Textbook - Areas Related to Circles Questions in English

(A) Length of the fence (in $m$) $= \frac{\text{Total cost}}{\text{Rate}} = \frac{5280}{24} = 220$.
So,the circumference of the field $= 220\, m$.
Let $r$ be the radius of the field. Then,$2\pi r = 220$.
$2 \times \frac{22}{7} \times r = 220$.
$r = \frac{220 \times 7}{2 \times 22} = 35\, m$.
Area of the field $= \pi r^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3850\, m^2$.
Cost of ploughing $1\, m^2$ of the field $= Rs. 0.50$.
Total cost of ploughing the field $= 3850 \times 0.50 = Rs. 1925$.

(B) Radius $(r_1)$ of the $1^{st}$ circle $= 19\, cm$.
Radius $(r_2)$ of the $2^{nd}$ circle $= 9\, cm$.
Let the radius of the $3^{rd}$ circle be $r$.
Circumference of the $1^{st}$ circle $= 2\pi r_1 = 2\pi(19) = 38\pi$.
Circumference of the $2^{nd}$ circle $= 2\pi r_2 = 2\pi(9) = 18\pi$.
Circumference of the $3^{rd}$ circle $= 2\pi r$.
According to the problem,the circumference of the $3^{rd}$ circle is equal to the sum of the circumferences of the first two circles:
$2\pi r = 38\pi + 18\pi$
$2\pi r = 56\pi$
Dividing both sides by $2\pi$:
$r = \frac{56\pi}{2\pi} = 28\, cm$.
Therefore,the radius of the required circle is $28\, cm$.

(C) Radius $(r_1)$ of the $1^{st}$ circle $= 8 \, cm$.
Radius $(r_2)$ of the $2^{nd}$ circle $= 6 \, cm$.
Let the radius of the $3^{rd}$ circle be $r$.
Area of the $1^{st}$ circle $= \pi r_1^2 = \pi(8)^2 = 64\pi \, cm^2$.
Area of the $2^{nd}$ circle $= \pi r_2^2 = \pi(6)^2 = 36\pi \, cm^2$.
According to the problem,the area of the $3^{rd}$ circle is equal to the sum of the areas of the two circles:
$\pi r^2 = \pi r_1^2 + \pi r_2^2$
$\pi r^2 = 64\pi + 36\pi$
$\pi r^2 = 100\pi$
$r^2 = 100$
$r = \sqrt{100} = 10 \, cm$.
Since the radius cannot be negative,the radius of the $3^{rd}$ circle is $10 \, cm$.

(N/A) Radius $(r_1)$ of the gold region (i.e.,$1^{st}$ circle) $= \frac{21}{2} = 10.5\, cm$.
Given that each circle is $10.5\, cm$ wider than the previous circle.
Therefore,radius $(r_2)$ of the $2^{nd}$ circle $= 10.5 + 10.5 = 21\, cm$.
Radius $(r_3)$ of the $3^{rd}$ circle $= 21 + 10.5 = 31.5\, cm$.
Radius $(r_4)$ of the $4^{th}$ circle $= 31.5 + 10.5 = 42\, cm$.
Radius $(r_5)$ of the $5^{th}$ circle $= 42 + 10.5 = 52.5\, cm$.
Area of gold region $= \pi r_1^2 = \frac{22}{7} \times (10.5)^2 = 346.5\, cm^2$.
Area of red region $= \pi r_2^2 - \pi r_1^2 = \frac{22}{7} \times (21^2 - 10.5^2) = 1039.5\, cm^2$.
Area of blue region $= \pi r_3^2 - \pi r_2^2 = \frac{22}{7} \times (31.5^2 - 21^2) = 1732.5\, cm^2$.
Area of black region $= \pi r_4^2 - \pi r_3^2 = \frac{22}{7} \times (42^2 - 31.5^2) = 2425.5\, cm^2$.
Area of white region $= \pi r_5^2 - \pi r_4^2 = \frac{22}{7} \times (52.5^2 - 42^2) = 3118.5\, cm^2$.
Thus,the areas of the $Gold, Red, Blue, Black,$ and $White$ regions are $346.5\, cm^2, 1039.5\, cm^2, 1732.5\, cm^2, 2425.5\, cm^2,$ and $3118.5\, cm^2$ respectively.

(A) Diameter of the wheel $= 80\, cm$.
Radius $(r)$ of the wheel $= 40\, cm$.
Circumference of the wheel $= 2\pi r = 2 \times \frac{22}{7} \times 40 = \frac{1760}{7}\, cm$.
Speed of the car $= 66\, km/h = \frac{66 \times 100000\, cm}{60\, min} = 110000\, cm/min$.
Distance travelled in $10\, minutes = 110000 \times 10 = 1100000\, cm$.
Let the number of revolutions be $n$.
$n \times \text{Circumference} = \text{Total distance}$.
$n \times \frac{1760}{7} = 1100000$.
$n = \frac{1100000 \times 7}{1760} = \frac{110000 \times 7}{176} = 4375$.
Thus,each wheel makes $4375$ complete revolutions.

