Mix Examples - Areas Related to Circles Questions in English

(C) The length of a minor arc of a circle with radius $r$ and central angle $\theta$ (in degrees) is given by the formula $l_{minor} = \frac{\theta}{360} \times 2 \pi r = \frac{\pi r \theta}{180}$.
Since the total circumference of the circle is $2 \pi r$,the length of the major arc is the total circumference minus the length of the minor arc.
Therefore,the length of the major arc $l_{major} = 2 \pi r - \frac{\pi r \theta}{180}$.

(B) The circumference of the circle $\odot(O, 7)$ is $C = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44.$
The length of a semicircle is $\pi r = \frac{22}{7} \times 7 = 22.$
Given that the length of $\widehat{ABC}$ is $14,$ and since $14 < 22,$ the arc $\widehat{ABC}$ is shorter than a semicircle.
Therefore,$\widehat{ABC}$ is a minor arc.

(B) sector of a circle is defined as the portion of a circle enclosed by two radii and the corresponding arc.
$1$. $A$ segment of a circle is the region bounded by a chord and an arc.
$2$. $A$ circular ring is the region between two concentric circles.
$3$. The circumference is the boundary or the perimeter of the circle.
Therefore,the figure formed by an arc and the two radii joining the center to the endpoints of the arc is a sector of a circle.

(C) $1$. $A$ circle is defined as $\odot(O, r)$ with center $O$ and radius $r$.
$2$. Points $A$ and $B$ are on the circle. The line segment $\overline{AB}$ is a chord of the circle.
$3$. The arc $\widehat{ACB}$ is the arc connecting points $A$ and $B$ passing through $C$. Since $C$ lies in the interior of $\angle AOB$,$\widehat{ACB}$ represents the minor arc.
$4$. The union of a chord $\overline{AB}$ and the minor arc $\widehat{ACB}$ defines the region known as the minor segment of the circle.
$5$. Therefore,$\overline{AB} \cup \widehat{ACB}$ is a minor segment.

(B) chord of a circle is a line segment whose endpoints lie on the circle.
An arc is a portion of the circumference of a circle.
The region enclosed by a chord and the corresponding arc is known as a circular segment,or simply a segment of the circle.
Therefore,the union of a chord and its corresponding arc is a segment.

(D) For the circular ground,the diameter $d = 105 \, m$.
The length of the fence is equal to the circumference of the circular ground.
Circumference $= \pi d = \frac{22}{7} \times 105$.
$= 22 \times 15 = 330 \, m$.
Therefore,the length of the fence is $330 \, m$.

(C) The area of a circle is given by the formula $A = \pi r^{2}$.
Given that the area $A = 154\,cm^{2}$ and taking $\pi = \frac{22}{7}$,we have:
$154 = \frac{22}{7} \times r^{2}$
$r^{2} = 154 \times \frac{7}{22}$
$r^{2} = 7 \times 7 = 49$
$r = \sqrt{49} = 7\,cm$.
The diameter $d$ of the circle is $2 \times r$.
$d = 2 \times 7 = 14\,cm$.

(B) The formula for the circumference of a circle is $C = 2 \pi r$,where $C$ is the circumference and $r$ is the radius.
Given that $C = 176\,cm$ and taking $\pi = \frac{22}{7}$,we have:
$176 = 2 \times \frac{22}{7} \times r$
To find $r$,we rearrange the equation:
$r = \frac{176 \times 7}{2 \times 22}$
$r = \frac{176 \times 7}{44}$
$r = 4 \times 7$
$r = 28\,cm$
Thus,the radius of the circle is $28\,cm$.

(D) The perimeter of a semicircular garden is the sum of its semi-circumference and its diameter.
Perimeter $= \pi r + 2r$
Given radius $r = 35 \, m$.
Perimeter $= \frac{22}{7} \times 35 + 2 \times 35$
Perimeter $= 22 \times 5 + 70$
Perimeter $= 110 + 70$
Perimeter $= 180 \, m$.
Therefore,one has to walk $180 \, m$ to make one complete round of the garden.

