Measurement of Area Questions in English

(A) Let the length of the field be $6x \, m$ and the breadth be $5x \, m$.
Given that the area of the rectangular field is $27000 \, m^2$.
Since the area of a rectangle is $\text{length} \times \text{breadth}$,we have:
$6x \times 5x = 27000$
$30x^2 = 27000$
$x^2 = \frac{27000}{30} = 900$
$x = \sqrt{900} = 30$
Therefore,the length is $6 \times 30 = 180 \, m$ and the breadth is $5 \times 30 = 150 \, m$.

(C) The area of a sector of a circle can be calculated using the formula: $\text{Area} = \frac{1}{2} \times \text{arc length} \times \text{radius}$.
Given,$\text{arc length} = 3.5 \, cm$ and $\text{radius} = 5 \, cm$.
Substituting these values into the formula:
$\text{Area} = \frac{1}{2} \times 3.5 \times 5 \, cm^2$.
$\text{Area} = \frac{17.5}{2} \, cm^2$.
$\text{Area} = 8.75 \, cm^2$.

(C) Let the width of the plot be $w \, m$. Then the length of the plot is $l = 2w \, m$.
Area of the plot $= l \times w = (2w) \times w = 2w^2 \, m^2$.
Given that a square piece of land of area $150 \, m^2$ occupies $\frac{1}{3}$ of the total area of the plot.
Therefore,$\frac{1}{3} \times (\text{Area of the plot}) = 150 \, m^2$.
Area of the plot $= 150 \times 3 = 450 \, m^2$.
Equating the two expressions for the area: $2w^2 = 450$.
$w^2 = 225$.
$w = \sqrt{225} = 15 \, m$.
The length of the plot is $l = 2w = 2 \times 15 = 30 \, m$.

(B) The length of a wire bent into a semi-circle is equal to the perimeter of the semi-circle.
The perimeter of a semi-circle is given by the formula: $P = \pi r + 2r$,where $r$ is the radius.
Given $r = 7 \, cm$ and taking $\pi = \frac{22}{7}$:
$P = (\frac{22}{7} \times 7) + (2 \times 7)$
$P = 22 + 14$
$P = 36 \, cm$.
Therefore,the length of the wire is $36 \, cm$.

(B) Let the radius of the outer circle be $R$ and the radius of the inner circle be $r$.
The width of the road is the difference between the radii,which is $(R - r)$.
The circumference of the outer circle is $2 \pi R$ and the circumference of the inner circle is $2 \pi r$.
Given that the difference between the circumferences is $66 \, m$:
$2 \pi R - 2 \pi r = 66$
$2 \pi (R - r) = 66$
$2 \times \frac{22}{7} \times (R - r) = 66$
$(R - r) = 66 \times \frac{7}{22} \times \frac{1}{2}$
$(R - r) = 3 \times 7 \times \frac{1}{2} = \frac{21}{2} = 10.5 \, m$.
Thus,the width of the road is $10.5 \, m$.

(A) Let the side of the square be $a$.
Given that the area of the square is $a^2 = 50$.
The diagonal of the square $d$ is given by $d = a\sqrt{2}$.
Substituting $a^2 = 50$,we get $a = \sqrt{50} = 5\sqrt{2}$.
Thus,the diagonal $d = (5\sqrt{2}) \times \sqrt{2} = 10$ units.
The circle is drawn on the diagonal as its diameter,so the diameter of the circle is $10$ units.
The radius $r$ of the circle is $d/2 = 10/2 = 5$ units.
The area of the circle is $\pi r^2 = \pi (5)^2 = 25\pi$ $sq$ units.

(C) The area of the rectangular sheet is given by $Area_{rect} = length \times width = 4 \, cm \times 2 \, cm = 8 \, cm^{2}$.
To cut a circle of the greatest possible area from a rectangle,the diameter of the circle must be equal to the smaller side of the rectangle.
Here,the smaller side is $2 \, cm$,so the diameter $d = 2 \, cm$,which means the radius $r = 1 \, cm$.
The area of this circle is $Area_{circle} = \pi r^{2} = \pi \times (1)^{2} = \pi \, cm^{2}$.
The area of the remaining portion is the difference between the area of the rectangle and the area of the circle.
$Area_{remaining} = Area_{rect} - Area_{circle} = 8 - \pi \, cm^{2}$.

