Measurement of Area Questions in English

(B) To draw the largest circle inside a rectangle,the diameter of the circle must be equal to the smaller side of the rectangle.
Here,the sides of the rectangle are $18 \, cm$ and $14 \, cm$.
Therefore,the diameter of the largest circle $d = 14 \, cm$.
The radius of the circle $r = \frac{d}{2} = \frac{14}{2} = 7 \, cm$.
The area of the circle is given by the formula $A = \pi r^2$.
Substituting the value of $r$,we get $A = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \, cm^2$.

(D) Given: Radius $R = 5 \, cm$ and arc length $S = 3.5 \, cm$.
The formula for the angle $\theta$ (in radians) subtended by an arc at the center is $\theta = \frac{S}{R}$.
Substituting the values: $\theta = \frac{3.5}{5} = 0.7 \, \text{rad}$.
The area of a sector is given by the formula $\text{Area} = \frac{1}{2} R^2 \theta$.
Substituting the values: $\text{Area} = \frac{1}{2} \times (5)^2 \times 0.7$.
$\text{Area} = \frac{1}{2} \times 25 \times 0.7 = 12.5 \times 0.7 = 8.75 \, cm^2$.

(D) The perimeter of a semi-$circle$ is given by the formula $P = \pi r + 2r$,where $r$ is the radius.
Given $P = 36 \, cm$,we have $(\pi + 2)r = 36$.
Substituting $\pi = \frac{22}{7}$,we get $(\frac{22}{7} + 2)r = 36$.
$(\frac{22 + 14}{7})r = 36 \implies \frac{36}{7}r = 36$.
Thus,$r = 7 \, cm$.
The area of a semi-$circle$ is given by $A = \frac{1}{2} \pi r^2$.
$A = \frac{1}{2} \times \frac{22}{7} \times 7^2 = \frac{1}{2} \times 22 \times 7 = 11 \times 7 = 77 \, cm^2$.

(B) Let the radius of the inner circle be $r$ and the radius of the outer circle be $R$.
Let the uniform width of the path be $x$.
Then,$R = r + x$,which implies $R - r = x$.
The difference between the outer and inner circumferences is given by $2 \pi R - 2 \pi r = 132 \, m$.
Factoring out $2 \pi$,we get $2 \pi (R - r) = 132$.
Since $R - r = x$,the equation becomes $2 \pi x = 132$.
Substituting $\pi = \frac{22}{7}$,we have $2 \times \frac{22}{7} \times x = 132$.
Solving for $x$: $x = \frac{132 \times 7}{2 \times 22} = \frac{132 \times 7}{44} = 3 \times 7 = 21 \, m$.
Therefore,the width of the path is $21 \, m$.

(B) Let the diameter of the wheel be $x \, m$.
The circumference of the wheel is given by $\pi x \, m$.
In $113$ revolutions,the total distance covered is $113 \times \pi x$.
Given that the total distance is $2 \, km$ $26 \, decametres$.
Convert the distance into metres: $2 \, km = 2000 \, m$ and $26 \, decametres = 260 \, m$.
Total distance $= 2000 + 260 = 2260 \, m$.
Equating the distance: $113 \times \frac{22}{7} \times x = 2260$.
Solving for $x$: $x = \frac{2260 \times 7}{113 \times 22}$.
Since $2260 / 113 = 20$,we have $x = \frac{20 \times 7}{22} = \frac{140}{22} = \frac{70}{11} = 6 \frac{4}{11} \, m$.

(C) Let the length of the rope be $l$ metres. The area grazed by the cow forms a circle with radius $l$.
Given,Area $= \pi l^{2} = 9856 \, m^{2}$.
$\Rightarrow l^{2} = \frac{9856}{\pi} = \frac{9856 \times 7}{22}$.
$\Rightarrow l^{2} = 448 \times 7 = 3136$.
$\Rightarrow l = \sqrt{3136} = 56 \, m$.
Thus,the length of the rope is $56 \, m$.

(D) Given,radius of the wheel,$r = 0.25\, m$.
Total distance to be covered,$D = 11\, km = 11000\, m$.
The distance covered in one revolution is equal to the circumference of the wheel,which is $2\pi r$.
Let $N$ be the number of revolutions.
Then,$N \times (2\pi r) = D$.
Substituting the values: $N \times 2 \times \frac{22}{7} \times 0.25 = 11000$.
$N = \frac{11000 \times 7}{2 \times 22 \times 0.25}$.
$N = \frac{77000}{11} = 7000$.
Thus,the wheel will make $7000$ revolutions.

