Measurement of Area Questions in English

(B) The area of a rectangle is given by the formula: $\text{Area} = \text{length} \times \text{breadth}$.
Given,$\text{Area} = 180 \, m^2$ and $\text{length} = 18 \, m$.
Therefore,$\text{breadth} = \frac{\text{Area}}{\text{length}} = \frac{180}{18} = 10 \, m$.
The perimeter of a rectangle is calculated as: $\text{Perimeter} = 2 \times (\text{length} + \text{breadth})$.
Substituting the values: $\text{Perimeter} = 2 \times (18 + 10) = 2 \times 28 = 56 \, m$.

(C) Let the side of the first square be $x\, cm$.
Then,the side of the second square is $(x + 4)\, cm$.
According to the problem,the sum of their areas is $400\, cm^2$:
$x^2 + (x + 4)^2 = 400$
$x^2 + x^2 + 8x + 16 = 400$
$2x^2 + 8x - 384 = 0$
Dividing by $2$,we get:
$x^2 + 4x - 192 = 0$
Factoring the quadratic equation:
$x^2 + 16x - 12x - 192 = 0$
$x(x + 16) - 12(x + 16) = 0$
$(x - 12)(x + 16) = 0$
Since the side length cannot be negative,$x = 12$.
Thus,the side of the first square is $12\, cm$ and the side of the second square is $12 + 4 = 16\, cm$.

(C) Area of the floor $= 960\, m^{2}$.
Area of one carpet $= 6\, m \times 4\, m = 24\, m^{2}$.
Therefore,the number of carpets required $= \frac{\text{Area of floor}}{\text{Area of one carpet}}$.
Number of carpets $= \frac{960}{24} = 40$.

(B) Length of the outer rectangle $= 60 \ m$.
Breadth of the outer rectangle $= 40 \ m$.
Area of the outer rectangle $= 60 \ m \times 40 \ m = 2400 \ m^2$.
Width of the path $= 1 \ m$.
Since the path is inside the lawn,the length of the inner rectangle $= 60 \ m - (1 \ m + 1 \ m) = 58 \ m$.
The breadth of the inner rectangle $= 40 \ m - (1 \ m + 1 \ m) = 38 \ m$.
Area of the inner rectangle $= 58 \ m \times 38 \ m = 2204 \ m^2$.
Area of the path $= \text{Area of outer rectangle} - \text{Area of inner rectangle}$.
Area of the path $= 2400 \ m^2 - 2204 \ m^2 = 196 \ m^2$.

(A) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{height}$.
In the given figure,the base $CD$ is $3 \text{ cm}$ and the corresponding height $BD$ is $4 \text{ cm}$.
Therefore,$\text{Area} = 3 \text{ cm} \times 4 \text{ cm} = 12 \text{ cm}^2$.
Alternatively,the parallelogram is composed of two congruent triangles,$\Delta ABD$ and $\Delta BDC$.
$\text{Area of } \Delta BDC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ cm}^2$.
$\text{Total Area} = 2 \times 6 \text{ cm}^2 = 12 \text{ cm}^2$.

(B) The sides of the triangle are $a = 9\, cm$,$b = 12\, cm$,and $c = 15\, cm$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{9 + 12 + 15}{2} = \frac{36}{2} = 18\, cm$.
Using Heron's formula,Area $= \sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{18(18 - 9)(18 - 12)(18 - 15)}$.
Area $= \sqrt{18 \times 9 \times 6 \times 3}$.
Area $= \sqrt{2916} = 54\, cm^2$.
Alternatively,since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$,it is a right-angled triangle.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \times 12 = 54\, cm^2$.

(C) The formula for the area of an equilateral triangle is $\text{Area} = \frac{\sqrt{3}}{4} \times a^2$,where $a$ is the side length.
Given $\text{Area} = 4 \sqrt{3} \text{ cm}^2$.
So,$\frac{\sqrt{3}}{4} \times a^2 = 4 \sqrt{3}$.
Dividing both sides by $\sqrt{3}$,we get $\frac{a^2}{4} = 4$.
$a^2 = 16$,which implies $a = 4 \text{ cm}$.
The perimeter of an equilateral triangle is $3 \times a$.
Perimeter $= 3 \times 4 = 12 \text{ cm}$.

(C) Let the radius of the circle be $r\, m$.
The area of the circle is given by the formula $A = \pi r^{2}$.
Given $A = 24.64\, m^{2}$,we have $\frac{22}{7} \times r^{2} = 24.64$.
Solving for $r^{2}$: $r^{2} = \frac{24.64 \times 7}{22} = 1.12 \times 7 = 7.84$.
Taking the square root: $r = \sqrt{7.84} = 2.8\, m$.
The circumference of the circle is given by $C = 2 \pi r$.
Substituting the value of $r$: $C = 2 \times \frac{22}{7} \times 2.8 = 2 \times 22 \times 0.4 = 17.6\, m$.

