Areas of Sector and Segment of a Circle Questions in English

(A) Given: Radius $r = 42 \, cm$ and central angle $\theta = 120^{\circ}$.
Length of the arc $l = \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{120^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 42 = \frac{1}{3} \times 2 \times 22 \times 6 = 88 \, cm$.
Area of the sector $A = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{120^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 42 \times 42 = \frac{1}{3} \times 22 \times 6 \times 42 = 22 \times 2 \times 42 = 1848 \, cm^2$.

(A) Given: Radius $r = 28 \ cm$,Arc length $l = 22 \ cm$.
$1$. To find the angle $\theta$ subtended at the centre:
The formula for arc length is $l = \frac{\theta}{360^{\circ}} \times 2\pi r$.
$22 = \frac{\theta}{360^{\circ}} \times 2 \times \frac{22}{7} \times 28$.
$22 = \frac{\theta}{360^{\circ}} \times 176$.
$\theta = \frac{22 \times 360^{\circ}}{176} = \frac{360^{\circ}}{8} = 45^{\circ}$.
$2$. To find the area of the sector:
The formula for the area of a sector is $A = \frac{\theta}{360^{\circ}} \times \pi r^2$.
$A = \frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 28 \times 28$.
$A = \frac{1}{8} \times 22 \times 4 \times 28 = 11 \times 28 = 308 \ cm^2$.

(D) The area of a sector of a circle is given by the formula: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$.
Since the two radii are perpendicular to each other,the central angle $\theta = 90^\circ$.
Given,$\text{Area} = 38.5 \, cm^2$ and $\pi = \frac{22}{7}$.
Substituting the values: $38.5 = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times r^2$.
$38.5 = \frac{1}{4} \times \frac{22}{7} \times r^2$.
$38.5 = \frac{11}{14} \times r^2$.
$r^2 = \frac{38.5 \times 14}{11}$.
$r^2 = 3.5 \times 14 = 49$.
$r = \sqrt{49} = 7 \, cm$.

(A) Given: Radius $r = 14 \,cm$,Angle $\theta = 90^\circ$.
$1$. Length of arc = $\frac{\theta}{360^\circ} \times 2\pi r = \frac{90}{360} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22 \,cm$.
$2$. Area of minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 22 \times 28 = 154 \,cm^2$.
$3$. Area of minor segment = Area of sector - Area of triangle = $154 - \frac{1}{2} \times r^2 \times \sin(90^\circ) = 154 - \frac{1}{2} \times 14 \times 14 \times 1 = 154 - 98 = 56 \,cm^2$.

(N/A) Given: Radius $r = 21 \, cm$,Arc length $l = 33 \, cm$.
$1$. Angle subtended at the centre $(\theta)$: The formula for arc length is $l = \frac{\theta}{360^\circ} \times 2\pi r$. Substituting the values: $33 = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 21$. Solving for $\theta$: $33 = \theta \times \frac{44}{60} \implies \theta = \frac{33 \times 60}{44} = 45^\circ$.
$2$. Area of the minor sector: $A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{45}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{8} \times 22 \times 3 \times 21 = 346.5 \, cm^2$.
$3$. Area of the minor segment: $A_{segment} = A_{sector} - A_{triangle} = 346.5 - \frac{1}{2} r^2 \sin(\theta) = 346.5 - \frac{1}{2} \times 21 \times 21 \times \sin(90^\circ)$. Note: The central angle is $90^\circ$ if calculated as $l = \frac{\theta}{360} \times 2\pi r \implies 33 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 21 \implies 33 = \theta \times \frac{132}{360} \implies \theta = 90^\circ$. Area of triangle $= \frac{1}{2} \times 21 \times 21 = 220.5 \, cm^2$. Area of segment $= 346.5 - 220.5 = 126 \, cm^2$.

(A) Given radius $r = 56 \, cm$. Since the two radii are perpendicular,the central angle $\theta = 90^\circ$.
Area of the minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 56 \times 56 = \frac{1}{4} \times 22 \times 8 \times 56 = 2464 \, cm^2$.
Area of the major sector = Total area of circle - Area of minor sector = $\pi r^2 - 2464 = \frac{22}{7} \times 56 \times 56 - 2464 = 9856 - 2464 = 7392 \, cm^2$.
Area of the minor segment = Area of minor sector - Area of triangle = $2464 - \frac{1}{2} \times r^2 \times \sin(90^\circ) = 2464 - \frac{1}{2} \times 56 \times 56 \times 1 = 2464 - 1568 = 896 \, cm^2$.