(B) Let the radius of the circle be $r$ $units$.
The circumference (perimeter) of the circle is given by the formula $C = 2\pi r$.
The area of the circle is given by the formula $A = \pi r^2$.
According to the problem,the numerical values of the perimeter and the area are equal,so we set $C = A$.
$2\pi r = \pi r^2$
Dividing both sides by $\pi r$ (assuming $r \neq 0$):
$2 = r$
Therefore,the radius of the circle is $2$ $units$.

(N/A) Given radius $r = 4\, cm$ and central angle $\theta = 30^{\circ}$.
Area of the sector $= \frac{\theta}{360^{\circ}} \times \pi r^{2}$
$= \frac{30}{360} \times 3.14 \times 4 \times 4\, cm^{2}$
$= \frac{1}{12} \times 3.14 \times 16\, cm^{2} = \frac{50.24}{12}\, cm^{2} \approx 4.19\, cm^{2}$.
Area of the corresponding major sector $= \pi r^{2} - \text{Area of minor sector}$
$= (3.14 \times 4^{2}) - 4.19\, cm^{2}$
$= (3.14 \times 16) - 4.19\, cm^{2}$
$= 50.24 - 4.19 = 46.05\, cm^{2} \approx 46.1\, cm^{2}$.

(N/A) Area of the segment $AYB = \text{Area of sector } OAYB - \text{Area of } \triangle OAB$ ......$(1)$
Now,area of the sector $OAYB = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \, cm^2 = 462 \, cm^2$ ......$(2)$
For finding the area of $\triangle OAB$,draw $OM \perp AB$ as shown in the figure.
Note that $OA = OB$. Therefore,by $RHS$ congruence,$\triangle AMO \cong \triangle BMO$.
So,$M$ is the mid-point of $AB$ and $\angle AOM = \angle BOM = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.
Let $OM = x \, cm$.
So,from $\triangle OMA$,$\frac{OM}{OA} = \cos 60^{\circ}$.
$\frac{x}{21} = \frac{1}{2} \implies x = \frac{21}{2}$.
So,$OM = \frac{21}{2} \, cm$.
Also,$\frac{AM}{OA} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
So,$AM = \frac{21\sqrt{3}}{2} \, cm$.
Therefore,$AB = 2 \times AM = 2 \times \frac{21\sqrt{3}}{2} = 21\sqrt{3} \, cm$.
So,area of $\triangle OAB = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441\sqrt{3}}{4} \, cm^2$ ......$(3)$
Therefore,area of the segment $AYB = \left( 462 - \frac{441\sqrt{3}}{4} \right) cm^2$ [From $(1), (2)$ and $(3)$].
$= \frac{21}{4} (88 - 21\sqrt{3}) \, cm^2$.

(A) Let $OACB$ be a sector of the circle making a $60^{\circ}$ angle at the centre $O$ of the circle.
The formula for the area of a sector with angle $\theta$ is given by:
$\text{Area of sector} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$
Given:
Radius $r = 6 \, cm$
Angle $\theta = 60^{\circ}$
$\pi = \frac{22}{7}$
Substituting the values into the formula:
$\text{Area} = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (6)^{2}$
$\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 36$
$\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 6 \times 6$
$\text{Area} = \frac{22 \times 6}{7} = \frac{132}{7} \, cm^{2}$
Therefore,the area of the sector is $\frac{132}{7} \, cm^{2}$.

(N/A) Let the radius of the circle be $r$.
Circumference $= 2 \pi r = 22 \, cm$.
$r = \frac{22}{2 \pi} = \frac{11}{\pi} = \frac{11}{22/7} = \frac{11 \times 7}{22} = 3.5 \, cm$.
$A$ quadrant of a circle subtends an angle of $90^{\circ}$ at the centre.
Area of the quadrant $= \frac{90^{\circ}}{360^{\circ}} \times \pi r^{2} = \frac{1}{4} \times \frac{22}{7} \times (3.5)^{2}$.
Area $= \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = \frac{1}{4} \times 22 \times 0.5 \times 3.5$.
Area $= \frac{11 \times 3.5}{4} = \frac{38.5}{4} = 9.625 \, cm^{2}$.

(N/A) We know that in $1$ hour (i.e.,$60$ minutes),the minute hand rotates $360^{\circ}$.
In $5$ minutes,the minute hand will rotate $= \frac{360^{\circ}}{60} \times 5 = 30^{\circ}$.
Therefore,the area swept by the minute hand in $5$ minutes will be the area of a sector of $30^{\circ}$ in a circle of radius $14 \, cm$.
Area of sector of angle $\theta = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Area $= \frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14$.
Area $= \frac{1}{12} \times 22 \times 2 \times 14$.
Area $= \frac{11 \times 14}{3} = \frac{154}{3} \, cm^{2}$.
Therefore,the area swept by the minute hand in $5$ minutes is $\frac{154}{3} \, cm^{2}$.