(B) Let the radius of the new circle be $R$. The area of the new circle is equal to the sum of the areas of the three given circles.
$\pi R^{2} = \pi(22)^{2} + \pi(19)^{2} + \pi(8)^{2}$
Dividing both sides by $\pi$,we get:
$R^{2} = 22^{2} + 19^{2} + 8^{2}$
$R^{2} = 484 + 361 + 64$
$R^{2} = 909$
$R = \sqrt{909} \approx 30.15$
Rounding to the nearest $cm$,we get $R = 30\, cm$.

(A) Let the original length be $l$ and the original width be $b$. The original area is $A = l \times b$.
The length is increased by $33.33 \%$,which is equivalent to an increase of $\frac{1}{3}$.
New length $l' = l + \frac{1}{3}l = \frac{4}{3}l$.
Let the new width be $b'$. To maintain the same area,$l \times b = l' \times b'$.
$l \times b = \frac{4}{3}l \times b'$.
$b' = \frac{3}{4}b = 0.75b$.
The reduction in width is $b - b' = b - 0.75b = 0.25b$.
Percentage decrease in width $= \frac{0.25b}{b} \times 100 \% = 25 \%$.

(D) The area of a rectangle is calculated as $\text{Length} \times \text{Width}$.
Given,$\text{Length} = 6.4 \, m$ and $\text{Width} = 2.5 \, m$.
$\text{Area of rectangle} = 6.4 \, m \times 2.5 \, m = 16 \, m^2$.
According to the problem,the area of the square is equal to the area of the rectangle.
Therefore,$\text{Area of square} = 16 \, m^2$.
Since the $\text{Area of a square} = \text{side}^2$,we have $\text{side}^2 = 16 \, m^2$.
Taking the square root of both sides,$\text{side} = \sqrt{16} \, m = 4 \, m$.
Thus,each side of the square measures $4 \, m$.

(D) The diameter of the larger circle is $10\, cm$,so its radius $R = 10/2 = 5\, cm$.
The diameter of the smaller circle is $8\, cm$,so its radius $r = 8/2 = 4\, cm$.
The area of the region between the two concentric circles (annulus) is given by the formula: $\text{Area} = \pi R^2 - \pi r^2$.
Substituting the values: $\text{Area} = \pi(5)^2 - \pi(4)^2$.
$\text{Area} = 25\pi - 16\pi = 9\pi\, cm^2$.

(B) The length of the rectangular room is $4\, m.$
Since the room is partitioned into two equal square rooms,the length of the rectangle must be divided into two equal segments to form the sides of the squares.
Each square will have a side length of $4\, m / 2 = 2\, m.$
The partition is the line segment that divides the rectangle,which corresponds to the side of the square.
Therefore,the length of the partition is $2\, m.$

(C) Let the length be $a \, m$ and the breadth be $b \, m$.
The perimeter of a rectangle is given by $2(a + b) = 46 \, m$.
Therefore,$a + b = 23 \, m$.
The area of the rectangle is given by $a \times b = 120 \, m^{2}$.
The length of the diagonal $d$ is given by the formula $d = \sqrt{a^{2} + b^{2}}$.
Using the algebraic identity $(a + b)^{2} = a^{2} + b^{2} + 2ab$,we can write $a^{2} + b^{2} = (a + b)^{2} - 2ab$.
Substituting the known values: $a^{2} + b^{2} = (23)^{2} - 2(120) = 529 - 240 = 289$.
Thus,the diagonal $d = \sqrt{289} = 17 \, m$.

(C) The perimeter of a square is given by the formula $P = 4 \times \text{side}$.
Given the side is $22 \, m$,the perimeter $P = 4 \times 22 \, m = 88 \, m$.
Let the radius of the circle be $r$. The circumference of the circle is given by $C = 2 \pi r$.
According to the problem,the circumference of the circle is equal to the perimeter of the square,so $2 \pi r = 88 \, m$.
Using $\pi \approx \frac{22}{7}$,we have $2 \times \frac{22}{7} \times r = 88$.
$r = 88 \times \frac{7}{2 \times 22} = 88 \times \frac{7}{44} = 2 \times 7 = 14 \, m$.
Therefore,the radius of the circle is $14 \, m$.