(A) Given: Radius $r = 42 \ cm$,Central angle $\theta = 60^{\circ}$.
$1$. Area of minor sector = $\frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 42 \times 42 = \frac{1}{6} \times 22 \times 6 \times 42 = 22 \times 42 = 924 \ cm^2$.
$2$. Area of minor segment = Area of minor sector - Area of $\triangle OAB$.
Area of $\triangle OAB = \frac{1}{2} r^2 \sin(\theta) = \frac{1}{2} \times 42 \times 42 \times \sin(60^{\circ}) = 882 \times \frac{\sqrt{3}}{2} = 441 \times 1.73 = 762.93 \ cm^2$.
Area of minor segment = $924 - 762.93 = 161.07 \ cm^2$.

(A) Given: Radius $r = 8.4 \, cm$,Central angle $\theta = 60^{\circ}$.
Area of minor sector = $\frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 8.4 \times 8.4 = \frac{1}{6} \times 22 \times 1.2 \times 8.4 = 36.96 \, cm^2$.
Area of circle = $\pi r^2 = \frac{22}{7} \times 8.4 \times 8.4 = 221.76 \, cm^2$.
Area of major sector = Area of circle - Area of minor sector = $221.76 - 36.96 = 184.8 \, cm^2$.

(B) Let the side of the square be $a$. The length of the diagonal of a square is $d = a\sqrt{2}$.
Given $d = 120\, m$,so $a\sqrt{2} = 120$,which implies $a = \frac{120}{\sqrt{2}} = 60\sqrt{2}\, m$.
The diagonals of a square bisect each other at $90^\circ$. Thus,the distance from the center $O$ to each vertex is $r = \frac{120}{2} = 60\, m$.
Each flower bed is a minor segment of a circle with radius $r = 60\, m$ and central angle $\theta = 90^\circ$.
The area of one minor segment is given by the formula: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 - \frac{1}{2} r^2 \sin\theta$.
Substituting the values: $\text{Area} = \frac{90}{360} \times 3.14 \times (60)^2 - \frac{1}{2} \times (60)^2 \times \sin(90^\circ)$.
$\text{Area} = \frac{1}{4} \times 3.14 \times 3600 - \frac{1}{2} \times 3600 \times 1$.
$\text{Area} = 3.14 \times 900 - 1800 = 2826 - 1800 = 1026\, m^2$.
Since there are two such flower beds,the total area is $2 \times 1026 = 2052\, m^2$.

(C) The perimeter of the sector is given by $P = \frac{\text{Total Cost}}{\text{Rate}} = \frac{5400}{30} = 180 \, m$.
The perimeter of a sector is $P = 2r + l$,where $r = 50 \, m$ and $l$ is the arc length.
$180 = 2(50) + l \implies 180 = 100 + l \implies l = 80 \, m$.
The area of the sector is $A = \frac{1}{2} \times l \times r = \frac{1}{2} \times 80 \times 50 = 2000 \, m^2$.
The cost of tilling is $\text{Area} \times \text{Rate} = 2000 \times 15 = ₹ 30,000$.

(A) The field is an equilateral triangle,so each interior angle is $60^{\circ}$.
The cow is tethered at one vertex,so the area it can graze is a sector of a circle with radius $r = 5\, m$ and central angle $\theta = 60^{\circ}$.
The formula for the area of a sector is $A = \frac{\theta}{360^{\circ}} \times \pi r^2$.
Substituting the values: $A = \frac{60^{\circ}}{360^{\circ}} \times 3.14 \times (5)^2$.
$A = \frac{1}{6} \times 3.14 \times 25$.
$A = \frac{78.5}{6} \approx 13.0833\, m^2$.
Rounding to two decimal places,the area is $13.08\, m^2$.

(B) The length of the minute hand is the radius of the circle,$r = 17.5\, cm$.
The minute hand completes a full circle $(360^{\circ})$ in $60$ minutes.
In $15$ minutes,the angle swept by the minute hand is $\theta = (360^{\circ} / 60) \times 15 = 90^{\circ}$.
The area swept by the minute hand is the area of a sector with angle $\theta = 90^{\circ}$ and radius $r = 17.5\, cm$.
Area of sector $= (\theta / 360^{\circ}) \times \pi r^2$.
Area $= (90^{\circ} / 360^{\circ}) \times (22 / 7) \times 17.5 \times 17.5$.
Area $= (1 / 4) \times (22 / 7) \times 17.5 \times 17.5$.
Area $= (1 / 4) \times 22 \times 2.5 \times 17.5$.
Area $= 0.25 \times 22 \times 43.75 = 240.625\, cm^2$.