(N/A) Let $AB$ be the chord of the circle subtending a $90^{\circ}$ angle at the centre $O$ of the circle.
Area of the major sector $OADB = \left(\frac{360^{\circ}-90^{\circ}}{360^{\circ}}\right) \times \pi r^{2} = \left(\frac{270^{\circ}}{360^{\circ}}\right) \times 3.14 \times 10^{2}$
$= \frac{3}{4} \times 3.14 \times 100 = 235.5\, cm^{2}$
Area of the minor sector $OACB = \frac{90^{\circ}}{360^{\circ}} \times \pi r^{2} = \frac{1}{4} \times 3.14 \times 100 = 78.5\, cm^{2}$
Area of $\triangle OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 10 \times 10 = 50\, cm^{2}$
Area of the minor segment $ACB = \text{Area of minor sector } OACB - \text{Area of } \triangle OAB = 78.5 - 50 = 28.5\, cm^{2}$

(N/A) Radius $(r)$ of the circle $= 21\, cm$.
Angle subtended by the given arc at the centre $(\theta) = 60^{\circ}$.
$(i)$ Length of the arc $= \frac{\theta}{360^{\circ}} \times 2 \pi r$
$= \frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21$
$= \frac{1}{6} \times 2 \times 22 \times 3 = 22\, cm$.
$(ii)$ Area of the sector $= \frac{\theta}{360^{\circ}} \times \pi r^{2}$
$= \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21$
$= \frac{1}{6} \times 22 \times 3 \times 21 = 231\, cm^{2}$.
$(iii)$ In $\Delta OAB$,$OA = OB = 21\, cm$ and $\angle AOB = 60^{\circ}$.
Since the two sides are equal,$\angle OAB = \angle OBA$.
In $\Delta OAB$,$\angle OAB + \angle OBA + \angle AOB = 180^{\circ}$.
$2 \angle OAB + 60^{\circ} = 180^{\circ} \implies 2 \angle OAB = 120^{\circ} \implies \angle OAB = 60^{\circ}$.
Thus,$\Delta OAB$ is an equilateral triangle.
Area of $\Delta OAB = \frac{\sqrt{3}}{4} \times (side)^{2} = \frac{\sqrt{3}}{4} \times (21)^{2} = \frac{441\sqrt{3}}{4}\, cm^{2}$.
Area of the segment = Area of sector $OACB - \text{Area of } \Delta OAB$
$= (231 - \frac{441\sqrt{3}}{4})\, cm^{2}$.

(N/A) Radius $(r)$ of the circle $= 15 \, cm$.
Area of sector $OPRQ = \frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$
$= \frac{1}{6} \times 3.14 \times (15)^{2}$
$= \frac{1}{6} \times 3.14 \times 225 = 117.75 \, cm^{2}$.
In $\triangle OPQ$,since $OP = OQ = 15 \, cm$,the angles opposite to these sides are equal,i.e.,$\angle OPQ = \angle OQP$.
Since $\angle POQ = 60^{\circ}$,the sum of the other two angles is $180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,$\angle OPQ = \angle OQP = 60^{\circ}$.
Since all angles are $60^{\circ}$,$\triangle OPQ$ is an equilateral triangle.
Area of $\triangle OPQ = \frac{\sqrt{3}}{4} \times (\text{side})^{2}$
$= \frac{1.73}{4} \times (15)^{2} = \frac{1.73}{4} \times 225 = 97.3125 \, cm^{2}$.
Area of minor segment $PRQ = \text{Area of sector } OPRQ - \text{Area of } \triangle OPQ$
$= 117.75 - 97.3125 = 20.4375 \, cm^{2}$.
Area of major segment $PSQ = \text{Area of circle} - \text{Area of minor segment } PRQ$
$= \pi r^{2} - 20.4375$
$= 3.14 \times 225 - 20.4375 = 706.5 - 20.4375 = 686.0625 \, cm^{2}$.

(C) Let us draw a perpendicular $OV$ on chord $ST$. It will bisect the chord $ST$.
$SV = VT$
In $\triangle OVS$,
$\frac{OV}{OS} = \cos 60^{\circ}$
$\frac{OV}{12} = \frac{1}{2}$
$OV = 6 \, cm$
$\frac{SV}{SO} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$
$\frac{SV}{12} = \frac{\sqrt{3}}{2}$
$SV = 6\sqrt{3} \, cm$
$ST = 2SV = 2 \times 6\sqrt{3} = 12\sqrt{3} \, cm$
Area of $\triangle OST = \frac{1}{2} \times ST \times OV$
$= \frac{1}{2} \times 12\sqrt{3} \times 6$
$= 36\sqrt{3} = 36 \times 1.73 = 62.28 \, cm^2$
Area of sector $OSUT = \frac{120^{\circ}}{360^{\circ}} \times \pi(12)^2$
$= \frac{1}{3} \times 3.14 \times 144 = 150.72 \, cm^2$
Area of segment $SUT = \text{Area of sector } OSUT - \text{Area of } \triangle OST$
$= 150.72 - 62.28$
$= 88.44 \, cm^2$