(C) The perimeter of each shape is $P = 132 \, cm$.
$1$. For an equilateral triangle: $3a = 132 \implies a = 44 \, cm$. Area $= \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 44^2 \approx 0.433 \times 1936 \approx 838.31 \, cm^2$.
$2$. For a square: $4a = 132 \implies a = 33 \, cm$. Area $= a^2 = 33^2 = 1089 \, cm^2$.
$3$. For a regular hexagon: $6a = 132 \implies a = 22 \, cm$. Area $= \frac{3\sqrt{3}}{2} a^2 = \frac{3\sqrt{3}}{2} \times 22^2 \approx 2.598 \times 484 \approx 1257.47 \, cm^2$.
$4$. For a circle: $2\pi r = 132 \implies r = \frac{132}{2 \times (22/7)} = 21 \, cm$. Area $= \pi r^2 = \frac{22}{7} \times 21^2 = 1386 \, cm^2$.
Comparing the areas: $838.31 < 1089 < 1257.47 < 1386$. Thus,the circle has the largest surface area.

(C) Let $r$ be the radius of the smaller circle.
Then,the radius of the larger circle is $R = 12r$.
The area of the smaller circle is $A_1 = \pi r^2$.
The area of the larger circle is $A_2 = \pi R^2 = \pi (12r)^2 = 144\pi r^2$.
To find how many times the area of the larger circle contains the area of the smaller circle,we calculate the ratio:
$\frac{A_2}{A_1} = \frac{144\pi r^2}{\pi r^2} = 144$.
Therefore,the area of the larger circle is $144$ times the area of the smaller circle.

(B) Let $a$ be the side of the square and $r$ be the radius of the circle.
Given that the circumference of the circle is equal to the perimeter of the square:
$2 \pi r = 4a$
$r = \frac{4a}{2 \pi} = \frac{2a}{\pi}$
Now,the area of the circle is $A_c = \pi r^2$ and the area of the square is $A_s = a^2$.
The ratio of the area of the circle to the area of the square is:
$\frac{A_c}{A_s} = \frac{\pi r^2}{a^2} = \frac{\pi (\frac{2a}{\pi})^2}{a^2}$
$= \frac{\pi \cdot \frac{4a^2}{\pi^2}}{a^2} = \frac{4}{\pi}$
Substituting $\pi = \frac{22}{7}$:
$= \frac{4}{22/7} = \frac{4 \times 7}{22} = \frac{28}{22} = \frac{14}{11}$
Thus,the ratio is $14: 11$.

(C) Let the breadth of the rectangular plot be $x \ ft$. Then,the length of the plot is $3x \ ft$.
The area of the plot is given by $\text{Length} \times \text{Breadth} = 3x \times x = 3x^2 \ ft^2$.
It is given that a playground of $1200 \ ft^2$ occupies $\frac{1}{4}$ of the total area of the plot.
Therefore,$\frac{1}{4} \times \text{Total Area} = 1200 \ ft^2$.
$\text{Total Area} = 1200 \times 4 = 4800 \ ft^2$.
Equating the two expressions for the area: $3x^2 = 4800$.
$x^2 = \frac{4800}{3} = 1600$.
$x = \sqrt{1600} = 40 \ ft$.
Since the length is $3x$,the length $= 3 \times 40 = 120 \ ft$.

(B) The area of a square is given by the formula $\text{Area} = \text{side}^2$.
Let the sides of squares $s_{1}$ and $s_{2}$ be $a_{1}$ and $a_{2}$ respectively.
The ratio of their areas is given as $\frac{a_{1}^2}{a_{2}^2} = \frac{4}{25}$.
Taking the square root of both sides,we get the ratio of their sides: $\frac{a_{1}}{a_{2}} = \sqrt{\frac{4}{25}} = \frac{2}{5}$.
Given that the side of $s_{1}$ $(a_{1})$ is $6 \text{ cm}$,we can set up the equation: $\frac{6}{a_{2}} = \frac{2}{5}$.
Solving for $a_{2}$: $a_{2} = \frac{6 \times 5}{2} = 3 \times 5 = 15 \text{ cm}$.
Therefore,the side of $s_{2}$ is $15 \text{ cm}$.