(B) Given,the perimeter of the semicircular table-top is $3.60 \, m$.
Converting meters to centimeters: $3.60 \, m = 360 \, cm$.
The perimeter of a semicircle is given by the formula: $P = \pi r + 2r = r(\pi + 2)$.
Substituting the given values:
$360 = r \left( \frac{22}{7} + 2 \right)$
$360 = r \left( \frac{22 + 14}{7} \right)$
$360 = r \left( \frac{36}{7} \right)$
Solving for $r$:
$r = 360 \times \frac{7}{36}$
$r = 10 \times 7$
$r = 70 \, cm$.
Thus,the radius of the semicircular table-top is $70 \, cm$.

(C) The formula for the circumference of a circle is $C = 2\pi r$.
Given the radius $r = 8.4\,cm$.
Substituting the values,we get $C = 2 \times \frac{22}{7} \times 8.4$.
$C = 2 \times 22 \times 1.2$.
$C = 44 \times 1.2 = 52.8\,cm$.

(A) The area of a circle is given by the formula $A = \pi r^2$,where $r$ is the radius.
Given,$A = 38.5 \, m^2$.
Substituting the value of $\pi = \frac{22}{7}$:
$38.5 = \frac{22}{7} \times r^2$
$r^2 = 38.5 \times \frac{7}{22}$
$r^2 = \frac{385}{10} \times \frac{7}{22} = \frac{77 \times 5}{2 \times 5} \times \frac{7}{22} = \frac{77 \times 7}{22 \times 2} = \frac{7 \times 7}{2 \times 2} = \frac{49}{4}$
$r = \sqrt{\frac{49}{4}} = \frac{7}{2} = 3.5 \, m$.
The diameter $d$ is given by $d = 2r$.
$d = 2 \times 3.5 = 7 \, m$.

(B) For the running track in the shape of a circular ring,let the inner radius be $r_{1} \,m$ and the outer radius be $r_{2} \,m$.
Difference of circumferences $= 44 \,m$
$\therefore 2 \pi r_{2} - 2 \pi r_{1} = 44$
$\therefore 2 \pi (r_{2} - r_{1}) = 44$
$\therefore 2 \times \frac{22}{7} (r_{2} - r_{1}) = 44$
$\therefore r_{2} - r_{1} = \frac{44 \times 7}{2 \times 22}$
$\therefore r_{2} - r_{1} = 7 \,m$
$\therefore$ The width of the track $= 7 \,m$.

(D) The circumference of a circle is given by the formula $C = 2 \pi r$.
Let the radii of the two concentric circles be $r_1 = 14 \, cm$ and $r_2 = 10.5 \, cm$.
The difference between their circumferences is given by $|2 \pi r_1 - 2 \pi r_2| = 2 \pi (r_1 - r_2)$.
Substituting the values,we get:
Difference $= 2 \times \frac{22}{7} \times (14 - 10.5) \, cm$.
Difference $= 2 \times \frac{22}{7} \times 3.5 \, cm$.
Difference $= 2 \times 22 \times 0.5 \, cm$.
Difference $= 22 \, cm$.

(C) Radius of the garden $r_2 = \frac{\text{diameter}}{2} = \frac{210}{2} = 105 \,m$.
Radius of the garden excluding the path $r_1 = 105 - 7 = 98 \,m$.
Area of the path in the form of a circular ring is given by the difference between the area of the outer circle and the inner circle.
Area of the path $= \pi r_2^2 - \pi r_1^2$
$= \pi(r_2^2 - r_1^2)$
$= \pi(r_2 + r_1)(r_2 - r_1)$
$= \frac{22}{7} \times (105 + 98) \times (105 - 98)$
$= \frac{22}{7} \times 203 \times 7$
$= 22 \times 203$
$= 4466 \,m^2$.

(C) Distance covered in one rotation $= 440 \, cm$.
Distance covered in $250$ rotations $= 440 \times 250 \, cm$.
Since the wheel makes $250$ rotations per minute,the distance covered in $1$ minute is $440 \times 250 \, cm$.
Distance covered in $1$ hour ($60$ minutes) $= 440 \times 250 \times 60 \, cm$.
To convert this distance into kilometers,we divide by $100$ (to convert $cm$ to $m$) and then by $1000$ (to convert $m$ to $km$):
Distance in $km = \frac{440 \times 250 \times 60}{100 \times 1000} \, km$.
$= \frac{6,600,000}{100,000} \, km = 66 \, km$.
Therefore,the speed of the truck is $66 \, km/h$.