(N/A) From the figure,it can be observed that the horse can graze a sector of $90^{\circ}$ in a circle of $5 \, m$ radius.
Area that can be grazed by the horse $=$ Area of sector with radius $r = 5 \, m$ and angle $\theta = 90^{\circ}$.
Area $= \frac{\theta}{360^{\circ}} \times \pi r^{2}$
$= \frac{90^{\circ}}{360^{\circ}} \times 3.14 \times (5)^{2}$
$= \frac{1}{4} \times 3.14 \times 25$
$= 19.625 \, m^{2}$
Area that can be grazed by the horse when the length of the rope is $10 \, m$ long $(r = 10 \, m)$:
Area $= \frac{90^{\circ}}{360^{\circ}} \times \pi \times (10)^{2}$
$= \frac{1}{4} \times 3.14 \times 100$
$= 78.5 \, m^{2}$
Increase in grazing area $= (78.5 - 19.625) \, m^{2}$
$= 58.875 \, m^{2}$

(N/A) The total length of wire required is the sum of the lengths of $5$ diameters and the circumference of the brooch.
Radius of the circle $r = \frac{35}{2} \, mm$.
Circumference of the brooch $= 2 \pi r = 2 \times \frac{22}{7} \times \frac{35}{2} = 110 \, mm$.
Length of $5$ diameters $= 5 \times 35 = 175 \, mm$.
Total length of wire required $= 110 + 175 = 285 \, mm$.
It can be observed from the figure that each of the $10$ sectors of the circle subtends an angle of $\frac{360^{\circ}}{10} = 36^{\circ}$ at the centre of the circle.
Therefore,the area of each sector $= \frac{36^{\circ}}{360^{\circ}} \times \pi r^{2} = \frac{1}{10} \times \frac{22}{7} \times \left(\frac{35}{2}\right) \times \left(\frac{35}{2}\right) = \frac{385}{4} \, mm^{2} = 96.25 \, mm^{2}$.

(N/A) An umbrella has $8$ ribs,which divide the circular area into $8$ equal sectors.
The angle subtended by each sector at the center is $\theta = \frac{360^{\circ}}{8} = 45^{\circ}$.
The radius of the circle is $r = 45 \, cm$.
The area of a sector is given by the formula: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Substituting the values:
$\text{Area} = \frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (45)^{2}$
$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times 2025$
$\text{Area} = \frac{11}{4 \times 7} \times 2025 = \frac{11 \times 2025}{28}$
$\text{Area} = \frac{22275}{28} \, cm^{2} \approx 795.54 \, cm^{2}$.

(N/A) It can be observed that each blade of the wiper sweeps an area of a sector of $115^{\circ}$ in a circle of radius $r = 25 \, cm$.
The area of a sector of a circle with angle $\theta$ and radius $r$ is given by the formula: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Area of one sector $= \frac{115^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (25)^{2}$
$= \frac{23}{72} \times \frac{22}{7} \times 625$
$= \frac{23 \times 11 \times 625}{36 \times 7} = \frac{158125}{252} \, cm^{2}$.
Since there are two wipers,the total area cleaned by both blades is:
Total Area $= 2 \times \left( \frac{158125}{252} \right)$
$= \frac{158125}{126} \, cm^{2} \approx 1254.96 \, cm^{2}$.

(D) The lighthouse spreads light over a sector of a circle with radius $r = 16.5 \, km$ and central angle $\theta = 80^{\circ}$.
The area of the sector is given by the formula:
$\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^2$
Substituting the given values:
$\text{Area} = \frac{80^{\circ}}{360^{\circ}} \times 3.14 \times (16.5)^2$
$= \frac{2}{9} \times 3.14 \times 272.25$
$= \frac{2}{9} \times 854.865$
$= 189.97 \, km^2$
Thus,the area of the sea over which the ships are warned is $189.97 \, km^2$.