(B) The circumference of the wheel is given by $C = 2 \pi r = 2 \times \frac{22}{7} \times 70\, cm = 440\, cm$.
The distance traveled in $10$ revolutions is $10 \times 440\, cm = 4400\, cm = 44\, m$.
The time taken is $5\, seconds$.
The speed in $m/s$ is $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{44}{5}\, m/s = 8.8\, m/s$.
To convert the speed from $m/s$ to $km/h$,we multiply by $\frac{18}{5}$:
$\text{Speed} = 8.8 \times \frac{18}{5}\, km/h = 31.68\, km/h$.

(D) Let the length of the carpet be $l$ and the breadth be $b$. Given,$l \times b = 60 \, m^{2}$.
By the Pythagorean theorem,the diagonal $d = \sqrt{l^{2} + b^{2}}$,so $d^{2} = l^{2} + b^{2}$.
According to the problem,$d + l = 5b$,which implies $d = 5b - l$.
Squaring both sides,$d^{2} = (5b - l)^{2} = 25b^{2} + l^{2} - 10bl$.
Substituting $d^{2} = l^{2} + b^{2}$,we get $l^{2} + b^{2} = 25b^{2} + l^{2} - 10bl$.
Simplifying,$b^{2} = 25b^{2} - 10(60)$,which gives $24b^{2} = 600$.
Thus,$b^{2} = 25$,so $b = 5 \, m$.
Since $l \times b = 60$,$l = 60 / 5 = 12 \, m$.

(B) The playground consists of a rectangle and two semi-circles on its smaller sides.
Length of the rectangle $(l)$ $= 36 \, m$.
Width of the rectangle $(b)$ $= 24.5 \, m$.
Diameter of each semi-circle $= 24.5 \, m$.
Radius of each semi-circle $(r)$ $= \frac{24.5}{2} = 12.25 \, m$.
Area of the playground $=$ Area of the rectangle $+$ Area of two semi-circles
Area of the playground $= (l \times b) + \pi r^2$
Area of the playground $= (36 \times 24.5) + \left( \frac{22}{7} \times 12.25 \times 12.25 \right)$
Area of the playground $= 882 + \left( \frac{22}{7} \times 150.0625 \right)$
Area of the playground $= 882 + 471.625 = 1353.625 \, m^2$.

(C) The radius of the circle is $r = 50\, m$.
The circumference of the circle is $C = 2 \pi r = 2 \times 3.14 \times 50 = 314\, m$.
The total distance covered for $10$ rounds is $D = 10 \times C = 10 \times 314 = 3140\, m$.
The speed of the man is $v = 12\, km/h$. Converting this to $m/s$:
$v = 12 \times \frac{5}{18} = \frac{60}{18} = \frac{10}{3}\, m/s$.
The time taken in seconds is $t = \frac{D}{v} = \frac{3140}{10/3} = 3140 \times \frac{3}{10} = 314 \times 3 = 942\, s$.
To convert the time into minutes, divide by $60$:
$t = \frac{942}{60} = 15.7\, \text{minutes}$.

(D) The dimensions of the room are $5\, m$ and $8\, m$.
$A$ margin of $10\, cm$ $(0.1\, m)$ is left from each of the four walls.
Therefore,the length of the carpeted area $= 8\, m - (0.1\, m + 0.1\, m) = 8 - 0.2 = 7.8\, m$.
The width of the carpeted area $= 5\, m - (0.1\, m + 0.1\, m) = 5 - 0.2 = 4.8\, m$.
Area to be carpeted $= 7.8\, m \times 4.8\, m = 37.44\, m^2$.
Total cost of carpeting $= 37.44\, m^2 \times ₹ 18/m^2 = ₹ 673.92$.

(B) Let the length of the big rectangle be $L$ and the width be $W_1 = 2 \ m$.
Area of the big rectangle $= L \times W_1 = L \times 2 = 2L \ m^2$.
Let the length of the small rectangle be $L$ (as given,lengths are equal) and the width be $W_2$.
Area of the small rectangle $= L \times W_2$.
According to the problem,the area of the big rectangle is equal to the area of the small rectangle.
Therefore,$2L = L \times W_2$.
Dividing both sides by $L$ (assuming $L \neq 0$),we get $W_2 = 2 \ m$.
Thus,the width of the small rectangle is $2 \ m$.