(C) The outer radius of the circular ground is $R = 56\,m$.
The width of the road is $7\,m$,so the inner radius is $r = 56 - 7 = 49\,m$.
The shaded region is a sector of an annulus with a central angle $\theta = 60^\circ$.
The area of the shaded region is given by the formula:
Area $= \frac{\theta}{360^\circ} \times \pi \times (R^2 - r^2)$
Area $= \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (56^2 - 49^2)$
Area $= \frac{1}{6} \times \frac{22}{7} \times (56 - 49)(56 + 49)$
Area $= \frac{1}{6} \times \frac{22}{7} \times 7 \times 105$
Area $= \frac{1}{6} \times 22 \times 105 = 11 \times 35 = 385\,m^2$.
The cost of repairing is at the rate of ₹ $40/m^2$.
Total cost $= 385 \times 40 = ₹ 15400$.

(N/A) Here,$PS = 12 \text{ cm}$ and $PQ = QR = RS$.
$\therefore PQ = QR = RS = \frac{12}{3} = 4 \text{ cm}$.
The radii of semicircles with diameters $\overline{PS}$,$\overline{QS}$,and $\overline{PQ}$ are:
$r_1 = \frac{PS}{2} = \frac{12}{2} = 6 \text{ cm}$,
$r_2 = \frac{QS}{2} = \frac{4+4}{2} = 4 \text{ cm}$,and
$r_3 = \frac{PQ}{2} = \frac{4}{2} = 2 \text{ cm}$ respectively.
Perimeter of the shaded region:
$= \text{Sum of the lengths of all the three semicircular arcs}$
$= \pi r_1 + \pi r_2 + \pi r_3$
$= \pi(r_1 + r_2 + r_3)$
$= 3.14(6 + 4 + 2)$
$= 37.68 \text{ cm}$.
Area of the shaded region:
$= \text{Area of the semicircle with radius } r_1 + \text{Area of the semicircle with radius } r_3 - \text{Area of the semicircle with radius } r_2$
$= \frac{1}{2} \pi r_1^2 + \frac{1}{2} \pi r_3^2 - \frac{1}{2} \pi r_2^2$
$= \frac{1}{2} \pi(r_1^2 + r_3^2 - r_2^2)$
$= \frac{1}{2} \times 3.14(6^2 + 2^2 - 4^2)$
$= \frac{1}{2} \times 3.14 \times (36 + 4 - 16)$
$= \frac{1}{2} \times 3.14 \times 24$
$= 37.68 \text{ cm}^2$.
Thus,the perimeter of the shaded region is $37.68 \text{ cm}$ and its area is $37.68 \text{ cm}^2$.

(B) $1$. The side of the equilateral triangle $\triangle ABC$ is $a = 70 \, cm$.
$2$. The area of $\triangle ABC = \frac{\sqrt{3}}{4} \times a^2 = \frac{1.73}{4} \times 70^2 = 0.4325 \times 4900 = 2119.25 \, cm^2$.
$3$. $P$ and $R$ are midpoints of $AB$ and $AC$,so $AP = AR = \frac{70}{2} = 35 \, cm$. This is the radius $r$ of the sector $APR$.
$4$. Since $\triangle ABC$ is equilateral,$\angle A = 60^\circ$.
$5$. The area of the sector $APR = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 35 \times 35 = \frac{1}{6} \times 22 \times 5 \times 35 = \frac{3850}{6} \approx 641.67 \, cm^2$.
$6$. The shaded area is the area of $\triangle ABC$ minus the area of the sector $APR$.
$7$. Shaded Area $= 2119.25 - 641.67 = 1477.58 \, cm^2$.

(D) The shaded region is the area of the quadrant $OAB$ minus the area of the triangle $OD A$.
Given radius $r = 21 \text{ cm}$.
Area of quadrant $OAB = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{4} \times 22 \times 3 \times 21 = \frac{1386}{4} = 346.5 \text{ cm}^2$.
Area of $\triangle ODA = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OD \times OA = \frac{1}{2} \times 10 \times 21 = 105 \text{ cm}^2$.
Area of shaded region = Area of quadrant $OAB - \text{Area of } \triangle ODA = 346.5 - 105 = 241.5 \text{ cm}^2$.