(A) The designs are segments of the circle.
Consider segment $APB$. Chord $AB$ is a side of the regular hexagon inscribed in the circle.
Each chord subtends an angle of $\frac{360^{\circ}}{6} = 60^{\circ}$ at the center $O$.
In $\triangle OAB$,$OA = OB = 28\, cm$ (radii).
Since $\angle AOB = 60^{\circ}$ and $OA = OB$,$\triangle OAB$ is an equilateral triangle.
Area of $\triangle OAB = \frac{\sqrt{3}}{4} \times (side)^2 = \frac{1.7}{4} \times 28 \times 28 = 1.7 \times 7 \times 28 = 333.2\, cm^2$.
Area of sector $OAPB = \frac{60^{\circ}}{360^{\circ}} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 28 \times 28 = \frac{1232}{3} \approx 410.67\, cm^2$.
Area of one segment = Area of sector $OAPB - \text{Area of } \triangle OAB = \frac{1232}{3} - 333.2 = 410.67 - 333.2 = 77.47\, cm^2$.
Total area of $6$ designs = $6 \times 77.47 = 464.82\, cm^2$.
Cost of making designs = $464.82 \times 0.35 = Rs.\, 162.687 \approx Rs.\, 162.68$.

(B) The area of a circle is $\pi R^{2}$.
The area of a sector with angle $\theta$ at the center is given by the formula:
$\text{Area of sector} = \frac{\theta}{360^{\circ}} \times \pi R^{2}$
Given the angle is $p$,the area of the sector is:
$\text{Area} = \frac{p}{360^{\circ}} \times \pi R^{2}$
To match this with the given options,we can multiply the numerator and denominator by $2$:
$\text{Area} = \frac{p \times 2}{360^{\circ} \times 2} \times \pi R^{2} = \frac{p}{720^{\circ}} \times 2 \pi R^{2}$
Thus,the correct option is $B$.

(C) The side of the square $ABCD$ is $a = 56 \, m$.
Area of the square lawn $ABCD = a^2 = 56 \times 56 = 3136 \, m^2$.
The diagonals of the square intersect at $O$. The distance from $O$ to the sides is half the side length,i.e.,$28 \, m$. The radius of the circular sectors forming the flower beds is $r = OA = OB = OC = OD$. In $\triangle OAB$,$OA^2 + OB^2 = AB^2$,so $2r^2 = 56^2$,which gives $r^2 = \frac{56 \times 56}{2} = 1568$.
The area of one circular sector (flower bed) with central angle $90^\circ$ is $\frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 1568 = 1232 \, m^2$.
There are two such flower beds,so their total area is $2 \times 1232 = 2464 \, m^2$.
However,the flower beds are segments of the circle. The area of one segment is (Area of sector $OAB$ - Area of $\triangle OAB$).
Area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 56 \times 28 = 784 \, m^2$.
Area of one flower bed segment = $1232 - 784 = 448 \, m^2$.
Total area = Area of square + $2$ $\times$ Area of segment = $3136 + 2 \times 448 = 3136 + 896 = 4032 \, m^2$.

(D) Area of square $ABCD = 14 \times 14 \, cm^2 = 196 \, cm^2$.
Since there are four circles in the square,the diameter of each circle is half the side of the square.
Diameter of each circle $= \frac{14}{2} \, cm = 7 \, cm$.
Radius of each circle $(r) = \frac{7}{2} \, cm = 3.5 \, cm$.
Area of one circle $= \pi r^2 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \, cm^2 = \frac{77}{2} \, cm^2 = 38.5 \, cm^2$.
Area of four circles $= 4 \times 38.5 \, cm^2 = 154 \, cm^2$.
Area of the shaded region = Area of square - Area of four circles.
Area of the shaded region $= 196 \, cm^2 - 154 \, cm^2 = 42 \, cm^2$.

(A) Let us mark the four unshaded regions as $I, II, III$ and $IV$ (see figure).
Area of $I +$ Area of $III = \text{Area of square } ABCD - \text{Area of two semicircles with diameter } 10 \, cm$.
Since the side of the square is $10 \, cm$,the radius of each semicircle is $r = 5 \, cm$.
Area of $I + III = (10 \times 10) - 2 \times (\frac{1}{2} \times \pi \times 5^2) \, cm^2$
$= 100 - (3.14 \times 25) \, cm^2 = 100 - 78.5 = 21.5 \, cm^2$.
Similarly,Area of $II + IV = 21.5 \, cm^2$.
Now,the area of the shaded design is the area of the square minus the sum of the areas of the four unshaded regions $(I, II, III, IV)$:
Area of shaded design $= \text{Area of } ABCD - (\text{Area of } I + II + III + IV)$
$= 100 - (21.5 + 21.5) \, cm^2 = 100 - 43 = 57 \, cm^2$.

(N/A) $1$. Since $RQ$ is a diameter of the circle,the angle in the semicircle is a right angle. Therefore,$\angle QPR = 90^{\circ}$.
$2$. In the right-angled triangle $\triangle QPR$,by Pythagoras theorem:
$RQ^2 = PQ^2 + PR^2$
$RQ^2 = 24^2 + 7^2 = 576 + 49 = 625$
$RQ = \sqrt{625} = 25 \, cm$.
$3$. The radius of the circle $r = \frac{RQ}{2} = \frac{25}{2} = 12.5 \, cm$.
$4$. The area of the semicircle is $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (12.5)^2 = \frac{11}{7} \times 156.25 = \frac{1718.75}{7} \approx 245.536 \, cm^2$.
$5$. The area of $\triangle QPR = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 24 = 84 \, cm^2$.
$6$. The area of the shaded region = Area of semicircle - Area of $\triangle QPR = \frac{1718.75}{7} - 84 = \frac{1718.75 - 588}{7} = \frac{1130.75}{7} \approx 161.54 \, cm^2$.