(B) Let the original radius of the circle be $r$.
The original circumference $C_1 = 2 \pi r$.
If the radius is reduced by $40 \%$,the new radius $r' = r - 0.40r = 0.60r$.
The new circumference $C_2 = 2 \pi (0.60r) = 1.2 \pi r$.
The reduction in circumference is $C_1 - C_2 = 2 \pi r - 1.2 \pi r = 0.8 \pi r$.
The percentage reduction in circumference is given by $\frac{C_1 - C_2}{C_1} \times 100$.
Percentage reduction $= \frac{0.8 \pi r}{2 \pi r} \times 100 = 0.4 \times 100 = 40 \%$.

(D) The total area is the sum of the area of the square and the areas of the four semi-circles.
Area of the square $= a^{2}$.
Each side of the square acts as the diameter of a semi-circle,so the radius $r = \frac{a}{2}$.
Area of one semi-circle $= \frac{1}{2} \pi r^{2} = \frac{1}{2} \pi (\frac{a}{2})^{2} = \frac{\pi a^{2}}{8}$.
Since there are four such semi-circles,the total area of the four semi-circles $= 4 \times \frac{\pi a^{2}}{8} = \frac{\pi a^{2}}{2}$.
Therefore,the total area of the figure $= a^{2} + \frac{\pi a^{2}}{2}$.

(A) The area of the square field is $6050\, m^2$. Let the side of the square be $a$. Then $a^2 = 6050$.
The diagonal of a square is given by $d = a\sqrt{2}$.
Thus,$d^2 = 2a^2 = 2 \times 6050 = 12100\, m^2$.
Therefore,the length of the diagonal $d = \sqrt{12100} = 110\, m$.
The speed is given as $10\, m$ per $30$ seconds,which is $10\, m$ per $0.5$ minutes.
Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{110\, m}{10\, m / 0.5\, \text{min}} = 11 \times 0.5 = 5.5$ minutes.
This is equal to $5 \frac{1}{2}$ minutes.

(B) regular hexagon inscribed in a circle of radius $r$ consists of $6$ equilateral triangles,each with a side length equal to the radius $r$ of the circle.
Therefore,the length of each side of the regular hexagon is $r$.
The perimeter of a regular hexagon is the sum of the lengths of its $6$ sides.
Perimeter $= 6 \times r = 6r$.
Thus,the perimeter is $6r$.

(B) Let $AL$ and $BM$ be the two poles with heights $15\, m$ and $30\, m$ respectively,standing on the ground $AB = 36\, m$.
Draw a line $LN$ parallel to $AB$ such that $LN = AB = 36\, m$ and $N$ lies on $BM$.
Then,$BN = AL = 15\, m$.
Therefore,$MN = BM - BN = 30\, m - 15\, m = 15\, m$.
In the right-angled triangle $\triangle LNM$,by the Pythagorean theorem:
$LM^2 = LN^2 + MN^2$
$LM^2 = 36^2 + 15^2$
$LM^2 = 1296 + 225 = 1521$
$LM = \sqrt{1521} = 39\, m$.
Thus,the distance between their tops is $39\, m$.

(D) Let the breadth of the rectangular hall be $x \, m$.
According to the problem,the length is $\frac{4}{3}$ of its width,so length $= \frac{4x}{3} \, m$.
The area of a rectangle is given by $\text{length} \times \text{breadth} = \text{Area}$.
Substituting the values: $\frac{4x}{3} \times x = 300$.
$\frac{4x^2}{3} = 300$.
$x^2 = 300 \times \frac{3}{4} = 75 \times 3 = 225$.
$x = \sqrt{225} = 15 \, m$.
So,the breadth is $15 \, m$.
The length is $\frac{4}{3} \times 15 = 20 \, m$.
The difference between the length and the breadth is $20 \, m - 15 \, m = 5 \, m$.