(A) The area of a circle is given by $A = \pi r^2$,so the area is proportional to the square of the radius $(A \propto r^2)$.
Given the ratio of areas $\frac{A_1}{A_2} = \frac{4}{9}$.
Therefore,the ratio of the radii is $\frac{r_1}{r_2} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
The circumference of a circle is given by $C = 2\pi r$,so the circumference is directly proportional to the radius $(C \propto r)$.
Thus,the ratio of their circumferences is $\frac{C_1}{C_2} = \frac{r_1}{r_2} = \frac{2}{3}$.
Hence,the ratio is $2:3$.

(D) The circumference of the circle is given by $C = 2 \pi r$.
Substituting $r = 42 \, cm$,we get $C = 2 \times \frac{22}{7} \times 42 = 264 \, cm$.
Since the wire is reshaped into a rectangle,the perimeter of the rectangle equals the circumference of the circle.
Let the sides of the rectangle be $6x$ and $5x$ (based on the ratio $6:5$).
The perimeter of the rectangle is $2(6x + 5x) = 2(11x) = 22x$.
Equating the perimeters: $22x = 264$.
Solving for $x$: $x = \frac{264}{22} = 12$.
The smaller side is $5x = 5 \times 12 = 60 \, cm$.

(C) For the square:
Perimeter $= 4 \times \text{side} = 44\, cm$.
Side $= 44 / 4 = 11\, cm$.
Area of square $= (\text{side})^2 = 11^2 = 121\, cm^2$.
For the circle:
Circumference $= 2\pi r = 44\, cm$.
$2 \times (22/7) \times r = 44$.
$r = (44 \times 7) / (2 \times 22) = 7\, cm$.
Area of circle $= \pi r^2 = (22/7) \times 7^2 = 22 \times 7 = 154\, cm^2$.
Comparing the areas:
Area of circle - Area of square $= 154 - 121 = 33\, cm^2$.
Thus,the circle has a larger area by $33\, cm^2$.

(C) Let the original length be $l$ and the original breadth be $b$. The original area is $A = l \times b$.
The new length $l' = l + 0.05l = 1.05l$.
The new breadth $b' = b - 0.04b = 0.96b$.
The new area $A' = l' \times b' = (1.05l) \times (0.96b) = 1.008lb$.
The change in area is $A' - A = 1.008lb - 1lb = 0.008lb$.
The error percent in the area is $\frac{A' - A}{A} \times 100 = \frac{0.008lb}{lb} \times 100 = 0.8 \%$.

(C) The area of a trapezium is given by the formula: $\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$.
Let the parallel sides be $5x$ and $3x$.
The height (perpendicular distance) is $24 \, m$.
Substituting the values into the formula:
$1440 = \frac{1}{2} \times (5x + 3x) \times 24$
$1440 = \frac{1}{2} \times (8x) \times 24$
$1440 = 4x \times 24$
$1440 = 96x$
$x = \frac{1440}{96} = 15$.
The larger parallel side is $5x = 5 \times 15 = 75 \, m$.

(C) Let the diagonals be $d_{1}$ and $d_{2}$.
Given that $d_{1} = 2d_{2}$.
The area of a rhombus is given by the formula $\text{Area} = \frac{1}{2} \times d_{1} \times d_{2}$.
Substituting the given values: $25 = \frac{1}{2} \times (2d_{2}) \times d_{2}$.
$25 = d_{2}^{2}$.
Taking the square root on both sides,$d_{2} = 5\, cm$.
Now,find $d_{1}$: $d_{1} = 2 \times 5 = 10\, cm$.
The sum of the diagonals is $d_{1} + d_{2} = 10 + 5 = 15\, cm$.

(B) The formula for the area of a rhombus is given by $\text{Area} = \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the lengths of the diagonals.
Given: $\text{Area} = 150 \, cm^2$ and $d_1 = 10 \, cm$.
Substituting the values into the formula:
$150 = \frac{1}{2} \times 10 \times d_2$
$150 = 5 \times d_2$
$d_2 = \frac{150}{5} = 30 \, cm$.
Therefore,the length of the other diagonal is $30 \, cm$.