(A) The side length of the square $ABCD$ is $s = 14 \, cm$.
The area of the square $ABCD = s^2 = 14^2 = 196 \, cm^2$.
Each vertex of the square is the centre of a circle with radius $r = 7 \, cm$.
Inside the square,there are four sectors,each with a central angle of $90^\circ$ (since it is a square).
The sum of the areas of these four sectors is $4 \times (\frac{90}{360} \times \pi r^2) = \pi r^2$.
Substituting $r = 7 \, cm$,the area of the four sectors $= \frac{22}{7} \times 7 \times 7 = 154 \, cm^2$.
The area of the shaded region = (Area of square) - (Sum of areas of four sectors).
Area of shaded region $= 196 - 154 = 42 \, cm^2$.

(B) The shaded region is the area between two sectors of circles with the same central angle $\theta = 90^\circ$.
Let $R$ be the radius of the outer sector $(OB = 21 \, \text{m})$ and $r$ be the radius of the inner sector $(OD = 14 \, \text{m})$.
The area of a sector is given by $\frac{\theta}{360^\circ} \times \pi R^2$.
The area of the shaded region = (Area of outer sector) - (Area of inner sector)
$= \frac{90^\circ}{360^\circ} \times \pi \times R^2 - \frac{90^\circ}{360^\circ} \times \pi \times r^2$
$= \frac{1}{4} \times \pi \times (R^2 - r^2)$
$= \frac{1}{4} \times \frac{22}{7} \times (21^2 - 14^2)$
$= \frac{1}{4} \times \frac{22}{7} \times (441 - 196)$
$= \frac{1}{4} \times \frac{22}{7} \times 245$
$= \frac{1}{4} \times 22 \times 35$
$= \frac{770}{4} = 192.5 \, \text{m}^2$.

(D) $1$. The side of the square $ABCD$ is $a = 60\, m$.
$2$. The diagonals of the square intersect at $O$. The distance from $O$ to any side is half the side length,i.e.,$h = 60/2 = 30\, m$.
$3$. The radius $r$ of the circle forming the segment is the distance from $O$ to a vertex (e.g.,$OA$). In a square of side $60\, m$,the diagonal is $60\sqrt{2}\, m$. Thus,$r = OA = (60\sqrt{2})/2 = 30\sqrt{2}\, m$.
$4$. The central angle $\theta$ subtended by the side $AD$ at the center $O$ is $90^\circ$ (since diagonals of a square are perpendicular).
$5$. The area of one circular segment is given by: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 - \text{Area of } \triangle OAD$.
$6$. $\text{Area of } \triangle OAD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 60 \times 30 = 900\, m^2$.
$7$. $\text{Area of segment} = (\frac{90}{360} \times 3.14 \times (30\sqrt{2})^2) - 900 = (0.25 \times 3.14 \times 1800) - 900 = 1413 - 900 = 513\, m^2$.
$8$. Since there are two such flower beds,the total area is $2 \times 513 = 1026\, m^2$.

(A) The side of the square $ABCD$ is $a = 42 \ cm$.
Area of the square $ABCD = a^2 = 42^2 = 1764 \ cm^2$.
There are four semicircles drawn on the sides of the square,each with a diameter equal to the side of the square,$d = 42 \ cm$. Therefore,the radius $r = 21 \ cm$.
The area of the four semicircles is $4 \times (\frac{1}{2} \pi r^2) = 2 \pi r^2$.
Area $= 2 \times \frac{22}{7} \times 21 \times 21 = 2 \times 22 \times 3 \times 21 = 2772 \ cm^2$.
The shaded area is the area of the four semicircles minus the area of the square,as the overlapping regions are counted twice.
Shaded Area $= 2772 - 1764 = 1008 \ cm^2$.