(B) The minute hand of a clock sweeps a circular sector in a given time interval.
Given radius $r = 12\,cm$.
In $60$ minutes,the minute hand completes one full rotation $(360^{\circ})$.
Therefore,in $5$ minutes,the angle swept $\theta = \frac{360^{\circ}}{60} \times 5 = 30^{\circ}$.
The area of the sector swept is given by the formula: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^2$.
Substituting the values: $\text{Area} = \frac{30}{360} \times 3.14 \times 12 \times 12$.
$\text{Area} = \frac{1}{12} \times 3.14 \times 144 = 3.14 \times 12 = 37.68\,cm^2$.

(C) The formula for the area of a minor sector is given by: $\text{Area} = \frac{\pi r^{2} \theta}{360^{\circ}}$.
Given radius $r = 42\,cm$ and central angle $\theta = 120^{\circ}$.
Substituting the values: $\text{Area} = \frac{22}{7} \times 42 \times 42 \times \frac{120^{\circ}}{360^{\circ}}$.
Simplifying the expression: $\text{Area} = \frac{22}{7} \times 42 \times 42 \times \frac{1}{3}$.
$\text{Area} = 22 \times 6 \times 14 = 1848\,cm^{2}$.

(D) For the minor sector $OACB$,the central angle $\theta = 90^{\circ}$ and the radius is $r\,cm$.
The perimeter of the minor sector $OACB$ is given by the sum of the two radii and the length of the arc $\widehat{ACB}$.
Perimeter $= 2r + \text{length of arc } \widehat{ACB}$
Since the length of an arc is $\frac{\pi r \theta}{180}$,we have:
$20 = 2r + \frac{\pi r \times 90}{180}$
Using $\pi \approx \frac{22}{7}$:
$20 = 2r + \frac{22}{7} \times \frac{r}{2}$
$20 = 2r + \frac{11r}{7}$
$20 = r \left( 2 + \frac{11}{7} \right)$
$20 = r \left( \frac{14 + 11}{7} \right)$
$20 = r \left( \frac{25}{7} \right)$
$r = \frac{20 \times 7}{25}$
$r = \frac{4 \times 7}{5} = \frac{28}{5} = 5.6\,cm$.

(C) Given that in $\odot(O, 5.6)$,$\overline{OA} \perp \overline{OB}$.
Thus,$OA = OB = 5.6 \text{ cm}$ and $m\angle AOB = 90^\circ$.
The area of the minor sector formed by minor $\widehat{AB}$ is given by $\frac{\theta}{360^\circ} \times \pi r^2$.
The area of the corresponding minor segment is given by $(\text{Area of minor sector}) - (\text{Area of } \Delta OAB)$.
Therefore,the difference between the area of the minor sector and the area of the minor segment is equal to the area of $\Delta OAB$.
Area of $\Delta OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 5.6 \times 5.6 = 15.68 \text{ cm}^2$.

(A) Given that the radii $\overline{ OA }$ and $\overline{ OB }$ are perpendicular to each other,the central angle $\theta = 90^\circ$.
The radius of the circle is $r = 5.6 \, cm$.
The formula for the area of a minor sector is $\text{Area} = \frac{\pi r^2 \theta}{360^\circ}$.
Substituting the values: $\text{Area} = \frac{22}{7} \times (5.6)^2 \times \frac{90^\circ}{360^\circ}$.
$\text{Area} = \frac{22}{7} \times 31.36 \times \frac{1}{4}$.
$\text{Area} = \frac{22}{7} \times 7.84 = 22 \times 1.12 = 24.64 \, cm^2$.

