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

Let measure of minor arc = x⁰
∴ Measure of major arc = 5x⁰
⇒ x^o + 5x⁰ = 360⁰ ⇒ 6x⁰ = 360⁰ ⇒ x = 60⁰
∴ Measure of major arc = 5 x 60⁰ = 300⁰
Area of sector = \({300^0\over 360^0}\times{22\over7}\times(10.5)2\)
= 288.75cm²

Length of the rectangle = 20 cm
Breadth of the rectangle = 14 cm
d of the semicircle = 14 cm
r of the semicircle = \(14\over2\) cm = 7 cm
Area of the remaining portion = area of the rectangle ABCD - area of the semicircle 
\(=AB\times BC-{1\over 2}\times\pi\times\left(BC\over2\right)^2\)
\(=20\times14-{1\over 2}\times{22\over 7}\times\left(14\over 2\right)^2\)
\(=(280-{11\over7}\times7\times7)cm^2\)
=(280-77)cm²
=203cm²

l=length of the arc=\({\theta\pi r\over 180^0}\)
\(={45^0\over 18060}\times{22\over 7}\times7={11\over 2}cm\)

Permiter of the sector
\(=(r+r+l)=\left(7+7+{11\over2}\right)cm\)
\(\left(14+{11\over2}\right)cm=\left(28+11\over 2\right)cm\)
\(={39\over 2}cm=19.5cm\)

Given, length of wiper blade = 25 cm = r (say)
and angle made by this blade, \(\theta\) = 115°
\(\therefore\) Area cleaned by one blade = Area of sector formed by blade
\(\begin{aligned} & =\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{115^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(25)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{23 \times 22}{72 \times 7} \times 625=\frac{23 \times 11 \times 625}{36 \times 7}=\frac{158125}{252} \mathrm{~cm}^2 \end{aligned}\)
\(\therefore\) Total area cleaned by both blades
= 2 \(\times\) Area cleaned by one blade
\(=\frac{2 \times 158125}{252}=\frac{158125}{126} \mathrm{~cm}^2\)

We know that area of sector of a circle =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)
Given, radius of circle, r = 6 cm
and angle of sector, \(\theta\)= 60°
\(\therefore\) Area of sector of a circle \(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^2=\frac{132}{7} \mathrm{~cm}^2\)

Given sector is OAPB 
Area of the sector \(=\frac{\theta}{360} \times \pi r^{2}\)
\(=\frac{30}{360} \times 3.14 \times 4 \times 4 \mathrm{~cm}^{2}\)
\(=\frac{12.56}{3} \mathrm{~cm}^{2}=4.19 \mathrm{~cm}^{2}(\text { approx. })\)
Area of the corresponding major sector
\(\begin{aligned} &=\pi r^{2}-\text { area of sector } \mathrm{OAPB}\\ &=(3.14 \times 16-4.19) \mathrm{cm}^{2}\\ &=46.05 \mathrm{~cm}^{2}=46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{aligned}\)
Alternatively, area of the major sector  \(=\frac{(360-\theta)}{360} \times \pi r^{2}\)
\(\begin{array}{l} =\left(\frac{360-30}{360}\right) \times 3.14 \times 16 \mathrm{~cm}^{2} \\ =\frac{330}{360} \times 3.14 \times 16 \mathrm{~cm}^{2}=46.05 \mathrm{~cm}^{2} \\ =46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{array}\)

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA
\(\begin{array}{l} =106+\frac{1}{2} \times 2 \pi r+106+\frac{1}{2} \times 2 \pi r \\ =212+\frac{1}{2} \times 2 \times \frac{22}{7} \times 30+\frac{1}{2} \times 2 \times \frac{22}{7} \times 30 \\ =212+2 \times \frac{22}{7} \times 30 \\ =212+\frac{1320}{7} \\ =\frac{1484+1320}{7}=\frac{2804}{7} m \end{array}\)
Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI − Area of semi-circle BEC) + (Area of semi-circle GLJ − Area of semi-circle AFD)
\(\begin{array}{l} =106 \times 80-106 \times 60+\frac{1}{2} \times \frac{22}{7} \times(40)^{2}-\frac{1}{2} \times \frac{22}{7} \times(30)^{2}+\frac{1}{2} \times \frac{22}{7} \times(40)^{2}-\frac{1}{2} \times \frac{22}{7} \times(30)^{2} \\ =106(80-60)+\frac{22}{7} \times(40)^{2}-\frac{22}{7} \times(30)^{2} \\ =106(20)+\frac{22}{7}\left[(40)^{2}-(30)^{2}\right] \\ =2120+\frac{22}{7}(40-30)(40+30) \\ =2120+\left(\frac{22}{7}\right)(10)(70) \end{array}\)

=2120+2200
= 4320 m²
Therefore, the area of the track is 4320 m².