(C) The shaded region is a sector of an annulus (a circular ring). The area of a sector of a circle with radius $r$ and central angle $\theta$ is given by $\frac{\theta}{360^\circ} \times \pi r^2$.
Here,the shaded region is the difference between the area of the sector of the larger circle (radius $R = 28 \, cm$) and the area of the sector of the smaller circle (radius $r = 21 \, cm$),both with the same central angle $\theta = 40^\circ$.
Area of shaded region = $\frac{\theta}{360^\circ} \times \pi R^2 - \frac{\theta}{360^\circ} \times \pi r^2$
Area = $\frac{\theta}{360^\circ} \times \pi (R^2 - r^2)$
Area = $\frac{40}{360} \times \frac{22}{7} \times (28^2 - 21^2)$
Area = $\frac{1}{9} \times \frac{22}{7} \times (784 - 441)$
Area = $\frac{1}{9} \times \frac{22}{7} \times 343$
Area = $\frac{1}{9} \times 22 \times 49$
Area = $\frac{1078}{9} \approx 119.777... \, cm^2$
Rounding to two decimal places,we get $119.78 \, cm^2$.

(C) The area of the shaded region is equal to the area of the semicircle minus the area of $\Delta ABC$.
Since $\overline{AC}$ is the diameter,$\angle ABC = 90^{\circ}$.
Radius $r = \frac{AC}{2} = \frac{35}{2} = 17.5 \, cm$.
Area of the semicircle $= \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 17.5 \times 17.5 = 481.25 \, cm^2$.
Area of $\Delta ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 21 \times 28 = 294 \, cm^2$.
Area of the shaded region $= 481.25 - 294 = 187.25 \, cm^2$.

(D) The radius of the outer circle is $R = 90\, m$.
The width of the road is $10\, m$,so the radius of the inner circle is $r = 90 - 10 = 80\, m$.
The area of the road is the difference between the area of the outer circle and the area of the inner circle.
Area of the road $= \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$.
Area $= 3.14 \times (90^2 - 80^2)$.
Area $= 3.14 \times (8100 - 6400)$.
Area $= 3.14 \times 1700$.
Area $= 5338\, m^2$.

(N/A) Given: Radius $r = 11.2 \, cm$,Central angle $\theta = 90^{\circ}$.
Area of the circle $A = \pi r^2 = \frac{22}{7} \times 11.2 \times 11.2 = 394.24 \, cm^2$.
Area of the minor sector $= \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{90}{360} \times 394.24 = \frac{1}{4} \times 394.24 = 98.56 \, cm^2$.
Area of the major sector $=$ Area of circle $-$ Area of minor sector $= 394.24 - 98.56 = 295.68 \, cm^2$.
Area of the minor segment $=$ Area of minor sector $-$ Area of $\triangle OAB = 98.56 - (\frac{1}{2} \times r^2 \times \sin 90^{\circ}) = 98.56 - (0.5 \times 11.2 \times 11.2 \times 1) = 98.56 - 62.72 = 35.84 \, cm^2$.

(B) The length of the minute hand $(r)$ is $10.5 \, cm$.
The time interval between $2.25 \, PM$ and $2.40 \, PM$ is $15 \, \text{minutes}$.
$A$ minute hand completes a full circle $(360^{\circ})$ in $60 \, \text{minutes}$.
Therefore,the angle swept in $15 \, \text{minutes}$ is $\theta = (15/60) \times 360^{\circ} = 90^{\circ}$.
The area swept by the minute hand is the area of a sector with radius $r = 10.5 \, cm$ and angle $\theta = 90^{\circ}$.
Area $= (\theta / 360^{\circ}) \times \pi r^2 = (90^{\circ} / 360^{\circ}) \times (22/7) \times 10.5 \times 10.5$.
Area $= (1/4) \times (22/7) \times 10.5 \times 10.5 = (1/4) \times 22 \times 1.5 \times 10.5$.
Area $= 0.25 \times 22 \times 15.75 = 5.5 \times 15.75 = 86.625 \, cm^2$.

(C) The area of a circle is given by $A = \pi r^{2}$.
The area of a minor sector subtending an angle $\theta$ at the centre is given by the formula $\frac{\theta}{360} \times \pi r^{2}$.
The major sector is the remaining part of the circle.
Therefore,the area of the major sector = (Area of the circle) - (Area of the minor sector).
Area of major sector = $\pi r^{2} - \frac{\pi r^{2} \theta}{360}$.