(D) The umbrella is assumed to be a flat circle with radius $r = 56 \, cm$.
Since there are $8$ equally spaced ribs,the circle is divided into $8$ equal sectors.
The central angle $\theta$ for each sector is $\frac{360^{\circ}}{8} = 45^{\circ}$.
The area between two consecutive ribs is the area of one sector,which is given by the formula:
Area $= \frac{\theta}{360^{\circ}} \times \pi r^{2}$
Area $= \frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (56)^{2}$
Area $= \frac{1}{8} \times \frac{22}{7} \times 56 \times 56$
Area $= \frac{1}{8} \times 22 \times 8 \times 56$
Area $= 22 \times 56 = 1232 \, cm^{2}$.

(A) The formula for the area of a sector is given by: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Given radius $r = 30\,cm$,angle $\theta = 60^{\circ}$,and $\pi = 3.14$.
Substituting the values into the formula:
$\text{Area} = \frac{60}{360} \times 3.14 \times (30)^{2}$
$\text{Area} = \frac{1}{6} \times 3.14 \times 900$
$\text{Area} = 3.14 \times 150$
$\text{Area} = 471\,cm^{2}$.

(C) The length of the minute hand represents the radius of the circle,so $r = 6 \, cm$.
In $60 \, \text{minutes}$,the minute hand sweeps an angle of $360^{\circ}$.
Therefore,in $10 \, \text{minutes}$,the angle swept by the minute hand is $\theta = \frac{360^{\circ}}{60} \times 10 = 60^{\circ}$.
The area of the region swept is the area of a sector with radius $r = 6 \, cm$ and central angle $\theta = 60^{\circ}$.
$\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60^{\circ}}{360^{\circ}} \times 3.14 \times 6 \times 6$.
$\text{Area} = \frac{1}{6} \times 3.14 \times 36 = 3.14 \times 6 = 18.84 \, cm^2$.

(A) The area of a minor sector is the sum of the area of the corresponding minor segment and the area of the triangle formed by the two radii and the chord.
Area of minor sector $OABC = \text{Area of minor segment} + \text{Area of } \Delta OAC$
Area of minor sector $OABC = 14.25\,cm^2 + 25\,cm^2$
Area of minor sector $OABC = 39.25\,cm^2$

(B) The diameter of the circle $= 2 \times \text{radius}$.
$= 2 \times 35 = 70 \, cm$.
Since the square is inscribed in the circle,the diagonal of the square is equal to the diameter of the circle.
Therefore,the diagonal of the square $= 70 \, cm$.
The area of a square $= \frac{(\text{diagonal})^{2}}{2}$.
$= \frac{70^{2}}{2} = \frac{4900}{2} = 2450 \, cm^{2}$.

(C) The area of a minor sector is the sum of the area of the corresponding minor segment and the area of the triangle formed by the chord and the centre of the circle.
Area of minor sector $OACB = \text{Area of minor segment } \overline{AB} \cup \widehat{ACB} + \text{Area of } \Delta OAB$
Given:
Area of minor segment $= 114 \, cm^2$
Area of $\Delta OAB = 200 \, cm^2$
Therefore,
Area of minor sector $OACB = 114 \, cm^2 + 200 \, cm^2 = 314 \, cm^2$.

(C) Given,radius $r = 14 \, cm$.
Since $\overline{OA} \perp \overline{OB}$,the central angle $\theta = 90^\circ$.
The formula for the area of a minor sector is $\frac{\theta}{360^\circ} \times \pi r^2$.
Substituting the values: $\text{Area} = \frac{90}{360} \times \frac{22}{7} \times 14 \times 14$.
$\text{Area} = \frac{1}{4} \times 22 \times 2 \times 14$.
$\text{Area} = \frac{1}{4} \times 616 = 154 \, cm^2$.

(C) The area of a sector of a circle is given by the formula $A = \frac{\theta}{360^\circ} \times \pi r^2$.
Let the radii of the two circles be $r_1$ and $r_2$ and the central angles be $\theta_1$ and $\theta_2$.
Given that $\theta_1 = \theta_2 = \theta$.
The ratio of the areas of the two sectors is given as $\frac{A_1}{A_2} = \frac{4}{9}$.
Substituting the formula,we get $\frac{\frac{\theta}{360^\circ} \times \pi r_1^2}{\frac{\theta}{360^\circ} \times \pi r_2^2} = \frac{4}{9}$.
Simplifying the expression,we have $\frac{r_1^2}{r_2^2} = \frac{4}{9}$.
Taking the square root on both sides,we get $\frac{r_1}{r_2} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
Therefore,the ratio of the radii is $2:3$.