(D) Let the base of both the triangle and the parallelogram be $b$.
Let the altitude of the triangle be $h_1$ and the altitude of the parallelogram be $h_2$.
The area of the triangle is given by $A_t = \frac{1}{2} \times b \times h_1$.
The area of the parallelogram is given by $A_p = b \times h_2$.
According to the problem,the areas are equal,so $A_t = A_p$.
Therefore,$\frac{1}{2} \times b \times h_1 = b \times h_2$.
Dividing both sides by $b$ (assuming $b \neq 0$),we get $\frac{1}{2} h_1 = h_2$,which implies $h_1 = 2 h_2$.
Given that the altitude of the parallelogram $h_2 = 100 \, m$.
Substituting this value,we get $h_1 = 2 \times 100 \, m = 200 \, m$.

(D) Let the sides of the original triangle be $a$,$b$,and $c$. The semi-perimeter is $s = \frac{a+b+c}{2}$.
According to Heron's formula,the area of the original triangle is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
If every side is doubled,the new sides are $2a$,$2b$,and $2c$.
The new semi-perimeter $s'$ is $s' = \frac{2a+2b+2c}{2} = 2s$.
The new area $\Delta'$ is $\sqrt{s'(s'-2a)(s'-2b)(s'-2c)}$.
Substituting the values: $\Delta' = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$.
$\Delta' = \sqrt{2s \cdot 2(s-a) \cdot 2(s-b) \cdot 2(s-c)}$.
$\Delta' = \sqrt{16 \cdot s(s-a)(s-b)(s-c)}$.
$\Delta' = 4 \sqrt{s(s-a)(s-b)(s-c)} = 4\Delta$.
Thus,the new area is $4$ times the old area,so $K = 4$.

(C) Given: Perimeter $P = 30 \text{ cm}$,so semi-perimeter $s = P/2 = 15 \text{ cm}$.
Let the sides be $a, b, c$ where $a = 13 \text{ cm}$.
Perimeter $a + b + c = 30 \implies 13 + b + c = 30 \implies b + c = 17$.
Using Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = 30$.
$\sqrt{15(15-13)(15-b)(15-c)} = 30$.
$\sqrt{15 \times 2 \times (15-b)(15-c)} = 30$.
$30(15-b)(15-c) = 30^2 = 900$.
$(15-b)(15-c) = 30$.
$225 - 15(b+c) + bc = 30$.
Substitute $b+c = 17$: $225 - 15(17) + bc = 30$.
$225 - 255 + bc = 30 \implies bc - 30 = 30 \implies bc = 60$.
We have $b+c = 17$ and $bc = 60$. The quadratic equation $x^2 - 17x + 60 = 0$ gives the sides.
$(x-5)(x-12) = 0$,so the sides are $5 \text{ cm}$ and $12 \text{ cm}$.
The sides of the triangle are $5 \text{ cm}, 12 \text{ cm}, 13 \text{ cm}$.
The smallest side is $5 \text{ cm}$.

(D) The ratio of the sides is given as $a: b: c = \frac{1}{2}: \frac{1}{3}: \frac{1}{4}$.
To simplify the ratio,multiply each term by the least common multiple $(LCM)$ of the denominators $(2, 3, 4)$,which is $12$.
$a: b: c = (\frac{1}{2} \times 12) : (\frac{1}{3} \times 12) : (\frac{1}{4} \times 12) = 6: 4: 3$.
Let the sides be $6x, 4x,$ and $3x$ centimeters.
The perimeter of the triangle is the sum of its sides,which is given as $52 \text{ cm}$.
$6x + 4x + 3x = 52$.
$13x = 52$.
$x = \frac{52}{13} = 4$.
The smallest side corresponds to the smallest ratio value,which is $3x$.
Smallest side $= 3 \times 4 = 12 \text{ cm}$.

(D) Let the side of the square be $a \text{ cm}$.
Therefore,the length of the rectangle $l = (a + 5) \text{ cm}$ and the breadth of the rectangle $b = (a - 3) \text{ cm}$.
Given that the area of the square is equal to the area of the rectangle:
$a^2 = (a + 5)(a - 3)$
Expanding the right side:
$a^2 = a^2 - 3a + 5a - 15$
$a^2 = a^2 + 2a - 15$
Subtracting $a^2$ from both sides:
$0 = 2a - 15$
$2a = 15$
$a = 7.5 \text{ cm}$
Now,calculate the dimensions of the rectangle:
Length $l = 7.5 + 5 = 12.5 \text{ cm}$
Breadth $b = 7.5 - 3 = 4.5 \text{ cm}$
The perimeter of the rectangle is $P = 2(l + b)$:
$P = 2(12.5 + 4.5) = 2(17) = 34 \text{ cm}$.