(D) The area of a sector of a circle is given by the formula $A = \frac{\theta}{360^\circ} \times \pi r^2$,where $\theta$ is the central angle in degrees.
We know that the length of an arc $l$ is given by $l = \frac{\theta}{360^\circ} \times 2 \pi r$.
From this,we can express the term $\frac{\theta}{360^\circ}$ as $\frac{l}{2 \pi r}$.
Substituting this into the area formula: $A = \left( \frac{l}{2 \pi r} \right) \times \pi r^2$.
Simplifying the expression: $A = \frac{l \times \pi r^2}{2 \pi r} = \frac{1}{2} r l$.

(A) The formula for the area of a sector $(A)$ given the radius $(r)$ and the arc length $(l)$ is $A = \frac{1}{2} \times l \times r$.
Given:
Radius $(r)$ = $20\, cm$
Arc length $(l)$ = $10\, cm$
Substituting the values into the formula:
$A = \frac{1}{2} \times 10\, cm \times 20\, cm$
$A = 5\, cm \times 20\, cm$
$A = 100\, cm^2$.
Therefore,the area of the sector is $100\, cm^2$.

(B) Let the radii of the two circles be $r_1$ and $r_2$. Given that $r_1 : r_2 = 4 : 5$.
The area of a circle is given by the formula $A = \pi r^2$.
Therefore,the ratio of their areas is $\frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.
Substituting the given ratio: $\left(\frac{4}{5}\right)^2 = \frac{16}{25}$.
Thus,the ratio of their areas is $16 : 25$.

(C) The area of a sector of a circle with radius $r$ and central angle $\theta$ is given by the formula: $A = \frac{\theta}{360^{\circ}} \times \pi r^2$.
Let the two sectors have angles $\theta_1 = 15^{\circ}$ and $\theta_2 = 90^{\circ}$.
The ratio of the areas of the two sectors is:
$\frac{A_1}{A_2} = \frac{\frac{\theta_1}{360^{\circ}} \times \pi r^2}{\frac{\theta_2}{360^{\circ}} \times \pi r^2} = \frac{\theta_1}{\theta_2}$.
Substituting the given values:
$\frac{A_1}{A_2} = \frac{15^{\circ}}{90^{\circ}} = \frac{1}{6}$.
Thus,the ratio of the areas of the sectors is $1: 6$.

(D) Let the radius of the original circle be $r$.
The area of the original circle is $A_1 = \pi r^2$.
If the radius is doubled,the new radius becomes $R = 2r$.
The area of the new circle is $A_2 = \pi R^2 = \pi (2r)^2 = 4 \pi r^2$.
Comparing the two areas,$A_2 = 4 A_1$.
Therefore,the area becomes $4$ times the area of the original circle.

(A) Let the radii of the two circles be $r_1$ and $r_2$ respectively.
The area of a circle is given by the formula $A = \pi r^2$.
The ratio of the areas is given as $\frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{25}{36}$.
Taking the square root of both sides,we get $\frac{r_1}{r_2} = \sqrt{\frac{25}{36}} = \frac{5}{6}$.
The circumference of a circle is given by $C = 2\pi r$.
The ratio of their circumferences is $\frac{C_1}{C_2} = \frac{2\pi r_1}{2\pi r_2} = \frac{r_1}{r_2}$.
Therefore,the ratio of their circumferences is $5: 6$.

(B) The radius of the circle is $r = 8.4 \, cm$.
Since the two radii are perpendicular to each other,the central angle of the minor sector is $\theta = 90^\circ$.
The formula for the area of a sector is $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$.
Substituting the values: $\text{Area} = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (8.4)^2$.
$\text{Area} = \frac{1}{4} \times \frac{22}{7} \times 8.4 \times 8.4$.
$\text{Area} = \frac{1}{4} \times 22 \times 1.2 \times 8.4$.
$\text{Area} = 5.5 \times 1.2 \times 8.4 = 55.44 \, cm^2$.

(C) The length of the minute hand is the radius of the circle,$r = 14\, cm$.
In $60$ minutes,the minute hand completes one full rotation $(360^{\circ})$.
In $1$ minute,the angle swept is $\frac{360^{\circ}}{60} = 6^{\circ}$.
In $10$ minutes,the angle swept $(\theta)$ is $10 \times 6^{\circ} = 60^{\circ}$.
The area of the sector swept is given by the formula: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^2$.
Substituting the values: $\text{Area} = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14$.
$\text{Area} = \frac{1}{6} \times 22 \times 2 \times 14 = \frac{1}{3} \times 22 \times 28 = \frac{616}{6} \approx 102.67\, cm^2$.