(C) Let the original side of the equilateral triangle be $a$.
The original area $A = \frac{\sqrt{3}}{4} a^2$.
The new side length $a' = a + 1.5\% \text{ of } a = a(1 + 0.015) = 1.015a$.
The new area $A' = \frac{\sqrt{3}}{4} (a')^2 = \frac{\sqrt{3}}{4} (1.015a)^2 = \frac{\sqrt{3}}{4} a^2 (1.015)^2$.
Since $A' = A \times (1.015)^2$,we calculate $(1.015)^2 = 1.030225$.
So,$A' = 1.030225 A$.
The percentage increase in area $= \frac{A' - A}{A} \times 100 = (1.030225 - 1) \times 100 = 0.030225 \times 100 = 3.0225 \%.$

(C) The area grazed by the horse is the area of a circle with the rope length as the radius.
Initial radius $r = 12 \ m$.
Final radius $R = 23 \ m$.
The additional area grazed is the area of the annulus (the region between two concentric circles) given by $\pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \pi(R+r)(R-r)$.
Using $\pi = \frac{22}{7}$:
Additional area $= \frac{22}{7} \times (23 + 12) \times (23 - 12) \ m^2$
$= \frac{22}{7} \times 35 \times 11 \ m^2$
$= 22 \times 5 \times 11 \ m^2$
$= 110 \times 11 \ m^2$
$= 1210 \ m^2$.

(A) Let the original diagonal of the square be $d$.
The area of a square in terms of its diagonal is given by the formula: $\text{Area} = \frac{1}{2} d^2$.
Original area $A_1 = \frac{1}{2} d^2$.
If the diagonal is doubled,the new diagonal $d' = 2d$.
New area $A_2 = \frac{1}{2} (d')^2 = \frac{1}{2} (2d)^2 = \frac{1}{2} (4d^2) = 2d^2$.
Comparing the new area with the original area:
$A_2 = 4 \times (\frac{1}{2} d^2) = 4 A_1$.
Therefore,the area of the square becomes $4$-fold.

(C) Let the length of the rectangle be $L$ and its width be $B$. The original perimeter is $P_1 = 2(L + B)$.
When the sides are increased by $20 \%$,the new length becomes $L' = L + 0.2L = 1.2L$ and the new width becomes $B' = B + 0.2B = 1.2B$.
The new perimeter is $P_2 = 2(L' + B') = 2(1.2L + 1.2B)$.
Simplifying this,we get $P_2 = 2 \times 1.2(L + B) = 2.4(L + B)$.
The increase in perimeter is $\Delta P = P_2 - P_1 = 2.4(L + B) - 2(L + B) = 0.4(L + B)$.
The percentage increase in perimeter is $\frac{\Delta P}{P_1} \times 100 = \frac{0.4(L + B)}{2(L + B)} \times 100 = 0.2 \times 100 = 20 \%$.

(B) Let the radius of the outer circle be $R$ and the radius of the inner circle be $r$.
The width of the road is given by $(R - r)$.
The circumference of the outer circle is $2 \pi R$ and the circumference of the inner circle is $2 \pi r$.
Given that the difference between the circumferences is $44\, m$:
$2 \pi R - 2 \pi r = 44$
$2 \pi (R - r) = 44$
$R - r = \frac{44}{2 \pi}$
$R - r = \frac{44}{2 \times (22/7)}$
$R - r = \frac{44 \times 7}{44} = 7\, m$.
Thus,the width of the road is $7\, m$.

(A) The area of the square-shaped pond is given by $side^2 = 8 \, m \times 8 \, m = 64 \, m^2$.
Given that the area of the pond is $\frac{1}{8}$ of the area of the field,the area of the field is $8 \times 64 \, m^2 = 512 \, m^2$.
Let the length of the rectangular field be $L$ and its width be $W$.
According to the problem,$L = 2W$,which implies $W = \frac{L}{2}$.
The area of the rectangular field is $L \times W = L \times \frac{L}{2} = \frac{L^2}{2}$.
Equating this to the calculated area: $\frac{L^2}{2} = 512$.
$L^2 = 512 \times 2 = 1024$.
$L = \sqrt{1024} = 32 \, m$.

