times 22}=35 m$
Thus, radius of inner semicircle is 35 m.

Now,
radius of outer semi circle r' = 35 + 14 = 49 m

Area of running track = $2\times \mathrm{Area} \mathrm{of} \mathrm{rectangle}+2\times \mathrm{Area} \mathrm{of} \mathrm{outer} \mathrm{semi} \mathrm{circle}-2\times \mathrm{Area} \mathrm{of} \mathrm{inner} \mathrm{semicircle}$
            $=2\times 90\times 14+2\times \frac{\mathrm{\pi }(49{)}^2}{2}-2\times \frac{\mathrm{\pi }(35{)}^2}{2}$
            $=2520+\mathrm{\pi }\times \left(49+35\right)\left(49-35\right)$
            $=2520+\frac{22}{7}\times 84\times 14$
            $=2520+3696=6216 m^2$
Hence, the area of running track = 6216 m²

Now, length L of outer running track is
L = 2 × l + 2$\mathrm{\pi }$r'
   $=2\times 90+2\mathrm{\pi }\times 49$
   $=180+2\times \frac{22}{7}\times 49$
   $=180+308=488 m$
   
Hence, the length L of outer running track is 488 m.
 

Answer:

We have a square ABCD.

We have,

(i)We have to find the perimeter of the triangle. We have a relation as,

So,

So perimeter of the circular region,

(ii)We have to find the perimeter of ABEF. Let O be the centre of the circular region.

Use Pythagoras theorem to get,

Similarly,

Now length of arc EF,

So, perimeter of ABFE,

Answer:

Area of shaded region = Area of square − Area of 4 semicircle having diameter 4 cm − Area of square having side 4 cm

$$\begin{gathered} ={\left(\mathrm{Side}\right)}^2-4\times \frac{1}{2}\mathrm{\pi }{r}^2-{\left(\mathrm{side}\right)}^2 \\ ={\left(14\right)}^2-2\times 3.14\times {\left(\frac{4}{2}\right)}^2-4^2 \\ =196-25.12-16 \\ =154.88 {\mathrm{cm}}^2 \end{gathered}$$
=24.5−14π(AB)2=24.5−14×227(3.5)2=24.5−9.625=14.875 cm2
Hence, the area of shaded region is 154.88 cm²

Answer:

We have to find the area of the shaded portion. We havewhich is an equilateral triangle and.

We have O as the incentre and OP, OQ and OR are equal.

So,

Thus,

So area of the shaded region,

Answer:

We have to find the area of the shaded portion. We havewhich is an equilateral triangle and. Let r be the radius of the circle.

We have O as the circumcentre.

So,

Thus,

So area of the shaded region,

Answer:

In equilateral traingle all the angles are of  60°
∴ ∠BAC =  60°
 Area of the shaded region = (Area of triangle  ABC − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)

$$\begin{gathered} =\frac{\sqrt{3}}{4}{\left(\mathrm{AB}\right)}^2-\frac{60°}{360°}\mathrm{\pi }{\left(7\right)}^2+\frac{300°}{360°}\mathrm{\pi }{\left(7\right)}^2 \\ =\frac{\sqrt{3}}{4}{\left(14\right)}^2-\frac{1}{6}\times \frac{22}{7}{\left(7\right)}^2+\frac{5}{6}\times \frac{22}{7}{\left(7\right)}^2 \\ =84.77-25.67+128.35 \\ =187.45 {\mathrm{cm}}^2 \end{gathered}$$
=3√4(AB)2−60°360°π(6)2+300°360°π(6)2=1.734(12)2−16×3.14(6)2+56×3.14(6)2=62.28−18.84+94.2=137.64 cm2
Hence, the area of shaded region is 187.45 cm²

Answer:

We have a square ABCD having. From the given diagram we can observe that,

Radius of incircle

Radius of circumcircle

(i) We have to find the ratio of the circumferences of the two circles. So the required ratio is,

(ii) We have to find the ratio of the areas of the two circles. So the required ratio is,

Answer:

(i) Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of the semi-circle with diameter of 9 cm − area of 2 semi-circles with radius 3cm − area of the circle with centre D + area of semi-circle with radius 3 cm

Substituting we get,

Therefore, area of the shaded region is.