Let radius of semicircular region be r units.
Perimeter = 2r+πr
let side of square be x units units.
Perimeter = 4xunits
A.T.Q. \(4x=2r+\pi r^2\ \Rightarrow x={2r+\pi r\over 4}\)
Area of semicircle =\(={1\over 2}\pi r^2\)
 Area of square = x²
A.T.Q. \(x^2={1\over 2}\pi r^2+4\)

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ΔOVS,
OV/OS = cos 60º
OV/12 = 1/2
OV = 6 cm
\(S \frac{V}{S} O=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)
\(\frac{S V}{12}=\frac{\sqrt{3}}{2} \)
\(S V=6 \sqrt{3} \mathrm{~cm} \)
\(S T=2 S V=2 \times 6 \sqrt{3}=12 \sqrt{3} \mathrm{~cm}\)
Area of ΔOST = 1/2 x ST x OV
\(\frac{1}{2} \times 12 \sqrt{3} \times 6 \)
\(=36 \sqrt{3}=36 \times 1.73=62.28 \mathrm{~cm}^{2}\)
Area of sector OSUT \(=\frac{120^{\circ}}{360^{\circ}} \times \pi(12)^{2}\)
Area of segment SUT = Area of sector OSUT − Area of ΔOST
= 150.72 − 62.28
= 88.44 cm²

Radius of circle = 7cm
Side of equilateral triangle = 12cm

∴ Area of shaded portion = (Area of a circle — area of sector of central angle 60⁰) + area of equilateral triangle OAB
\(=\left[\left\{\pi(7)^2-{60^0\over 360^0}\times\pi(7)^2\right\}+{\sqrt3\over 4}(12)^2\right]cm^2\)
\(=\left[{5\over6}\times{22\over 7}\times49+{\sqrt3\over4}\times144\right]cm^2\)
=(128.33+62.35)cm²=190.68cm²

O is the centre of the circle
⇒ QR is the diameter of the circle
⇒ QPR is in the semicircle
⇒ ㄥQPR - 90⁰
Area of the ΔPQR=\(={1\over2}PR\times QP\)
\(={1\over 2}\times24\times7=84cm^2\)
In ΔPQR, QR²+PR²
=7²+24²=49+576=625cm²
\(⇒\ QR=\sqrt{625}cm=25=\)Diameter
⇒Radius \(={d\over 2}={25\over 2}cm\)
Area of the semicircle QPR=\({1\over 2}\pi r^2={1\over 2}\pi\times\left(25\over 2\right)^2={1\over 2}\times \pi\times {625\over4}cm^2={625\over 8}\pi cm^2\)
∴ Area of the shaded region \(=\left({625\over 8}\pi-84\right)cm^2\)
 

Area of square ABCD = 14 × 14 cm² = 196 cm²
Diameter of each circle \(=\frac{14}{2} \mathrm{~cm}=7 \mathrm{~cm}\)
So, radius of each circle \(=\frac{7}{2} \mathrm{~cm}\)
So, area of one circle \(=\pi r^{2}=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{2}\)
\(=\frac{154}{4} \mathrm{~cm}=\frac{77}{2} \mathrm{~cm}^{2}\)
Therefore, area of the four circles \(=4 \times \frac{77}{2} \mathrm{~cm}^{2}=154 \mathrm{~cm}^{2}\)
Hence, area of the shaded region = (196 – 154) cm² = 42 cm²

Given, diameter of circle, d = 35 mm
\(\therefore\) Circumference of circle = \(\pi\)d      [\(\because\) d = 2r]
\(=\frac{22}{7} \times 35=110 \mathrm{~mm}^2\)
Now, length of 5 diameters = 5 \(\times\) 35 = 175 mm
(i) Total length of the silver wire = \(\pi\)d + 5d
= 110 + 175 = 285 mm²
(ii) Here, we see that total circle is divided into 10 sectors.
\(\therefore\) Angle of each sector = \(\frac{360^{\circ}}{10}=36^{\circ}\)
Then, area of each sector ofthe brooch = \(=\frac{\theta}{360^{\circ}} \times \pi r^2\)
\(\begin{aligned} & =\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7}\left(\frac{35}{2}\right)^2 \quad\left[\because r=\frac{d}{2}=\frac{35}{2} \mathrm{~mm}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{10} \times \frac{22}{1} \times \frac{5}{2} \times \frac{35}{2}=\frac{11 \times 35}{2 \times 2}=\frac{385}{4} \mathrm{~mm}^2 \end{aligned}\)