(C) Let the side of the square be $x \, cm$.
When one pair of opposite sides is increased by $5 \, cm$,the new dimensions of the rectangle become $(x + 5) \, cm$ and $x \, cm$.
According to the problem,the ratio of the length to the breadth is $3:2$:
$\frac{x + 5}{x} = \frac{3}{2}$
Cross-multiplying gives:
$2(x + 5) = 3x$
$2x + 10 = 3x$
$x = 10 \, cm$
The area of the original square is $x^2 = 10^2 = 100 \, cm^2$.

(B) Let the original side of the square be $s$. The original area is $A_1 = s^2$.
Let the new side be $s'$. The new area is $A_2 = A_1 + 0.69 A_1 = 1.69 s^2$.
Since $A_2 = (s')^2$,we have $(s')^2 = 1.69 s^2$.
Taking the square root of both sides,$s' = \sqrt{1.69} s = 1.3 s$.
The increase in the side is $s' - s = 1.3 s - s = 0.3 s$.
Percentage increase in the side = $(0.3 s / s) \times 100 = 30 \%$.

(B) Let the side length of the outer square be $x \, m$.
Since the road is $3 \, m$ wide and runs along the inside of the square,the side length of the inner square is $(x - 3 - 3) = (x - 6) \, m$.
The area of the road is the difference between the area of the outer square and the inner square:
$\text{Area of road} = x^{2} - (x - 6)^{2} = 1764 \, m^{2}$.
Using the identity $a^{2} - b^{2} = (a - b)(a + b)$:
$(x - (x - 6))(x + (x - 6)) = 1764$
$(6)(2x - 6) = 1764$
$2x - 6 = \frac{1764}{6} = 294$
$2x = 294 + 6 = 300$
$x = 150 \, m$.
The perimeter along the outer edge of the road is the perimeter of the outer square:
$\text{Perimeter} = 4 \times x = 4 \times 150 = 600 \, m$.

(B) Perimeter of square $= 48\, cm$.
Side of square $= \frac{48}{4} = 12\, cm$.
Area of square $= (12)^2 = 144\, cm^2$.
Area of rectangle $= 144 - 4 = 140\, cm^2$.
Given length of rectangle $= 14\, cm$.
Breadth of rectangle $= \frac{\text{Area}}{\text{Length}} = \frac{140}{14} = 10\, cm$.
Perimeter of rectangle $= 2(l + b) = 2(14 + 10) = 2(24) = 48\, cm$.

(D) Let the length and breadth of the rectangular plot be $l$ and $b$ respectively.
Given,the area of the plot is $l \times b = 96 \text{ m}^2$.
$A$ pathway of width $2 \text{ m}$ is constructed inside the plot.
The dimensions of the inner rectangle (excluding the path) will be $(l - 4)$ and $(b - 4)$ because the path reduces the length and breadth from both sides.
The area of the path is calculated as: $\text{Area of path} = \text{Total Area} - \text{Inner Area} = lb - (l - 4)(b - 4)$.
Substituting $lb = 96$: $\text{Area of path} = 96 - (lb - 4l - 4b + 16) = 96 - (96 - 4(l + b) + 16) = 4(l + b) - 16$.
Since the values of $l$ and $b$ are not individually known,the sum $(l + b)$ cannot be determined from the given area $lb = 96$ alone.
Therefore,the area of the path cannot be calculated,and consequently,the total cost of construction cannot be determined.
Thus,the data provided is inadequate.

(C) Let the length and breadth of the rectangular plot be $L$ and $B$ respectively.
Perimeter $= 2(L + B) = 340 \text{ m}$,so $L + B = 170 \text{ m}$.
The area of the $1 \text{ m}$ broad boundary (path) around the plot is given by the formula: $\text{Area} = 2(L + B) \times w + 4w^2$,where $w$ is the width of the path.
Here,$w = 1 \text{ m}$.
$\text{Area} = 340 \times 1 + 4(1)^2 = 340 + 4 = 344 \text{ m}^2$.
The cost of gardening is at the rate of $₹ 10$ per square metre.
$\text{Total Cost} = 344 \text{ m}^2 \times 10 \text{ ₹/m}^2 = ₹ 3440$.