Now we will find the cost of painting the shaded region at the rate of 25 paise per cm².

paise

Therefore, cost of painting the shaded region to the nearest rupee is.

Answer:

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AC + area of right angled triangle ABC − area of sector

First we will find the hypotenuse of right angled triangle ABC.

Substituting we get,

Therefore, area of shaded region is.

Answer:

(i) We will first find the length of the boundary.

Length of the boundary perimeter of semi-circle with diameter AB + boundary of semi-circle with diameter 7 cm

Therefore, length of the boundary is.

Now we will find the area of the shaded region as shown below,

Area of the shaded region = Area of the semi-circle with AB as a diameter − area of the semi-circle with radius AE − area of the semi-circle with radius BC + area of the semi-circle with diameter 7 cm.

Therefore, area of the shaded region is.

Answer:

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AB − area of two semicircles with diameters AM and MB - area of circle ……..(1)

Let us calculate the area of the semi-circle with AB as a diameter.

Now we will find the area of the semi-circle with AM as a diameter.

Area of the semi-circle with MB as a diameter is same as the area of the semi-circle with diameter with AM as a diameter.

Now we will find the area of the circle with centre C.

We know that radius of the circle is one sixth of AB.

Now we will substitute all these values in equation (1).

Therefore, area of shaded region is.

Answer:

We have given three semi-circles and one right angled triangle.

     ……..(1)

Let us calculate the area of the semi-circle with AB as a diameter.

Now we will find the area of the semi-circle with AC as a diameter.

Now we will find the length of BC.

In right angled triangle ABC, we will use Pythagoras theorem,

Now we will calculate the area of the right angled triangle ABC.

Now we will find the area of the semi-circle with BC as a diameter.

Now we will substitute all these values in equation (1).

Therefore, area of shaded region is.

Answer:

We have a cross section of a railway tunnel. is a right angled isosceles triangle, right angled at O. let OM be perpendicular to AB.

(i) We have to find the height of the tunnel. We have,

Use Pythagoras theorem into get,

Let the height of the tunnel be h. So,

Thus,

Therefore,

$h=\left(2+\sqrt{2}\right) \mathrm{m}$

(ii)

Perimeter of cross-section is,

$=\left(3\mathrm{\pi }+2\sqrt{2}\right) m$

(iii)

$\left(3\mathrm{\pi }+2\right) m^2$

Answer:

Answer:

The area grazed by the cow, buffalo and the horse are in the form of sectors of the circle with radius 7 m. 
Let the angle formed in the three sectors be ${\theta }_1,{\theta }_2,{\theta }_3$
Now the area of these sectors will be 
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{with} \mathrm{angle} {\theta }_1=\frac{\mathrm{\pi } {\mathrm{\theta }}_1{\left(7\right)}^2}{360} \\ \mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{with} \mathrm{angle} {\theta }_2=\frac{\mathrm{\pi } {\mathrm{\theta }}_2{\left(7\right)}^2}{360} \\ \mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{with} \mathrm{angle} {\theta }_3=\frac{\mathrm{\pi } {\mathrm{\theta }}_3{\left(7\right)}^2}{360} \end{gathered}$$
Area of the triangle ABC wil be
$$\begin{gathered} s=\frac{a+b+c}{2}=\frac{15+16+17}{2}=\frac{48}{2}=24 \\ Area=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{24\left(24-15\right)\left(24-16\right)\left(24-17\right)} \\ =24\sqrt{21} {\mathrm{m}}^2 \end{gathered}$$
Area of the field not grazed by the animals = Area of the triangle ABC − Area of the three sectors
$$\begin{gathered} =24\sqrt{21}-\left[\left(\frac{{\mathrm{\pi \theta }}_1{\left(7\right)}^2}{360}\right)+\left(\frac{{\mathrm{\pi \theta }}_2{\left(7\right)}^2}{360}\right)+\left(\frac{{\mathrm{\pi \theta }}_3{\left(7\right)}^2}{360}\right)\right] \\ =24\sqrt{21}-\left(\frac{\pi {\left(7\right)}^2}{360}\right)\left({\theta }_1+{\theta }_2+{\theta }_3\right) \\ =24\sqrt{21}-\left(\frac{\pi {\left(7\right)}^2}{360}\right)\left(180\right) \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ =\left(24\sqrt{21}-77\right) {\mathrm{m}}^2 \end{gathered}$$