(D) The area of a sector of a circle with radius $r$ and central angle $\theta$ is given by the formula $A = \frac{\theta}{360^\circ} \times \pi r^2$.
For two sectors with central angles $\theta_1$ and $\theta_2$ in the same circle,the ratio of their areas is:
$\frac{A_1}{A_2} = \frac{(\frac{\theta_1}{360^\circ} \times \pi r^2)}{(\frac{\theta_2}{360^\circ} \times \pi r^2)} = \frac{\theta_1}{\theta_2}$.
Given that the ratio of the areas is $1:4$,we have $\frac{A_1}{A_2} = \frac{1}{4}$.
Therefore,$\frac{\theta_1}{\theta_2} = \frac{1}{4}$.
Thus,the ratio of the angles at the centre is $1:4$.

(B) In $\Delta ABC$,the base $BC$ is the diameter of the semicircle.
Given radius $r = 10 \, cm$,so the base $BC = 2r = 2 \times 10 = 20 \, cm$.
The maximum height of the triangle inscribed in a semicircle is equal to the radius of the semicircle.
Therefore,the maximum height $OA = 10 \, cm$.
The maximum area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{maximum height}$.
Area $= \frac{1}{2} \times 20 \times 10 = 100 \, cm^2$.

(A) When a square is inscribed in a circle,the diagonal of the square passes through the center of the circle.
Therefore,the length of the diagonal of the square is equal to the diameter of the circle.
Given,the radius of the circle $r = 10\, cm$.
Diameter of the circle $= 2 \times r = 2 \times 10\, cm = 20\, cm$.
Thus,the length of the diagonal of the square is $20\, cm$.

(C) The length of the minute hand is the radius of the circle,$r = 10.5 \, cm = \frac{21}{2} \, cm$.
The minute hand completes $360^{\circ}$ in $60$ minutes.
In $20$ minutes,the angle swept by the minute hand is $\theta = \frac{360^{\circ}}{60} \times 20 = 120^{\circ}$.
The area of the region swept is the area of a sector with radius $r$ and central angle $\theta$.
Area $= \frac{\theta}{360^{\circ}} \times \pi r^2$
Area $= \frac{120}{360} \times \frac{22}{7} \times \left(\frac{21}{2}\right)^2$
Area $= \frac{1}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$
Area $= \frac{1}{3} \times 11 \times 3 \times \frac{21}{2} = \frac{231}{2} = 115.5 \, cm^2$.

(B) Given,the radius of the circle $r = 3.5\,cm = \frac{7}{2}\,cm$.
Since the two radii are perpendicular,the central angle $\theta = 90^\circ$.
The area of a minor sector is given by the formula:
Area $= \frac{\theta}{360^\circ} \times \pi r^2$
Area $= \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (3.5)^2$
Area $= \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5$
Area $= \frac{1}{4} \times 22 \times 0.5 \times 3.5$
Area $= \frac{1}{4} \times 38.5 = 9.625\,cm^2$.

(A) The perimeter of a minor sector is given by the sum of two radii and the length of the arc.
Perimeter $= 2r + l$,where $r$ is the radius and $l$ is the arc length.
Given $r = 21 \ cm$ and Perimeter $= 64 \ cm$.
Substituting the values: $64 = 2(21) + l$.
$64 = 42 + l$.
$l = 64 - 42$.
$l = 22 \ cm$.
Therefore,the length of the arc is $22 \ cm$.

(D) The perimeter of a minor sector is given by the formula $P = 2r + l$,where $r$ is the radius and $l$ is the arc length.
Given $r = 7 \, cm$ and $P = \frac{86}{3} \, cm$.
Substituting the values: $\frac{86}{3} = 2(7) + l$.
$\frac{86}{3} = 14 + l \implies l = \frac{86}{3} - 14 = \frac{86 - 42}{3} = \frac{44}{3} \, cm$.
The area of a minor sector is given by $A = \frac{1}{2} r l$.
$A = \frac{1}{2} \times 7 \times \frac{44}{3} = \frac{154}{3} \, cm^2$.