(C) Total area of the sheet $= 20\, cm \times 30\, cm = 600\, cm^2$.
The width available for typing $= 20\, cm - (2\, cm + 2\, cm) = 20\, cm - 4\, cm = 16\, cm$.
The height available for typing $= 30\, cm - (3\, cm + 3\, cm) = 30\, cm - 6\, cm = 24\, cm$.
Area used for typing $= 16\, cm \times 24\, cm = 384\, cm^2$.
Percentage of the page used for typing $= \left( \frac{384}{600} \right) \times 100 = \frac{384}{6} = 64\%$.

(C) Let the parallel sides of the trapezium be $l_1$ and $l_2$,where $l_1 > l_2$.
Given that the difference between the parallel sides is $4\, cm$,so $l_1 - l_2 = 4$,which implies $l_1 = l_2 + 4$.
The perpendicular distance (height) between the parallel sides is $h = 19\, cm$.
The area of a trapezium is given by the formula: $\text{Area} = \frac{1}{2} \times (l_1 + l_2) \times h$.
Substituting the given values: $475 = \frac{1}{2} \times (l_2 + 4 + l_2) \times 19$.
$475 = \frac{19}{2} \times (2l_2 + 4)$.
$475 = 19 \times (l_2 + 2)$.
$l_2 + 2 = \frac{475}{19} = 25$.
$l_2 = 25 - 2 = 23\, cm$.
Now,find $l_1$: $l_1 = l_2 + 4 = 23 + 4 = 27\, cm$.
Thus,the lengths of the parallel sides are $27\, cm$ and $23\, cm$.

(A) Let the original length be $L$ and the original breadth be $B$.
Original Area $A_1 = L \times B$.
New length $L' = L + 0.50L = 1.5L$.
New breadth $B' = B + 0.20B = 1.2B$.
New Area $A_2 = L' \times B' = (1.5L) \times (1.2B) = 1.8 \times (L \times B) = 1.8 A_1$.
The increase in area is $A_2 - A_1 = 1.8 A_1 - A_1 = 0.8 A_1$.
Therefore,the area increases by $0.8$ times the original area.

(C) Let the initial length be $l$ and the initial breadth be $b$. The initial area is $A_1 = l \times b$.
If each side is increased by $20 \%$,the new length $l' = l + 0.20l = 1.2l$ and the new breadth $b' = b + 0.20b = 1.2b$.
The new area is $A_2 = l' \times b' = (1.2l) \times (1.2b) = 1.44lb = 1.44A_1$.
The percentage increase in area is given by $\frac{A_2 - A_1}{A_1} \times 100 = \frac{1.44A_1 - A_1}{A_1} \times 100 = 0.44 \times 100 = 44 \%$.
Alternatively,using the formula for successive percentage change: $x + y + \frac{xy}{100} = 20 + 20 + \frac{20 \times 20}{100} = 40 + 4 = 44 \%$.

(A) Let $l$ and $b$ be the length and breadth of the rectangular paper.
When folded along the breadth,the new dimensions are $l/2$ and $b$. The perimeter is $2(l/2 + b) = 34$,which simplifies to $l + 2b = 34$ $....(1)$
When folded along the length,the new dimensions are $l$ and $b/2$. The perimeter is $2(l + b/2) = 38$,which simplifies to $2l + b = 38$ $....(2)$
Multiplying equation $(1)$ by $2$,we get $2l + 4b = 68$ $....(3)$
Subtracting equation $(2)$ from equation $(3)$,we get $(2l + 4b) - (2l + b) = 68 - 38$,which gives $3b = 30$,so $b = 10\, cm$.
Substituting $b = 10$ into equation $(1)$,we get $l + 2(10) = 34$,so $l + 20 = 34$,which gives $l = 14\, cm$.
The area of the paper is $l \times b = 14 \times 10 = 140\, cm^2$.

(D) Let the length of the rectangle be $l$ and the width be $w$.
Diagonal $d = \sqrt{l^2 + w^2}$.
Distance covered by $A$ diagonally $= \frac{52 \text{ m}}{60 \text{ s}} \times 15 \text{ s} = 13 \text{ m}$.
So,$l^2 + w^2 = 13^2 = 169$.
Distance covered by $B$ along the sides $= l + w = \frac{68 \text{ m}}{60 \text{ s}} \times 15 \text{ s} = 17 \text{ m}$.
We know that $(l + w)^2 = l^2 + w^2 + 2lw$.
Substituting the values: $17^2 = 169 + 2lw$.
$289 = 169 + 2lw$.
$2lw = 289 - 169 = 120$.
Area of the field $= l \times w = \frac{120}{2} = 60 \text{ m}^2$.