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:

Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylinderical tank, V_c = $\mathrm{\pi }{r}^{2}H$ = $\mathrm{\pi }\times {\left(1\right)}^2\times 5=5\mathrm{\pi } \mathrm{m}$

Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = h​
Volume of water in the park = lbh = $25\times 20\times h$

Now water from the tank is used to irrigate the park. So, 
Volume of cylinderical tank = Volume of water in the park

$$\begin{gathered} \Rightarrow 5\mathrm{\pi }=25\times 20\times h \\ \Rightarrow \frac{5\mathrm{\pi }}{25\times 20}=h \\ \Rightarrow h=\frac{\mathrm{\pi }}{100} \mathrm{m} \\ \Rightarrow h=0.0314 \mathrm{m} \end{gathered}$$
 

Answer:

We are given that diameter and side of an equilateral triangle are equal.

Let d and a are the diameter and side of circle and equilateral triangle respectively.

Therefore d = a

We know that area of the circle

Area of the equilateral triangle

Now we will find the ratio of the areas of circle and equilateral triangle.

We know that radius is half of the diameter of the circle.

Now we will substitute in the above equation,

Therefore, ratio of the areas of circle and equilateral triangle is.

Answer:

We are given ratio of circumferences of two circles. If and are circumferences of two circles such that

…… (1)

Simplifying equation (1) we get,

Let and are the areas of the respective circles and we are asked to find their ratio.

…… (2)

We know that substituting this value in equation (2) we get,

Therefore, ratio of their areas is.

Answer:

We know that area of the sector of the circle of radius r

Length of the arc

But we have given that length of the arc

…… (1)

Area of the sector

Now we will adjust 2 in the following way,

Area of the sector

Area of the sector

From equation (1) we will substitute

Area of the sector

Area of the sector

Therefore, area of the sector =.

Answer:

We have

We have to find the length of the arc.

Substituting the values we get,

……….(1)

Now we will simplify the equation (1) as below,

Therefore, length of the arc is

Answer:

We have

We will find the angle subtended at the centre of a circle.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, angle subtended at the centre of the circle is.

Answer:

We have

Now we will find the area of the sector.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, area of the sector is.

Answer:

We have given the radius of the circle and angle subtended at the centre of the circle.

Now we will find the area of the sector.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, area of the sector is.

Answer:

We have the following situation

Let BD be the diameter and diagonal of the circle and the square respectively.

We know that area of the circle

Area of the square

As we know that diagonal of the square is the diameter of the square.

…… (1)

Substituting in equation (1) we get,

Now we will find the ratio of the areas of circle and square.

Now we will simplify the above equation as below,

Therefore, ratio of areas of circle and square is.

Answer:

Let r be the radius of the circle and angle θ subtended at the centre of the circle.

Area of the sector of the circle

Therefore, area of the sector is.

Answer:

In this figure, centre of the circle is O, radius OA = r and

We are going to find the area of the segment AXB.

…… (1)

We know that

We also know that

Substituting these values in equation (1) we get,

Therefore, area of the segment is.

Answer:

Given figure is a quadrant of a circle. We have given radius of sector that is 10.5 cm. Arc AB subtended an angle of 60° at the centre of the circle.

Perimeter of the sector

Substituting the values we get,

Perimeter of the sector …… (1)

Now we will simplify equation (1) as shown below,

Now we will substitute.

Therefore, perimeter of the given sector is.

Answer:

Let AB be the diameter of the semi-circular protractor.

We know that perimeter of the semicircle ……(1)

We have given the diameter of the protractor.

Therefore, radius of the protractor

So, radius of the protractor

Substituting the value of r in equation (1) we get,

Substituting we get,

Therefore, perimeter of the semi-circular protractor is.