(C) The formula for the area of a sector given the radius $r$ and arc length $l$ is given by: $\text{Area} = \frac{1}{2} \times r \times l$.
Given: $r = 6 \, cm$ and $l = 12 \, cm$.
Substituting the values: $\text{Area} = \frac{1}{2} \times 6 \times 12 = 36 \, cm^2$.

(A) The formula for the area of a sector $(A)$ in terms of the radius $(r)$ and the arc length $(l)$ is given by $A = \frac{1}{2} \times r \times l$.
Given that the radius $r = 10\,cm$ and the area $A = 40\,cm^2$.
Substituting these values into the formula:
$40 = \frac{1}{2} \times 10 \times l$
$40 = 5 \times l$
$l = \frac{40}{5}$
$l = 8\,cm$.
Therefore,the length of the arc is $8\,cm$.

(B) The formula for the area of a sector given the radius $r$ and arc length $l$ is given by $\text{Area} = \frac{1}{2} \times r \times l$.
Given that the radius $r = 12\,cm$ and the arc length $l = 12\,cm$.
Substituting these values into the formula:
$\text{Area} = \frac{1}{2} \times 12\,cm \times 12\,cm = 6 \times 12\,cm^{2} = 72\,cm^{2}$.

(D) The area of a circle is given by the formula $A = \pi r^2$.
For $\odot(O, 6)$,the radius $r_1 = 6$. Thus,Area $A_1 = \pi(6)^2 = 36\pi$.
For $\odot(P, 12)$,the radius $r_2 = 12$. Thus,Area $A_2 = \pi(12)^2 = 144\pi$.
The ratio of the areas is $\frac{A_1}{A_2} = \frac{36\pi}{144\pi} = \frac{36}{144} = \frac{1}{4}$.
Therefore,the ratio is $1:4$.

(B) The area of a circle is given by the formula $A = \pi r^2$,where $r$ is the radius.
Let the radii of the two circles be $r_1 = 8 \, cm$ and $r_2 = 12 \, cm$.
The ratio of their areas is $\frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2}$.
Substituting the values,we get $\frac{8^2}{12^2} = \frac{64}{144}$.
Simplifying the fraction by dividing both numerator and denominator by $16$,we get $\frac{64 \div 16}{144 \div 16} = \frac{4}{9}$.
Thus,the ratio of the areas is $4:9$.

(C) Actual area of the circle with radius $r_1 = 5 \, cm$ is $A_1 = \pi r_1^2 = \pi(5)^2 = 25\pi \, cm^2$.
Calculated area of the circle with radius $r_2 = 6 \, cm$ is $A_2 = \pi r_2^2 = \pi(6)^2 = 36\pi \, cm^2$.
The increase in area is $A_2 - A_1 = 36\pi - 25\pi = 11\pi \, cm^2$.
The percentage increase in the calculated area is given by $\frac{\text{Increase in area}}{\text{Actual area}} \times 100 = \frac{11\pi}{25\pi} \times 100$.
$= \frac{11}{25} \times 100 = 11 \times 4 = 44 \%$.

(A) Given that the radius $r = 4$ and the central angle $\theta = 45^\circ$.
The formula for the length of a minor arc is given by $L = \frac{\theta}{360^\circ} \times 2\pi r$.
Substituting the values into the formula:
$L = \frac{45^\circ}{360^\circ} \times 2 \times \pi \times 4$
$L = \frac{1}{8} \times 8\pi$
$L = \pi$.
Therefore,the length of the minor arc $\widehat{ACB}$ is $\pi$.

