Mix Examples - Areas Related to Circles Questions in English

(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 $\text{Area} \times \text{Rate} = 2000 \times 15 = ₹ 30,000$.

(D) The area of the circular sheet is given by $A_{circle} = \pi r^2 = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386\, cm^2$.
$A$ regular hexagon with side length $a = 21\, cm$ consists of $6$ equilateral triangles with side $21\, cm$.
The area of the regular hexagon is $A_{hexagon} = 6 \times \frac{\sqrt{3}}{4} a^2 = \frac{3\sqrt{3}}{2} \times (21)^2$.
Substituting $\sqrt{3} = 1.73$ and $a = 21$:
$A_{hexagon} = \frac{3 \times 1.73}{2} \times 441 = 1.5 \times 1.73 \times 441 = 2.595 \times 441 = 1144.395\, cm^2$.
The area of the remaining sheet is $A_{remaining} = A_{circle} - A_{hexagon} = 1386 - 1144.395 = 241.605\, cm^2$.

(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 $37.68 \text{ cm}^2$.

(A) $1$. The diagonal of the rectangle $ABCD$ is the diameter of the circle.
$2$. Using the Pythagorean theorem in $\triangle ABC$,the diagonal $AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, cm$.
$3$. The diameter of the circle is $10 \, cm$,so the radius $r = 5 \, cm$.
$4$. The area of the circle is $\pi r^2 = 3.14 \times 5^2 = 3.14 \times 25 = 78.5 \, cm^2$.
$5$. The area of the rectangle $ABCD$ is $AB \times BC = 8 \times 6 = 48 \, cm^2$.
$6$. The area of the shaded region is the area of the circle minus the area of the rectangle: $78.5 - 48 = 30.5 \, cm^2$.

(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 $= 2119.25 - 641.67 = 1477.58 \, cm^2$.

(C) $1$. Area of rectangle $ABCD = \text{length} \times \text{breadth} = 20 \times 14 = 280 \, cm^2$.
$2$. Area of the semicircle with diameter $BC = 14 \, cm$ (radius $r = 7 \, cm$): $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77 \, cm^2$.
$3$. Area of the sector with centre $A$ and radius $AD = 14 \, cm$ (angle $\theta = 90^{\circ}$): $\frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \, cm^2$.
$4$. Area of the remaining sheet = Area of rectangle - (Area of semicircle + Area of sector) = $280 - (77 + 154) = 280 - 231 = 49 \, cm^2$.

(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)
$= \frac{90^\circ}{360^\circ} \times \pi \times R^2 - \frac{90^\circ}{360^\circ} \times \pi \times r^2$
$= \frac{1}{4} \times \pi \times (R^2 - r^2)$
$= \frac{1}{4} \times \frac{22}{7} \times (21^2 - 14^2)$
$= \frac{1}{4} \times \frac{22}{7} \times (441 - 196)$
$= \frac{1}{4} \times \frac{22}{7} \times 245$
$= \frac{1}{4} \times 22 \times 35$
$= \frac{770}{4} = 192.5 \, \text{m}^2$.

(C) Given: $AB = BC = 14 \text{ cm}$ and $\angle B = 90^{\circ}$.
First,find the length of the hypotenuse $AC$ using the Pythagorean theorem:
$AC^2 = AB^2 + BC^2 = 14^2 + 14^2 = 196 + 196 = 392$.
$AC = \sqrt{392} = 14\sqrt{2} \text{ cm}$.
The radius of the semicircle with diameter $AC$ is $r = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \text{ cm}$.
Area of the semicircle with diameter $AC = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 = \frac{1}{2} \times \frac{22}{7} \times 98 = 11 \times 14 = 154 \text{ cm}^2$.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2$.
Area of the segment of the circle (bounded by chord $AC$ and arc $APC$) = Area of sector $BAPC$ - Area of $\Delta ABC$.
Area of sector $BAPC = \frac{90}{360} \times \pi \times (14)^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = 154 \text{ cm}^2$.
Area of segment $APC = 154 - 98 = 56 \text{ cm}^2$.
The shaded region is the area of the semicircle minus the area of the segment $APC$.
Area of shaded region = $154 - 56 = 98 \text{ cm}^2$.

(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 $= 2772 - 1764 = 1008 \ cm^2$.