(N/A) The shaded region is the area of the sector of the larger circle minus the area of the sector of the smaller circle.
Let $R = 14\, cm$ be the radius of the larger circle and $r = 7\, cm$ be the radius of the smaller circle.
The central angle $\theta = 40^{\circ}$.
The area of a sector is given by the formula $\frac{\theta}{360^{\circ}} \times \pi r^2$.
Area of the shaded region = Area of sector $OAC$ - Area of sector $OBD$
$= \frac{40^{\circ}}{360^{\circ}} \times \pi R^2 - \frac{40^{\circ}}{360^{\circ}} \times \pi r^2$
$= \frac{1}{9} \times \pi (R^2 - r^2)$
$= \frac{1}{9} \times \frac{22}{7} \times (14^2 - 7^2)$
$= \frac{1}{9} \times \frac{22}{7} \times (196 - 49)$
$= \frac{1}{9} \times \frac{22}{7} \times 147$
$= \frac{1}{9} \times 22 \times 21$
$= \frac{462}{9} = \frac{154}{3} \approx 51.33\, cm^2$.

(D) It can be observed from the figure that the side of the square is $14 \, cm$. Therefore,the diameter of each semicircle is $14 \, cm$,and the radius $(r)$ of each semicircle is $7 \, cm$.
Area of one semicircle $= \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (7)^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = 77 \, cm^2$.
Area of two semicircles $= 2 \times 77 = 154 \, cm^2$.
Area of square $ABCD = (\text{side})^2 = (14)^2 = 196 \, cm^2$.
Area of the shaded region $=$ Area of square $ABCD -$ Area of two semicircles.
Area of the shaded region $= 196 \, cm^2 - 154 \, cm^2 = 42 \, cm^2$.

(N/A) The shaded region consists of the area of the equilateral triangle $OAB$ and the area of the major circle excluding the sector $OCDE$ that overlaps with the triangle.
$1$. Area of equilateral triangle $OAB = \frac{\sqrt{3}}{4} \times (side)^2 = \frac{\sqrt{3}}{4} \times (12)^2 = 36\sqrt{3} \, cm^2$.
$2$. Area of the circle with radius $r = 6 \, cm = \pi r^2 = \frac{22}{7} \times 6^2 = \frac{792}{7} \, cm^2$.
$3$. The angle of the equilateral triangle is $60^{\circ}$. The area of the sector $OCDE$ (which is the overlapping part) $= \frac{60^{\circ}}{360^{\circ}} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7} \, cm^2$.
$4$. Total area of the shaded region = Area of $\triangle OAB +$ Area of circle $-$ Area of sector $OCDE$.
$= 36\sqrt{3} + \frac{792}{7} - \frac{132}{7} = 36\sqrt{3} + \frac{660}{7} \, cm^2$.

(N/A) $1$. Area of the square $= (\text{side})^2 = (4\, cm)^2 = 16\, cm^2$.
$2$. Area of each quadrant of a circle with radius $r = 1\, cm$ is $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (1)^2 = \frac{22}{28} = \frac{11}{14}\, cm^2$.
$3$. Total area of $4$ quadrants $= 4 \times \frac{11}{14} = \frac{22}{7}\, cm^2$.
$4$. Area of the central circle with diameter $2\, cm$ (radius $r = 1\, cm$) $= \pi r^2 = \frac{22}{7} \times (1)^2 = \frac{22}{7}\, cm^2$.
$5$. Area of the remaining portion $= \text{Area of square} - (\text{Total area of } 4 \text{ quadrants} + \text{Area of central circle})$.
$6$. Area $= 16 - (\frac{22}{7} + \frac{22}{7}) = 16 - \frac{44}{7} = \frac{112 - 44}{7} = \frac{68}{7}\, cm^2$.

(N/A) Radius $(r)$ of the circle $= 32 \, cm$.
$AD$ is the median of $\triangle ABC$.
Since $O$ is the centroid of the equilateral triangle,$AO = \frac{2}{3} AD = 32 \, cm$.
Therefore,$AD = 32 \times \frac{3}{2} = 48 \, cm$.
In $\triangle ABD$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2$
$AB^2 = (48)^2 + \left(\frac{AB}{2}\right)^2$
$AB^2 - \frac{AB^2}{4} = 2304$
$\frac{3 AB^2}{4} = 2304$
$AB^2 = \frac{2304 \times 4}{3} = 3072$
$AB = \sqrt{3072} = 32 \sqrt{3} \, cm$.
Area of equilateral triangle $\triangle ABC = \frac{\sqrt{3}}{4} \times (side)^2 = \frac{\sqrt{3}}{4} \times (32 \sqrt{3})^2$
$= \frac{\sqrt{3}}{4} \times 1024 \times 3 = 768 \sqrt{3} \, cm^2$.
Area of circle $= \pi r^2 = \frac{22}{7} \times (32)^2 = \frac{22}{7} \times 1024 = \frac{22528}{7} \, cm^2$.
Area of the design = Area of circle $-$ Area of $\triangle ABC$
$= \left(\frac{22528}{7} - 768 \sqrt{3}\right) \, cm^2$.