(D) Given, the radius $r = 6$ and the central angle $\theta = 60^{\circ}$.
The circumference of the circle is $C = 2 \pi r = 2 \pi(6) = 12 \pi$.
The length of the minor arc $\widehat{AC}$ is given by the formula $L_{minor} = \frac{\theta}{360^{\circ}} \times 2 \pi r = \frac{60^{\circ}}{360^{\circ}} \times 12 \pi = \frac{1}{6} \times 12 \pi = 2 \pi$.
The length of the major arc $\widehat{ABC}$ is the circumference minus the length of the minor arc.
Length of major $\widehat{ABC} = 12 \pi - 2 \pi = 10 \pi$.

(C) The circumference of a circle is given by $C = 2\pi r$.
For a circle with radius $r = 7$,the circumference is $C = 2 \times \frac{22}{7} \times 7 = 44$ units.
The semi-circumference of the circle is $\frac{C}{2} = \frac{44}{2} = 22$ units.
$A$ minor arc is defined as an arc whose length is less than the semi-circumference of the circle.
Therefore,the length of the minor arc must be less than $22$ units.
Among the given options,only $12$ is less than $22$.
Hence,the length of a minor arc of $\odot(O, 7)$ can be $12$ units.

(B) The length of an arc is given by the formula $L = \frac{\theta}{360^{\circ}} \times 2\pi r$,where $\theta$ is the angle subtended at the centre.
Given that the length of minor $\widehat{ACB} = \frac{1}{6} \times \text{circumference}$.
Since the circumference of the circle is $2\pi r$,we have:
$\frac{\theta}{360^{\circ}} \times 2\pi r = \frac{1}{6} \times 2\pi r$
Dividing both sides by $2\pi r$,we get:
$\frac{\theta}{360^{\circ}} = \frac{1}{6}$
$\theta = \frac{360^{\circ}}{6}$
$\theta = 60^{\circ}$

(A) The length of an arc subtending an angle $\theta$ at the centre is given by $L = \frac{\theta}{360^{\circ}} \times 2\pi r$.
The circumference of the circle is $C = 2\pi r$.
Therefore,the ratio of the length of the minor arc $\widehat{ACB}$ to the circumference is $\frac{L}{C} = \frac{(\frac{\theta}{360^{\circ}}) \times 2\pi r}{2\pi r} = \frac{\theta}{360^{\circ}}$.
Given $\theta = 72^{\circ}$,the ratio is $\frac{72^{\circ}}{360^{\circ}} = \frac{1}{5}$.
Thus,the ratio is $1:5$.

(C) The area of a circle is given as $200 \, cm^{2}$.
$A$ minor sector is a portion of the circle bounded by two radii and an arc,where the central angle is less than $180^{\circ}$.
Consequently,the area of a minor sector must be strictly less than the area of a semicircle.
The area of the semicircle is $\frac{200}{2} = 100 \, cm^{2}$.
Among the given options,only $75 \, cm^{2}$ is less than $100 \, cm^{2}$.
Therefore,the area of the minor sector can be $75 \, cm^{2}$.

(D) The area of a circle with radius $r = 7 \, cm$ is given by $A = \pi r^{2}$.
$A = \frac{22}{7} \times 7 \times 7 = 154 \, cm^{2}$.
$A$ minor sector is a sector with a central angle $\theta < 180^{\circ}$.
Therefore,the area of a minor sector must be less than half the area of the circle (i.e.,less than the area of a semicircle).
Area of a semicircle $= \frac{154}{2} = 77 \, cm^{2}$.
Among the given options,only $55 \, cm^{2}$ is less than $77 \, cm^{2}$.
Thus,the area of the minor sector can be $55 \, cm^{2}$.

(D) The area of a sector is given by the formula: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$.
Given: Radius $r = 14 \, cm$,Area $= 77 \, cm^2$,and $\pi = \frac{22}{7}$.
Substituting the values into the formula: $77 = \frac{\theta}{360^\circ} \times \frac{22}{7} \times 14 \times 14$.
$77 = \frac{\theta}{360^\circ} \times 22 \times 2 \times 14$.
$77 = \frac{\theta}{360^\circ} \times 616$.
$\theta = \frac{77 \times 360^\circ}{616}$.
$\theta = \frac{27720}{616} = 45^\circ$.
Therefore,the angle subtended at the centre is $45^\circ$.