Answer:

Length of the wire = 22 cm
Angle subtended at the centre = $\theta$ = 60º
Length of the arc = $\frac{\theta }{360}\times 2\mathrm{\pi }r$
$$\begin{gathered} \Rightarrow \frac{60}{360}\times 2\times \frac{22}{7}\times r=22 \\ \Rightarrow \frac{22}{7}r=22\times 3 \\ \Rightarrow r=21 \mathrm{cm} \end{gathered}$$

Answer:

Radius of the semi circle r units

When a largest triangle inscribed in a semicircle, then base = r + r = 2r units

Thus, Area of triangle is given by
$$\begin{gathered} \frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times \left(2r\right)\times r \\ =r^2 \mathrm{square} \mathrm{units} \end{gathered}$$

Answer:

We have given an angle subtended by an arc at the centre of the circle and radius of the circle.

Now we will find the area of the minor sector.

Area of the minor sector =

Substituting the values we get,

Area of the minor sector = …… (1)

Now we will simplify the equation (1) as below,

Area of the minor sector =

Area of the minor sector =

Area of the minor sector = π × 7 × 7

Area of the minor sector = 49π

Therefore, area of the minor sector is.

Answer:

Area of a sector of a circle = $\frac{\theta }{360}\times {\mathrm{\pi r}}^2$
$$\begin{gathered} =\frac{120}{360}\times \pi {\left(21\right)}^2 \\ =462 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

The diameter of the circle = Diagonal of the square = p
We know that diagonal of a square = $\sqrt{2}a$
$$\begin{gathered} \Rightarrow \sqrt{2}a=p \\ \Rightarrow a=\frac{p}{\sqrt{2}} \end{gathered}$$
Thus, the area of the square = $a^2={\left(\frac{p}{\sqrt{2}}\right)}^2=\frac{p^2}{2} {\mathrm{cm}}^2$

Answer:

A circle has two segments, a major segment and a minor segment.
The area of the minor segment is less than the area of the corresponding sector. 
But same is not true for the major segment.

Answer:

Given that area of the circle = circumference 
$$\begin{gathered} {\mathrm{\pi r}}^2=2\mathrm{\pi r} \\ \Rightarrow {\mathrm{\pi r}}^2-2\mathrm{\pi r}=0 \\ \Rightarrow \mathrm{\pi r}\left(\mathrm{r}-2\right)=0 \\ \Rightarrow \mathrm{r}=0 \mathrm{or} \mathrm{r}=2 \end{gathered}$$
Since radius cannot be equal to 0 so, r = 2

Answer:

In one revolution, the wheel covers a distance of $2\mathrm{\pi r}$
So, to cover a distance of s meters, the wheel has to make $\frac{s}{2\mathrm{\pi r}}$ revolutions.

Answer:

Let the side of the square inscribed in a square be a units.
Diameter of the circle outside the square = Diagonal of the square = $\sqrt{2}a$
Radius = $\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}$                              
So, the area of the circle circumscribing the square = $\mathrm{\pi }{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}^2$                         .....(i)
Now, the radius of the circle inscribed in a square = $\frac{a}{2}$
Hence, area of the circle inscribed in a square = $\mathrm{\pi }{\left(\frac{\mathrm{a}}{2}\right)}^2$                                     .....(ii)
From (i) and (ii)
$$\begin{gathered} \frac{\mathrm{Area} \mathrm{of} \mathrm{circle} \mathrm{circumscribing} \mathrm{a} \mathrm{square}}{\mathrm{Area} \mathrm{of} \mathrm{circle} \mathrm{inscribed} \mathrm{in} \mathrm{a} \mathrm{square}}=\frac{\mathrm{\pi }{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}{\mathrm{\pi }{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}^2} \\ =\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{4}}} \\ =\frac{2}{1} \end{gathered}$$
Hence, the required ratio is 2 : 1.
 

Answer:

Let the radius of the circle be r cm.

Area of the circle = 154 cm²

$$\begin{gathered} \Rightarrow \mathrm{\pi }{r}^2=154 {\mathrm{cm}}^2 \\ \Rightarrow \frac{22}{7}\times r^2=154 \\ \Rightarrow r=\sqrt{\frac{154 \times 7}{22}}=7 \mathrm{cm} \end{gathered}$$

∴ Perimeter of the circle

= Circumference of the circle

$$\begin{gathered} =2\mathrm{\pi }r \\ =2\times \frac{22}{7}\times 7 \\ =44 \mathrm{cm} \end{gathered}$$

If the area of a circle is 154 cm², then its perimeter is ___44 cm___.

Answer:

Ans