(D) The area of each of the $4$ sectors is equal to each other,and each is a sector of $90^{\circ}$ in a circle of $7 \, cm$ radius.
Area of each sector $= \frac{90^{\circ}}{360^{\circ}} \times \pi(7)^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7$
$= \frac{77}{2} \, cm^{2} = 38.5 \, cm^{2}$
Area of square $ABCD = (\text{Side})^{2} = (14)^{2} = 196 \, cm^{2}$
Area of shaded portion = Area of square $ABCD - 4 \times$ (Area of each sector)
$= 196 - 4 \times \frac{77}{2}$
$= 196 - 154$
$= 42 \, cm^{2}$
Therefore,the area of the shaded portion is $42 \, cm^{2}$.

(N/A) Given:
Inner length of the straight parts $= 106 \, m$
Inner width (distance between parallel lines) $= 60 \, m$
Inner radius $(r) = \frac{60}{2} = 30 \, m$
Width of the track $= 10 \, m$
Outer radius $(R) = 30 + 10 = 40 \, m$
$(i)$ Distance around the track along its inner edge $= AB + \text{arc } BEC + CD + \text{arc } DFA$
$= 106 + (\pi r) + 106 + (\pi r)$
$= 212 + 2 \pi r$
$= 212 + 2 \times \frac{22}{7} \times 30$
$= 212 + \frac{1320}{7} = \frac{1484 + 1320}{7} = \frac{2804}{7} \, m \approx 400.57 \, m$
$(ii)$ Area of the track $= 2 \times (\text{Area of rectangle } 106 \times 10) + 2 \times (\text{Area of semi-circular ring})$
$= 2 \times (106 \times 10) + 2 \times \left[ \frac{1}{2} \pi (R^2 - r^2) \right]$
$= 2120 + \pi (40^2 - 30^2)$
$= 2120 + \frac{22}{7} (1600 - 900)$
$= 2120 + \frac{22}{7} \times 700$
$= 2120 + 2200 = 4320 \, m^2$

(B) Radius $(r_1)$ of the larger circle $= 7 \, cm$.
Radius $(r_2)$ of the smaller circle $= \frac{7}{2} \, cm = 3.5 \, cm$.
Area of the smaller circle $= \pi r_2^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5 \, cm^2$.
Area of the semi-circle $ACB$ of the larger circle $= \frac{1}{2} \pi r_1^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77 \, cm^2$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OC = \frac{1}{2} \times 14 \times 7 = 49 \, cm^2$.
Area of the shaded region $=$ Area of the smaller circle $+$ (Area of semi-circle $ACB - \text{Area of } \triangle ABC$).
Area of the shaded region $= 38.5 + (77 - 49) = 38.5 + 28 = 66.5 \, cm^2$.

(C) Let the side of the equilateral triangle be $a$.
Area of equilateral triangle $= 17320.5 \, cm^2$.
$\frac{\sqrt{3}}{4} a^2 = 17320.5$
$\frac{1.73205}{4} a^2 = 17320.5$
$a^2 = 4 \times 10000 = 40000$
$a = 200 \, cm$.
The radius of each circle is $r = \frac{a}{2} = \frac{200}{2} = 100 \, cm$.
Each interior angle of an equilateral triangle is $60^{\circ}$.
Area of one sector $= \frac{60^{\circ}}{360^{\circ}} \times \pi r^2 = \frac{1}{6} \times 3.14 \times (100)^2 = \frac{31400}{6} = \frac{15700}{3} \, cm^2$.
Area of three sectors $= 3 \times \frac{15700}{3} = 15700 \, cm^2$.
Area of the shaded region $=$ Area of equilateral triangle $-$ Area of three sectors
$= 17320.5 - 15700 = 1620.5 \, cm^2$.

(D) From the figure,it can be observed that the side of the square is $3 \times (2 \times 7) = 42\, cm$.
Area of square $= (\text{Side})^2 = (42)^2 = 1764\, cm^2$.
Area of one circle $= \pi r^2 = \frac{22}{7} \times (7)^2 = 154\, cm^2$.
Area of $9$ circles $= 9 \times 154 = 1386\, cm^2$.
Area of the remaining portion of the handkerchief $= \text{Area of square} - \text{Area of } 9 \text{ circles} = 1764 - 1386 = 378\, cm^2$.