(D) Let the length and breadth of the rectangular room be $l$ and $b$ respectively.
According to the first condition:
$(l+1)(b+1) - lb = 21$
$lb + l + b + 1 - lb = 21$
$l + b + 1 = 21$
$l + b = 20$ $...(1)$
According to the second condition:
$(l+1)(b-1) - lb = -5$
$lb - l + b - 1 - lb = -5$
$-l + b - 1 = -5$
$-l + b = -4$
$l - b = 4$ $...(2)$
Adding equations $(1)$ and $(2)$:
$(l + b) + (l - b) = 20 + 4$
$2l = 24$
$l = 12\, m$
Substituting $l = 12$ in equation $(1)$:
$12 + b = 20$
$b = 8\, m$
The perimeter of the floor is given by the formula $P = 2(l + b)$.
$P = 2(12 + 8) = 2(20) = 40\, m$.

(D) Let the sides of the triangle be $3x$,$4x$,and $5x$.
The semi-perimeter $s$ is given by $s = \frac{3x + 4x + 5x}{2} = \frac{12x}{2} = 6x$.
Using Heron's formula,the area $\Delta$ is $\sqrt{s(s-a)(s-b)(s-c)}$.
$\Delta = \sqrt{6x(6x-3x)(6x-4x)(6x-5x)} = \sqrt{6x \cdot 3x \cdot 2x \cdot x} = \sqrt{36x^{4}} = 6x^{2}$.
Given that the area is $216 \, cm^{2}$,we have $6x^{2} = 216$.
$x^{2} = 36$,which implies $x = 6 \, cm$.
The perimeter of the triangle is $3x + 4x + 5x = 12x$.
Perimeter $= 12 \times 6 = 72 \, cm$.

(C) Let the bases of the two triangles be $b_1$ and $b_2$,and their altitudes be $h_1$ and $h_2$ respectively.
Given that the ratio of their bases is $\frac{b_1}{b_2} = \frac{x}{y}$ and the ratio of their areas is $\frac{\Delta_1}{\Delta_2} = \frac{a}{b}$.
The area of a triangle is given by the formula $\Delta = \frac{1}{2} \times \text{base} \times \text{altitude}$.
Therefore,$\frac{\Delta_1}{\Delta_2} = \frac{\frac{1}{2} b_1 h_1}{\frac{1}{2} b_2 h_2} = \frac{b_1}{b_2} \times \frac{h_1}{h_2}$.
Substituting the given values,we get $\frac{a}{b} = \frac{x}{y} \times \frac{h_1}{h_2}$.
Solving for the ratio of altitudes $\frac{h_1}{h_2}$,we get $\frac{h_1}{h_2} = \frac{a}{b} \times \frac{y}{x} = \frac{ay}{bx}$.
Thus,the ratio of their corresponding altitudes is $ay: bx$.

(C) Area of the square $= 484 \, cm^2$.
Let the side of the square be $a$. Then $a^2 = 484$,which gives $a = \sqrt{484} = 22 \, cm$.
The perimeter of the square $= 4 \times a = 4 \times 22 = 88 \, cm$.
When the same wire is bent into a circle,the circumference of the circle equals the perimeter of the square.
Circumference $= 2\pi r = 88 \, cm$.
$2 \times \frac{22}{7} \times r = 88$.
$r = \frac{88 \times 7}{44} = 14 \, cm$.
The area of the circle $= \pi r^2 = \frac{22}{7} \times 14 \times 14$.
Area $= 22 \times 2 \times 14 = 616 \, cm^2$.

(A) The diameter of the wheel is $d = 40 \, cm$.
The circumference of the wheel is given by $C = \pi d = \frac{22}{7} \times 40 = \frac{880}{7} \, cm$.
The total distance to be covered is $176 \, m$. Converting this to centimeters,we get $176 \times 100 = 17600 \, cm$.
The number of revolutions is calculated by dividing the total distance by the circumference:
$\text{Number of revolutions} = \frac{\text{Total distance}}{\text{Circumference}} = \frac{17600}{\frac{880}{7}} = \frac{17600 \times 7}{880} = 20 \times 7 = 140$.