(B) The area of the circular disc is given by $A = \pi r^2 = 0.49 \pi \, m^2$.
By equating the two,we get $r^2 = 0.49$,which implies $r = 0.7 \, m$.
The circumference of the disc is $C = 2 \pi r = 2 \times \frac{22}{7} \times 0.7 = 4.4 \, m$.
The total distance covered is $1.76 \, km = 1760 \, m$.
The number of revolutions is given by the ratio of the total distance to the circumference: $\text{Number of revolutions} = \frac{1760}{4.4} = 400$.

(C) The radius of the circle is $r = 6\, cm$.
The area of the circle is given by the formula $A = \pi r^2$.
$A = \pi \times (6)^2 = 36\pi\, cm^2$.
Let the height of the triangle be $h\, cm$ and the base be $b = 6\, cm$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
According to the problem,the area of the triangle is equal to the area of the circle:
$\frac{1}{2} \times 6 \times h = 36\pi$.
$3 \times h = 36\pi$.
$h = \frac{36\pi}{3} = 12\pi\, cm$.

(B) Let the height of the wall be $h$ feet. The length of the ladder is also $h$ feet.
When the ladder is placed on a $2$-feet tall stool at a distance of $10$ feet from the wall,the vertical height covered by the ladder against the wall is $(h - 2)$ feet.
This forms a right-angled triangle where the base is $10$ feet,the perpendicular height is $(h - 2)$ feet,and the hypotenuse (the ladder) is $h$ feet.
According to the Pythagorean theorem: $h^2 = 10^2 + (h - 2)^2$.
$h^2 = 100 + (h^2 - 4h + 4)$.
$h^2 = 100 + h^2 - 4h + 4$.
Subtracting $h^2$ from both sides: $0 = 104 - 4h$.
$4h = 104$.
$h = 26$ feet.
Thus,the height of the wall is $26$ feet.

(A) Let the side of the original square be $x \text{ cm}$.
The diagonal of the original square is $d_1 = \sqrt{2}x \text{ cm}$.
The area of the original square is $A_1 = \frac{d_1^2}{2} = \frac{(\sqrt{2}x)^2}{2} = x^2 \text{ cm}^2$.
According to the question,the diagonal of the new square is doubled,so $d_2 = 2 \times d_1 = 2\sqrt{2}x \text{ cm}$.
The area of the new square is $A_2 = \frac{d_2^2}{2} = \frac{(2\sqrt{2}x)^2}{2} = \frac{8x^2}{2} = 4x^2 \text{ cm}^2$.
Comparing the areas,$A_2 = 4 \times A_1$.
Therefore,the area of the new square becomes $4$-fold.

(B) The area of the circle is given by $A = \pi r^{2} = 154 \, cm^{2}$.
Given $\pi = \frac{22}{7}$,we have $\frac{22}{7} \times r^{2} = 154$.
$r^{2} = \frac{154 \times 7}{22} = 7 \times 7 = 49$.
Thus,the radius $r = 7 \, cm$.
The circumference of the circle is $C = 2 \pi r = 2 \times \frac{22}{7} \times 7 = 44 \, cm$.
The length of an arc that subtends an angle $\theta = 45^{\circ}$ at the centre is given by the formula $L = \frac{\theta}{360^{\circ}} \times 2 \pi r$.
$L = \frac{45}{360} \times 44 = \frac{1}{8} \times 44 = 5.5 \, cm$.

(B) Let the base of the triangle be $2x\, m$ and the height be $3x\, m$.
The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Given that $1\, \text{hectare} = 10000\, m^2$,the area of the lawn is $\frac{1}{12} \times 10000 = \frac{10000}{12} = \frac{2500}{3}\, m^2$.
Setting up the equation: $\frac{1}{2} \times (2x) \times (3x) = \frac{2500}{3}$.
$3x^2 = \frac{2500}{3}$.
$x^2 = \frac{2500}{9}$.
$x = \sqrt{\frac{2500}{9}} = \frac{50}{3}$.
Base $= 2x = 2 \times \frac{50}{3} = \frac{100}{3} = 33\frac{1}{3}\, m$.
Height $= 3x = 3 \times \frac{50}{3} = 50\, m$.