(D) For a triangle inscribed in a semicircle,the base of the triangle is the diameter of the semicircle,and the height is the radius of the semicircle.
Given,diameter $d = 30 \, cm$.
Therefore,radius $r = \frac{d}{2} = \frac{30}{2} = 15 \, cm$.
The base of the triangle $b = 30 \, cm$.
The maximum height of the triangle is equal to the radius of the semicircle,$h = 15 \, cm$.
Area of the triangle $= \frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times 30 \times 15 = 15 \times 15 = 225 \, cm^{2}$.

(A) Given,the area of the circle $A = 3850 \, cm^2$.
We know that the area of a circle is given by $A = \pi r^2$.
So,$\pi r^2 = 3850$.
Taking $\pi = \frac{22}{7}$,we have $\frac{22}{7} \times r^2 = 3850$.
$r^2 = \frac{3850 \times 7}{22} = 175 \times 7 = 1225$.
$r = \sqrt{1225} = 35 \, cm$.
The length of an arc subtending an angle $\theta$ at the centre is given by $L = \frac{\theta}{360^\circ} \times 2 \pi r$.
Here,$\theta = 90^\circ$ (right angle).
$L = \frac{90^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 35$.
$L = \frac{1}{4} \times 2 \times 22 \times 5 = \frac{220}{4} = 55 \, cm$.

(B) Given: Radius $r = 20\, cm$ and Area of sector $A = 150\, cm^2$.
The formula for the area of a sector is $A = \frac{1}{2} \times l \times r$,where $l$ is the length of the arc.
Substituting the given values into the formula:
$150 = \frac{1}{2} \times l \times 20$
$150 = l \times 10$
$l = \frac{150}{10} = 15\, cm$.
Therefore,the length of the arc is $15\, cm$.

(C) Given,the circumference of the circle is $C = 88 \, cm$.
Using the formula $C = 2 \pi r$,we have $2 \times \frac{22}{7} \times r = 88$.
$r = \frac{88 \times 7}{44} = 14 \, cm$.
The diameter of the circle is $d = 2r = 28 \, cm$.
When a square is inscribed in a circle,the diagonal of the square is equal to the diameter of the circle.
Let the side of the square be $a$. Then,the diagonal is $a \sqrt{2}$.
Equating the diagonal to the diameter: $a \sqrt{2} = 28$.
$a = \frac{28}{\sqrt{2}} = \frac{28 \times \sqrt{2}}{2} = 14 \sqrt{2} \, cm$.

(D) When a square is inscribed in a circle,the diagonal of the square is equal to the diameter of the circle.
Given radius $r = 70 \, cm$.
Diameter of the circle $d = 2 \times r = 2 \times 70 = 140 \, cm$.
Let the side of the square be $a$.
The diagonal of the square is $a\sqrt{2}$.
Therefore,$a\sqrt{2} = 140$.
$a = \frac{140}{\sqrt{2}} = 70\sqrt{2} \, cm$.
The area of the square is $a^2 = (70\sqrt{2})^2 = 4900 \times 2 = 9800 \, cm^2$.

(C) The length of the minute hand is the radius of the circle,$r = 7\,cm$.
$A$ minute hand completes a full rotation $(360^\circ)$ in $60$ minutes.
In $1$ minute,the angle swept is $\frac{360^\circ}{60} = 6^\circ$.
In $20$ minutes,the angle swept by the minute hand is $\theta = 20 \times 6^\circ = 120^\circ$.
The area of the sector swept is given by the formula: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$.
Substituting the values: $\text{Area} = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 7 \times 7$.
$\text{Area} = \frac{1}{3} \times 22 \times 7 = \frac{154}{3}\,cm^2$.

(D) The area of a sector is given by the formula $A = \frac{1}{2} \times l \times r$,where $l$ is the arc length and $r$ is the radius of the circle.
Given,the radius $r = 30 \, cm$ and the area $A = 300 \, cm^2$.
Substituting the values into the formula:
$300 = \frac{1}{2} \times l \times 30$
$300 = 15 \times l$
$l = \frac{300}{15}$
$l = 20 \, cm$.
Thus,the length of the arc is $20 \, cm$.