(C) The area of the square is $81 \, cm^2$.
Since the area of a square is $side^2$,the side length is $\sqrt{81} = 9 \, cm$.
The perimeter (length) of the wire is $4 \times 9 = 36 \, cm$.
When this wire is bent into a semi-circular shape,its perimeter is given by the formula $P = \pi r + 2r = r(\pi + 2)$.
Setting the perimeter equal to $36 \, cm$:
$r(\frac{22}{7} + 2) = 36$
$r(\frac{22 + 14}{7}) = 36$
$r(\frac{36}{7}) = 36$
$r = 7 \, cm$.
The area of the semi-circle is $\frac{1}{2} \pi r^2$:
Area $= \frac{1}{2} \times \frac{22}{7} \times 7^2 = \frac{1}{2} \times 22 \times 7 = 77 \, cm^2$.

(B) The area of the square sheet is $784 \, cm^2$.
Side of the square sheet $= \sqrt{784} = 28 \, cm$.
Since four equal-sized circular plates of maximum diameter are cut from the square,they are arranged in a $2 \times 2$ grid.
The diameter of each circular plate is half the side of the square: $d = \frac{28}{2} = 14 \, cm$.
The circumference of each plate is $C = \pi d = \frac{22}{7} \times 14 = 44 \, cm$.
Converting the circumference into meters: $44 \, cm = 0.44 \, m$.

(A) Let the two equal sides of the isosceles right triangle be $x$.
Then,the hypotenuse is $\sqrt{x^2 + x^2} = x\sqrt{2}$.
The perimeter is given by $x + x + x\sqrt{2} = x(2 + \sqrt{2})$.
Given that the perimeter is $(6 + 3\sqrt{2}) \text{ m}$,we have:
$x(2 + \sqrt{2}) = 3(2 + \sqrt{2})$.
Therefore,$x = 3 \text{ m}$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times x = \frac{1}{2} \times 3 \times 3 = 4.5 \text{ m}^2$.

(D) Let the breadth of the hall be $x$ metres.
Then,the length of the hall is $(x + 5)$ metres.
The area of the rectangular hall is given by $\text{length} \times \text{breadth} = 750\, m^{2}$.
So,$x(x + 5) = 750$.
$x^{2} + 5x - 750 = 0$.
Factoring the quadratic equation: $x^{2} + 30x - 25x - 750 = 0$.
$x(x + 30) - 25(x + 30) = 0$.
$(x - 25)(x + 30) = 0$.
Since the breadth cannot be negative,$x = 25$.
Therefore,the length of the hall is $x + 5 = 25 + 5 = 30\, m$.

(D) Let the breadth of the rectangle be $b \, m$.
Since the length is $15 \%$ more than the breadth,the length $l = b + 0.15b = 1.15b \, m$.
The area of a rectangle is given by the formula: $\text{Area} = l \times b$.
Substituting the given values: $1.15b \times b = 460$.
$1.15b^2 = 460$.
$b^2 = \frac{460}{1.15} = 400$.
Taking the square root of both sides: $b = \sqrt{400} = 20 \, m$.
Therefore,the breadth of the rectangular field is $20 \, m$.

(C) The side length of a square is given by $s = \frac{\text{Perimeter}}{4}$.
The sides of the five squares are:
$s_1 = \frac{24}{4} = 6 \ cm$,$s_2 = \frac{32}{4} = 8 \ cm$,$s_3 = \frac{40}{4} = 10 \ cm$,$s_4 = \frac{76}{4} = 19 \ cm$,$s_5 = \frac{80}{4} = 20 \ cm$.
The sum of the areas of these squares is:
$A = (6^2 + 8^2 + 10^2 + 19^2 + 20^2) \ cm^2$
$A = (36 + 64 + 100 + 361 + 400) \ cm^2 = 961 \ cm^2$.
Let the side of the new square be $S$. Since its area is equal to the sum of the areas of the five squares:
$S^2 = 961 \ cm^2$
$S = \sqrt{961} = 31 \ cm$.
The perimeter of the new square is $P = 4 \times S = 4 \times 31 = 124 \ cm$.

(B) The side of the square room is $3 \text{ m}$.
Area of the room $= 3 \text{ m} \times 3 \text{ m} = 9 \text{ m}^2$.
Since $1 \text{ m} = 100 \text{ cm}$,$1 \text{ m}^2 = 10,000 \text{ cm}^2$.
Therefore,the area of the room in $\text{cm}^2 = 9 \times 10,000 = 90,000 \text{ cm}^2$.
The area of one marble slab $= 20 \text{ cm} \times 30 \text{ cm} = 600 \text{ cm}^2$.
The number of slabs required $= \frac{\text{Total area of the room}}{\text{Area of one slab}} = \frac{90,000 \text{ cm}^2}{600 \text{ cm}^2} = 150$.