(B) The area of the rectangular farm is given by $Area = \text{length} \times \text{width} = 1200 \ m^2$.
Given the short side (width) is $30 \ m$, the length of the longer side is $\frac{1200}{30} = 40 \ m$.
The diagonal of the rectangle forms a right-angled triangle with the sides. Using the Pythagorean theorem, the length of the diagonal is $\sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \ m$.
The total length of the fence required is the sum of one long side, one short side, and the diagonal: $40 + 30 + 50 = 120 \ m$.
Given the cost of fencing is ₹ $10$ per $m$, the total cost of the job is $120 \times 10 = ₹ 1200$.

(D) Let the diameter of the circle be $d$ and the circumference be $C$.
We know that the circumference of a circle is given by $C = \pi d$.
According to the problem,the diameter is $105 \, cm$ less than the circumference,so $d = C - 105$.
Substituting $C = \pi d$ into the equation,we get $d = \pi d - 105$.
Rearranging the terms,we have $\pi d - d = 105$,which implies $d(\pi - 1) = 105$.
Using $\pi = \frac{22}{7}$,we get $d(\frac{22}{7} - 1) = 105$.
$d(\frac{22 - 7}{7}) = 105 \Rightarrow d(\frac{15}{7}) = 105$.
$d = 105 \times \frac{7}{15} = 7 \times 7 = 49 \, cm$.
Therefore,the diameter of the circle is $49 \, cm$.

(C) Area of the garden $= 24\, m \times 14\, m = 336\, m^2$.
The path is $1\, m$ wide outside the garden,so the new length $= 24 + 1 + 1 = 26\, m$ and the new width $= 14 + 1 + 1 = 16\, m$.
Area of the (garden $+$ path) $= 26\, m \times 16\, m = 416\, m^2$.
Area of the path $= 416\, m^2 - 336\, m^2 = 80\, m^2$.
Area of $1$ tile $= 20\, cm \times 20\, cm = 400\, cm^2$.
Since $1\, m^2 = 10000\, cm^2$,the area of $1$ tile in $m^2 = \frac{400}{10000} = 0.04\, m^2$.
Number of tiles required $= \frac{\text{Area of the path}}{\text{Area of 1 tile}} = \frac{80}{0.04} = 2000$.

(B) Let the length of the longer diagonal be $d_1 = x$.
Then,the length of the shorter diagonal is $d_2 = 80\% \text{ of } x = 0.8x = \frac{4}{5}x$.
The area of a rhombus is given by the formula $A = \frac{1}{2} \times d_1 \times d_2$.
Substituting the values,$A = \frac{1}{2} \times x \times \frac{4}{5}x = \frac{2}{5}x^2$.
Thus,the area of the rhombus is $\frac{2}{5}$ times the square of the length of the longer diagonal.

(D) In a trapezium $ABCD$ with $AB \parallel CD$,triangles $\Delta AOB$ and $\Delta COD$ are similar by $AA$ similarity criterion because $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (alternate interior angles).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta AOB)}{\text{Area}(\Delta COD)} = \left( \frac{AB}{CD} \right)^2$.
Given $AB = 2CD$,we have $\frac{AB}{CD} = \frac{2}{1}$.
Thus,$\frac{\text{Area}(\Delta AOB)}{\text{Area}(\Delta COD)} = \left( \frac{2}{1} \right)^2 = \frac{4}{1}$.
The ratio is $4:1$.

(C) According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Let the sides be $3x$ and $4x$.
Ratio of areas $= \frac{(3x)^2}{(4x)^2} = \frac{9x^2}{16x^2} = \frac{9}{16}$.
Therefore,the ratio of their areas is $9:16$.

(C) For an equilateral triangle with side $a = 24 \, cm$,the height $h$ is given by $h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 24 = 12 \sqrt{3} \, cm$.
The radius $r$ of the inscribed circle (inradius) of an equilateral triangle is given by $r = \frac{h}{3}$.
Substituting the value of $h$,we get $r = \frac{12 \sqrt{3}}{3} = 4 \sqrt{3} \, cm$.
The area of the circle is given by $A = \pi r^2$.
Substituting the value of $r$,$A = \pi (4 \sqrt{3})^2 = \pi (16 \times 3) = 48 \pi \, cm^2$.