(B) Given: Radius of the large circle $R = 7\, cm$.
Diameter of the small circle is $OD = R = 7\, cm$,so its radius $r = 3.5\, cm$.
Area of the small shaded circle = $\pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5\, cm^2$.
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 semi-circle $ACB$ = $\frac{1}{2} \pi R^2 = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77\, cm^2$.
Area of the shaded segment (excluding the small circle) = Area of semi-circle $ACB$ - Area of $\triangle ABC = 77 - 49 = 28\, cm^2$.
Total shaded area = Area of small circle + Area of shaded segment = $38.5 + 28 = 66.5\, cm^2$.

(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 $119.78 \, cm^2$.

(D) The side length of the square $ABCD$ is $s = 35 \, cm$.
The area of the square $ABCD = s^2 = 35^2 = 1225 \, cm^2$.
There are two semicircles drawn on sides $\overline{AB}$ and $\overline{CD}$ with diameter $d = 35 \, cm$,so the radius $r = \frac{35}{2} = 17.5 \, cm$.
The area of two semicircles = $2 \times (\frac{1}{2} \pi r^2) = \pi r^2 = \frac{22}{7} \times 17.5 \times 17.5 = 22 \times 2.5 \times 17.5 = 962.5 \, cm^2$.
The area of the shaded region = (Area of square) - (Area of two semicircles) = $1225 - 962.5 = 262.5 \, cm^2$.

(A) The radius of each circle is $r = 21\, cm$.
Diameter of each circle is $d = 2r = 2 \times 21 = 42\, cm$.
Since there are $3$ circles along each side of the square,the side length of the square is $S = 3 \times d = 3 \times 42 = 126\, cm$.
Area of the square $ABCD = S^2 = 126^2 = 15876\, cm^2$.
Area of one circle = $\pi r^2 = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386\, cm^2$.
Area of $9$ circles = $9 \times 1386 = 12474\, cm^2$.
Area of the region without design = Area of square - Area of $9$ circles.
Area = $15876 - 12474 = 3402\, cm^2$.

(B) The figure consists of one large semicircle with diameter $AB$ and two smaller semicircles with diameters $OA$ and $OB$.
Given $OA = 70\, cm$ and $OB = 70\, cm$,the diameter of the large semicircle is $AB = OA + OB = 70 + 70 = 140\, cm$.
The radius of the large semicircle is $R = 140 / 2 = 70\, cm$.
The radius of each smaller semicircle is $r = 70 / 2 = 35\, cm$.
The area of the large semicircle is $\frac{1}{2} \pi R^2 = \frac{1}{2} \times \frac{22}{7} \times 70 \times 70 = 11 \times 10 \times 70 = 7700\, cm^2$.
The area of the two smaller semicircles is $2 \times (\frac{1}{2} \pi r^2) = \pi r^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3850\, cm^2$.
The total area of the figure is the sum of the area of the large semicircle and the two smaller semicircles: $7700 + 3850 = 11550\, cm^2$.

(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 $( heta)$ 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 $a^2 = (70\sqrt{2})^2 = 4900 \times 2 = 9800 \, cm^2$.

(A) Let the initial radius of the circle be $r$. The initial area is $A_1 = \pi r^2$.
If the radius is increased by $10 \%$,the new radius $r'$ becomes $r + 0.10r = 1.1r$.
The new area $A_2$ is $\pi (1.1r)^2 = \pi (1.21r^2) = 1.21 \pi r^2$.
The increase in area is $A_2 - A_1 = 1.21 \pi r^2 - \pi r^2 = 0.21 \pi r^2$.
The percentage increase in area is $\frac{0.21 \pi r^2}{\pi r^2} \times 100 \% = 21 \%$.

(B) The diameter of the semicircle is $d = 50 \, cm$,so the radius is $r = 25 \, cm$.
For a triangle inscribed in a semicircle,the base of the triangle is the diameter of the semicircle $(b = 50 \, cm)$.
The height of the triangle is the radius of the semicircle $(h = r = 25 \, cm)$ because the maximum height is achieved when the vertex is at the midpoint of the arc.
The area of the triangle is given by the formula $A = \frac{1}{2} \times \text{base} \times \text{height}$.
Substituting the values: $A = \frac{1}{2} \times 50 \times 25 = 25 \times 25 = 625 \, cm^{2}$.

(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 $l = \frac{\pi r \theta}{180^{\circ}}$.