(B) Let the length of the greater line segment be $x \, cm$.
Then,the length of the smaller line segment is $(x - 2) \, cm$.
The area of the square formed by the greater line segment is $x^2 \, cm^2$.
The area of the square formed by the smaller line segment is $(x - 2)^2 \, cm^2$.
According to the problem,the difference between the areas is $32 \, cm^2$:
$x^2 - (x - 2)^2 = 32$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$(x - (x - 2))(x + (x - 2)) = 32$
$(2)(2x - 2) = 32$
Dividing both sides by $2$:
$2x - 2 = 16$
$2x = 18$
$x = 9$
Therefore,the length of the greater line segment is $9 \, cm$.

(B) Let the radius of the circle be $r$.
The circumference of the circle is given by $C = 2 \pi r$.
The area of the circle is given by $A = \pi r^2$.
According to the problem,the circumference and the area are numerically equal:
$2 \pi r = \pi r^2$
Dividing both sides by $\pi r$ (assuming $r \neq 0$):
$2 = r$
Thus,the radius $r = 2$.
The diameter $d$ of the circle is $2r$:
$d = 2 \times 2 = 4$.
Therefore,the diameter is $4$.

(C) Let the sides of the triangle be $a = 15 \, m$,$b = 16 \, m$,and $c = 17 \, m$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{15 + 16 + 17}{2} = \frac{48}{2} = 24 \, m$.
Using Heron's formula,the area of the triangle is given by:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
Substitute the values:
$\text{Area} = \sqrt{24(24 - 15)(24 - 16)(24 - 17)}$
$\text{Area} = \sqrt{24 \times 9 \times 8 \times 7}$
$\text{Area} = \sqrt{12096} = \sqrt{576 \times 21} = 24 \sqrt{21} \, m^{2}$.

(A) Using the Pythagorean theorem,the height of the triangle is given by:
Height $= \sqrt{(\text{hypotenuse})^2 - (\text{base})^2}$
Height $= \sqrt{(6.5)^2 - 6^2}$
Height $= \sqrt{42.25 - 36}$
Height $= \sqrt{6.25} = 2.5\,m$
The area of a right-angled triangle is calculated as:
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times 6\,m \times 2.5\,m$
Area $= 3 \times 2.5 = 7.5\,m^2$

(B) An equilateral triangle has all sides equal. Given side $a = 12\, cm$.
The formula for the height $h$ of an equilateral triangle is $h = \frac{\sqrt{3}}{2} \times a$.
Substituting the value of $a$:
$h = \frac{\sqrt{3}}{2} \times 12\, cm$
$h = 6 \sqrt{3}\, cm$.
Alternatively,using the area formula:
Area $= \frac{\sqrt{3}}{4} \times a^2 = \frac{\sqrt{3}}{4} \times 144 = 36 \sqrt{3}\, cm^2$.
Also,Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times h = 6h$.
Equating the two: $6h = 36 \sqrt{3}$,so $h = 6 \sqrt{3}\, cm$.

(B) Let the length of each equal side be $x \, m$.
The sides of the triangle are $x, x,$ and $64$.
The semi-perimeter $s = \frac{x + x + 64}{2} = x + 32$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
$1600 = \sqrt{(x+32)(x+32-x)(x+32-x)(x+32-64)}$.
$1600 = \sqrt{(x+32)(32)(32)(x-32)}$.
$1600 = 32 \sqrt{(x+32)(x-32)}$.
$50 = \sqrt{x^{2} - 32^{2}}$.
Squaring both sides,$2500 = x^{2} - 1024$.
$x^{2} = 3524$.
$x = \sqrt{3524} \approx 59.36 \, m$.

(C) Let the sides of the triangle be $a = 20\, m$,$b = 21\, m$,and $c = 29\, m$.
First,calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{20 + 21 + 29}{2} = \frac{70}{2} = 35\, m$.
Using Heron's formula,the area of the triangle is given by:
Area $= \sqrt{s(s - a)(s - b)(s - c)}$.
Substituting the values:
Area $= \sqrt{35(35 - 20)(35 - 21)(35 - 29)}$
Area $= \sqrt{35 \times 15 \times 14 \times 6}$
Area $= \sqrt{(5 \times 7) \times (3 \times 5) \times (2 \times 7) \times (2 \times 3)}$
Area $= \sqrt{5^2 \times 7^2 \times 3^2 \times 2^2}$
Area $= 5 \times 7 \times 3 \times 2 = 210\, m^2$.