(A) The area of a sector of a circle is given by the formula $A = \frac{\theta}{360^\circ} \times \pi r^2$,where $r$ is the radius and $\theta$ is the central angle.
Let the radii of the two circles be $r_1$ and $r_2$,and the central angles be $\theta_1$ and $\theta_2$.
Given: $\frac{r_1}{r_2} = \frac{2}{3}$ and $\frac{\theta_1}{\theta_2} = \frac{5}{2}$.
The ratio of the areas of the two sectors is $\frac{A_1}{A_2} = \frac{\frac{\theta_1}{360^\circ} \times \pi r_1^2}{\frac{\theta_2}{360^\circ} \times \pi r_2^2} = \left( \frac{\theta_1}{\theta_2} \right) \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the given values: $\frac{A_1}{A_2} = \left( \frac{5}{2} \right) \times \left( \frac{2}{3} \right)^2 = \left( \frac{5}{2} \right) \times \left( \frac{4}{9} \right) = \frac{20}{18} = \frac{10}{9}$.
Thus,the ratio of the areas of the sectors is $10:9$.

(B) The formula for the area of a sector of a circle given the radius $r$ and arc length $l$ is given by $A = \frac{1}{2} \times l \times r$.
Given:
Radius $r = 50 \, cm$
Arc length $l = 20 \, cm$
Substituting the values into the formula:
$A = \frac{1}{2} \times 20 \times 50$
$A = 10 \times 50$
$A = 500 \, cm^2$.
Thus,the area of the sector is $500 \, cm^2$.

(C) The side length of the square plot $ABCD$ is $l = 50 \, m$.
Area of the square plot $ABCD = l^2 = 50 \times 50 = 2500 \, m^2$.
For the flower beds formed at each vertex,the radius is $r = 10 \, m$ and the angle of the sector is $\theta = 90^\circ$ (since it is a square).
Area of one flower bed (sector) $= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times 3.14 \times 10^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5 \, m^2$.
Total area of $4$ flower beds $= 4 \times 78.5 = 314 \, m^2$.
Area of the plot excluding the flower beds $=$ Area of the square plot $-$ Total area of the flower beds.
$= 2500 - 314 = 2186 \, m^2$.
Thus,the area of the plot excluding the flower beds is $2186 \, m^2$.

(D) The shaded portion is the union of two congruent segments formed by the arcs $\widehat{APC}$ and $\widehat{AQC}$ within the square $ABCD$.
Area of segment $\widehat{APC}$ (with respect to sector $BAPC$):
Area $= \text{Area of sector } BAPC - \text{Area of } \Delta ABC$
$= \frac{90^\circ}{360^\circ} \times \pi \times (21)^2 - \frac{1}{2} \times 21 \times 21$
$= \frac{1}{4} \times \frac{22}{7} \times 441 - 220.5$
$= 346.5 - 220.5 = 126 \ cm^2$.
Since the two segments are congruent,the total shaded area is:
Area $= 2 \times 126 = 252 \ cm^2$.

(A) For the sector $OACB$,the radius $r = 35 \text{ cm}$ and the measure of the angle $\theta = 90^{\circ}$ (since $\overline{OA} \perp \overline{OB}$).
Area of sector $OACB = \frac{\pi r^2 \theta}{360^{\circ}}$
$= \frac{22}{7} \times \frac{35 \times 35 \times 90^{\circ}}{360^{\circ}}$
$= 962.5 \text{ cm}^2$
In $\Delta ODA$,$\angle O = 90^{\circ}$.
Area of $\Delta ODA = \frac{1}{2} \times OD \times OA$
$= \frac{1}{2} \times 12 \times 35$
$= 210 \text{ cm}^2$
Area of the shaded region = Area of sector $OACB - \text{Area of } \Delta ODA$
$= 962.5 - 210$
$= 752.5 \text{ cm}^2$
Thus,the area of the shaded region is $752.5 \text{ cm}^2$.

(C) The circumference of a circle is $2\pi r$,which corresponds to an angle of $360^{\circ}$ at the center.
Using the unitary method,the length of an arc $l$ subtending an angle $\theta$ at the center is given by the ratio of the angle to the total angle multiplied by the total circumference.
$l = \frac{\theta}{360^{\circ}} \times 2\pi r$
$l = \frac{\theta \times 2\pi r}{360^{\circ}}$
$l = \frac{\pi r \theta}{180^{\circ}}$
Therefore,the correct formula is $l = \frac{\pi r \theta}{180^{\circ}}$.

(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) 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 $2 \times r$.
$d = 2 \times 7 = 14\,cm$.