(N/A) $(i)$ Since $OACB$ is a quadrant,it subtends a $90^{\circ}$ angle at the centre $O$.
Area of quadrant $OACB = \frac{90^{\circ}}{360^{\circ}} \times \pi r^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times (3.5)^{2} = \frac{1}{4} \times \frac{22}{7} \times \left( \frac{7}{2} \right)^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{11 \times 7}{8} = \frac{77}{8} \, cm^{2} = 9.625 \, cm^{2}$.
$(ii)$ Area of $\triangle BOD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OB \times OD$
$= \frac{1}{2} \times 3.5 \times 2 = 3.5 \, cm^{2} = \frac{7}{2} \, cm^{2}$.
Area of the shaded region = Area of quadrant $OACB - \text{Area of } \triangle BOD$
$= \frac{77}{8} - \frac{7}{2} = \frac{77 - 28}{8} = \frac{49}{8} \, cm^{2} = 6.125 \, cm^{2}$.

(B) In $\triangle OAB$,since $OABC$ is a square,$\angle OAB = 90^{\circ}$.
By Pythagoras theorem,$OB^2 = OA^2 + AB^2$.
Since $OA = AB = 20 \, cm$,we have $OB^2 = 20^2 + 20^2 = 400 + 400 = 800$.
Thus,the radius $(r)$ of the quadrant is $OB = \sqrt{800} = 20\sqrt{2} \, cm$.
The area of the quadrant $OPBQ = \frac{1}{4} \times \pi r^2 = \frac{1}{4} \times 3.14 \times (20\sqrt{2})^2$.
$= \frac{1}{4} \times 3.14 \times 800 = 3.14 \times 200 = 628 \, cm^2$.
The area of the square $OABC = (\text{side})^2 = 20^2 = 400 \, cm^2$.
The area of the shaded region = Area of quadrant $OPBQ$ - Area of square $OABC$.
$= 628 \, cm^2 - 400 \, cm^2 = 228 \, cm^2$.

(N/A) The area of the shaded region is the difference between the area of the larger sector $OAB$ and the smaller sector $OCD$.
Area of sector $OAB = \frac{\theta}{360^{\circ}} \times \pi R^2 = \frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (21)^2 = \frac{1}{12} \times \frac{22}{7} \times 441 = \frac{1}{12} \times 22 \times 63 = \frac{1386}{12} = 115.5\, cm^2$.
Area of sector $OCD = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (7)^2 = \frac{1}{12} \times \frac{22}{7} \times 49 = \frac{1}{12} \times 22 \times 7 = \frac{154}{12} = \frac{77}{6}\, cm^2$.
Area of the shaded region = Area of sector $OAB$ $-$ Area of sector $OCD = 115.5 - \frac{77}{6} = \frac{693 - 77}{6} = \frac{616}{6} = \frac{308}{3}\, cm^2$.

(D) Since $ABC$ is a quadrant of the circle,$\angle BAC = 90^{\circ}$.
In $\triangle ABC$,by Pythagoras theorem:
$BC^2 = AC^2 + AB^2 = (14)^2 + (14)^2 = 196 + 196 = 392$.
$BC = \sqrt{392} = 14\sqrt{2} \, cm$.
Radius $(r_1)$ of the semicircle drawn on $BC = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \, cm$.
Area of $\triangle ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 14 \times 14 = 98 \, cm^2$.
Area of quadrant $ABC = \frac{1}{4} \times \pi \times r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \, cm^2$.
Area of segment $BC$ (the region between chord $BC$ and arc $BC$) = Area of quadrant $ABC - \text{Area of } \triangle ABC = 154 - 98 = 56 \, cm^2$.
Area of semicircle on $BC = \frac{1}{2} \times \pi \times r_1^2 = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 = \frac{1}{2} \times \frac{22}{7} \times 98 = 154 \, cm^2$.
Area of shaded region = Area of semicircle on $BC - \text{Area of segment } BC = 154 - 56 = 98 \, cm^2$.

(N/A) The designed area is the common region between two sectors $BAEC$ and $DAFC$.
Area of sector $BAEC = \frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (8)^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times 64$
$= \frac{22 \times 16}{7} = \frac{352}{7} \, cm^{2}$
Area of $\triangle BAC = \frac{1}{2} \times BA \times BC$
$= \frac{1}{2} \times 8 \times 8 = 32 \, cm^{2}$
Area of the designed portion $= 2 \times (\text{Area of segment } AEC)$
$= 2 \times (\text{Area of sector } BAEC - \text{Area of } \triangle BAC)$
$= 2 \times \left(\frac{352}{7} - 32\right) = 2 \left(\frac{352 - 224}{7}\right)$
$= \frac{2 \times 128}{7} = \frac{256}{7} \, cm^{2}$