Cost of fencing 1 m =Rs.12
Total cost of fencing the field = Rs.2640
∴ Total length to be fenced =\(={2640\over 12}m\)
 = 220 m = circumference of the circle
⇒ Let radius of the circle be r cm 
⇒ 2πr=220
⇒ \(2\times{22\over 7}r=220\)
⇒   r=7x5m
=35m

Area of the field=\(\pi r^2={22\over 7}\times35\times35m^2\)
Cost of ploughing 1m²=Rs.0.5
∴ Total cost of ploughing=Rs.0.50 x 22 x 5 x 35 =Rs.1925

Here, l = 8m, b = 4m and r = 6-4 = 2m
Area of the shaded field = area of rectangle ABCD + Area of semicircle DEF 
\(=l\times b+{1\over 2}(\pi r^2)=8\times4+{1\over 2}[\pi(2)^2]\)
=(32+2π)m²

In equilateral triangle, each angle=60⁰
and radius of each sector
\(r={side\over 2}={10\over 2}=5cm\)
∴ Area of shaded region
=3 x area of one sector
\(=3\times{\theta\over 360^2}\times\pi r^2\)
\(=3\times{60^0\over 360^0}\times\pi(5)^2\)
\(=3\times{25\pi\over 6}={25\pi\over 2}=39.25cm^2\)

Let r cm be the radius of protractor
Perimeter of semicircle = 2r + πr
∴ 2r + πr = 66
\(⇒\ r\left(2+{22\over 7}\right)=66\)
\(⇒\ r\times{36\over 7}=660\)
\(⇒\ r={66\times7\over 36}={77\over 6}\)
=12.83cm

17.6 sq.cm (a) Pollution is very harmful
(1) Affects human life and surroundings.
(2) Causes diseases.
(3) Causes ecological unbalance.
(i) Save fuel use resources judiciously.
(ii) Use biodegradable and ecofriendly material
(iii) More plantation of trees.

4 : 9

8 units

Perimeter of the circle = area of the circle.
\(\because\) 2\(\pi\)r = \(\pi\)r²
\(\therefore\) r = 2 units

Rs. 82500

\(\frac { 4 }{ 3 } \left( 4\pi -3\sqrt { 3 } \right) { cm }^{ 2 }\)

364 cm²

(i) Given, total cost of fencing = Rs. 5500
and rate of fencing per metre = Rs. 25
Length of fencing = Total cost / Rate of fencing per metre
\(=\frac{5500}{25}=220 m\)
(ii) Circumference of the circular field = Length of the fence
\(\Rightarrow \quad 2\pi r=220\quad \Rightarrow 2\times \frac { 22 }{ 7 } \times r=220\)
\(\Rightarrow \quad r=\frac { 220\times 7 }{ 2\times 22 } \Rightarrow \quad r\quad =\quad 35\quad m\)
i.e. Radius of the circular field = 35 m
\(\therefore \)  Area of the circular field = \(\pi { r }^{ 2 }\)
\(=\frac { 22 }{ 7 } \times { \left( 35 \right) }^{ 2 }=\frac { 22 }{ 7 } \times 35\times 35\)
= 22 x 5 x 35 = 3850 m²
Now, cost of ploughing at the rate Rs. 0.75 per m²
= Rs. 3850 x 0.75 = 2887.5
Hence, total cost of ploughing the field is 2887.5
(iii) Security and separate the boundary of a field.
Artist believes in self employment, which gives self respect and dignity of labour.

There are four equally semi-circles and JKLM formed a square
\(\therefore\)  FH=14-(3+3)=8 cm
Let the side of square JKLM be x can
Then, FH=\(\frac { x }{ 2 } +x+\frac { x }{ 2 } \)
\(\Rightarrow\) 8=2x \(\Rightarrow\)  x=4
So, the side of square should be 4 cm and radius of semi-circle of both ends are 2 cm each.
\(\therefore\) Area of square JKLm=(4)²=16 cm²
Area of semi-circle JHM=\(\frac { \pi { r }^{ 2 } }{ 2 } \)
\(=\frac { \pi \times (2)^{ 2 } }{ 2 } =2\pi \quad { cm }^{ 2 }\)
\(\therefore\) Area of four semi-circles=\(4\times 2\pi \)
Now, area of square ABCD=(14)²=196 cm²
\(\therefore\) Area of shaded region=Area of square ABCD-(Area of four semi-circles + Area of square JKLM)
\(=196-(8\pi +16)=196-16-8\pi \)
\(=(180-8\pi ){ cm }^{ 2 }\)
Hence, the required of the shaded region is \((180-8\pi ){ cm }^{ 2 }\)

\(\frac {22}{7}\) cm²

Area of quadrant ABED = \(1\over4\) x \(\pi\) x 8²
= \({1\over4}\times{22\over7}\times8\times8\)
= \(352\over7\) cm²
Area \(\triangle\)ABD = \(1\over2\) x AB x AD
= \(1\over2\) x 8 x 8 = 32 cm²
Area of shaded region
= 2[Area of quadrant ABED - Area of \(\triangle\)ABD]
= \(2\left[ \frac { 352 }{ 7 } -32 \right] =2\left[ \frac { 352-224 }{ 7 } \right] =\frac { 2\times 128 }{ 7 } =\frac { 256 }{ 7 } \)
= 36.57 cm²

Perimeter of square and circle formed by wire will be equal.154 cm²

81.75 cm²

Given, side of square= 14 cm
Also, APD and BPC are semi-circles, therefore their radius, 
r=14/2=7 cm
Now, Area of semi-circle APD=Area of semi-circle BPC
=\(\frac { { \pi r }^{ 2 } }{ 2 } =\frac { 22 }{ 7\times 2 } { \left( 7 \right) }^{ 2 }=77\quad { cm }^{ 2 }\)
and area of square ABCD=(side)²=(14)²=196 cm²
Hence, area of shaded region=Area of square-(Area of semi-circle APD+ Area of semi-circle BPC)
=196-(77+77)=42 cm²

616 cm²

Let the lengths of the corresponding arcs be l₁ and l₂.

Given, radius of sector PO₁QP=7 cm and radius of sector AO₂BA=21 cm
Central angle of the sector PO₁QP=120°
and central angle of the sector AO₂BA = 40°
ஃ  Area of the sector with central angle O₁
\(=\frac { \pi { r }_{ 1 }^{ 2 } }{ { 360 }^{ O } } \times { \theta }_{ 1 }=\frac { \pi \times { 7 }^{ 2 }\times { 120 }^{ O } }{ { 360 }^{ O } } \)
=22/7 x 7 x 7/360° x 120°=22 x 7/3=154/3 cm²
and area of the sector with central angle ^O₂
\(=\frac { \pi { r }^{ 2 } }{ { 360 }^{ O } } \times { \theta }=\frac { 22 }{ 7 } \times \frac { 21\times 21 }{ { 360 }^{ O } } \times { 40 }^{ O }\)
22x3/21/9=22x7=154 cm²
Now, length of the arc of the sector PO₁QP=\(\frac { \pi { r }\theta }{ { 180 }^{ O } } \)
22/7 x 7 x 120^o/180^o=2/3 x 7 x 22/7 = 44/3 cm
and length of the arc of the sector AO₂BA=\(\frac { \pi { r }\theta }{ { 180 }^{ O } } \)
22/7x21x 40°/ 180°=2/9x21x22/7 = 2/3x22=44/3 cm
Hence, we observe that length of the arc of two sectors of two different circles may be equal but their area need not be equal.

(i) Since OACB is a quadrant, it will subtend 90° angle at O.
Area of quadrant OACB = \(=\frac{90^{\circ}}{360^{\circ}} \times \pi r^{2}\)
\(\begin{array}{l} =\frac{1}{4} \times \frac{22}{7} \times(3.5)^{2}=\frac{1}{4} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^{2} \\ =\frac{11 \times 7 \times 7}{2 \times 7 \times 2 \times 2}=\frac{77}{8} \mathrm{~cm}^{2} \end{array}\)
(ii) Area of ΔOBD = 1/2 x OB x OD
\(\begin{array}{l} =\frac{1}{2} \times 3.5 \times 2 \\ =\frac{1}{2} \times \frac{7}{2} \times 2 \\ =\frac{7}{2} \mathrm{~cm}^{2} \end{array}\)
Area of the shaded region = Area of quadrant OACB − Area of ΔOBD
\(\begin{array}{l} =\frac{77}{8}-\frac{7}{2} \\ =\frac{77-28}{8} \\ =\frac{49}{8} \mathrm{~cm}^{2} \end{array}\)

Area of major sector = area of circle - area of sector
\(=\pi { r }^{ 2 }\left( 1-\frac { \theta }{ 360 } \right) \)
\(=\frac { 22 }{ 7 } \times 28\times 28\left( 1-\frac { 45 }{ 360 } \right) \)
\(=22\times 4\times 28\times \frac { 7 }{ 8 } \)
= 2156 cm²