(C) Let the initial radius of the circle be $r$.
The new radius after a $20 \%$ decrease is $80 \%$ of $r$,which is $r \times \frac{80}{100} = \frac{4r}{5}$.
The initial area of the circle is $A_1 = \pi r^2$.
The new area of the circle is $A_2 = \pi \left(\frac{4r}{5}\right)^2 = \frac{16}{25} \pi r^2$.
The decrease in area is $A_1 - A_2 = \pi r^2 - \frac{16}{25} \pi r^2 = \frac{9}{25} \pi r^2$.
The percentage decrease in area is $\frac{\text{Decrease in Area}}{\text{Initial Area}} \times 100 = \frac{\frac{9}{25} \pi r^2}{\pi r^2} \times 100$.
Calculating this gives $\frac{9}{25} \times 100 = 9 \times 4 = 36 \%$.

(D) Let the radius of the semi-circular protractor be $r \, cm$.
The perimeter of a semi-circle is given by the sum of the arc length $(\pi r)$ and the diameter $(2r)$.
Perimeter $= \pi r + 2r = r(\pi + 2) = 36 \, cm$.
Substituting $\pi = \frac{22}{7}$,we get $r(\frac{22}{7} + 2) = 36$.
$r(\frac{22 + 14}{7}) = 36 \Rightarrow r(\frac{36}{7}) = 36$.
Solving for $r$,we get $r = 7 \, cm$.
The diameter of the protractor is $2r = 2 \times 7 = 14 \, cm$.

(C) Let the side of the square be $x$.
The radius of the incircle $(r)$ is half the side of the square: $r = \frac{x}{2}$.
The area of the incircle is $A_1 = \pi r^2 = \pi (\frac{x}{2})^2 = \frac{\pi x^2}{4}$.
The diagonal of the square is $\sqrt{2}x$. The radius of the circumcircle $(R)$ is half the diagonal: $R = \frac{\sqrt{2}x}{2} = \frac{x}{\sqrt{2}}$.
The area of the circumcircle is $A_2 = \pi R^2 = \pi (\frac{x}{\sqrt{2}})^2 = \frac{\pi x^2}{2}$.
The ratio of the area of the incircle to the circumcircle is $\frac{A_1}{A_2} = \frac{\pi x^2 / 4}{\pi x^2 / 2} = \frac{2}{4} = 1:2$.

(A) The diagonal of a square is given by the formula $d = a\sqrt{2}$,where $a$ is the side length of the square.
Given $d = 50 \, m$.
Therefore,$a\sqrt{2} = 50$,which implies $a = \frac{50}{\sqrt{2}} \, m$.
The area of a square is given by $Area = a^2$.
$Area = \left(\frac{50}{\sqrt{2}}\right)^2 = \frac{2500}{2} = 1250 \, m^2$.

(A) Given,circumference $(C) = 2 \pi r = 176\, m$.
$r = \frac{176}{2 \times \pi} = \frac{176 \times 7}{2 \times 22} = \frac{88 \times 7}{22} = 4 \times 7 = 28\, m$.
Area of the circle $(A) = \pi r^2 = \frac{22}{7} \times 28 \times 28$.
$A = 22 \times 4 \times 28 = 88 \times 28 = 2464\, m^2$.

(A) The formula for the length of an arc is given by $L = 2 \pi r \times \frac{\theta}{360^{\circ}}$.
Given radius $r = 42\, cm$ and central angle $\theta = 72^{\circ}$.
Substituting the values into the formula:
$L = 2 \times \frac{22}{7} \times 42 \times \frac{72}{360}$.
$L = 2 \times 22 \times 6 \times \frac{1}{5}$.
$L = 44 \times 6 \times 0.2$.
$L = 264 \times 0.2 = 52.8\, cm$.

(D) Let the two equal sides of the isosceles right-angled triangle be $a$.
The area of a right-angled triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Since it is an isosceles right-angled triangle,the base and height are equal,so $\text{Area} = \frac{1}{2} a^2$.
Given $\text{Area} = 200 \, cm^2$,we have:
$200 = \frac{1}{2} a^2$
$a^2 = 400$
$a = 20 \, cm$.
The hypotenuse $h$ of a right-angled triangle is given by $h = \sqrt{a^2 + a^2} = a\sqrt{2}$.
Substituting the value of $a$:
$h = 20 \sqrt{2} \, cm$.

(C) To find the least number of square slabs,we must find the side of the largest possible square tile that can fit perfectly into the dimensions of the room.
$1$. The side of the largest square tile is the Greatest Common Measure $(GCM)$ or Highest Common Factor $(HCF)$ of the length $(10.5 \text{ m})$ and the breadth $(3 \text{ m})$.
$2$. Convert $10.5$ and $3$ to a common unit or fraction: $10.5 = \frac{21}{2}$ and $3 = \frac{6}{2}$.
$3$. The $HCF$ of $\frac{21}{2}$ and $\frac{6}{2}$ is $\frac{\text{HCF}(21, 6)}{\text{LCM}(2, 2)} = \frac{3}{2} = 1.5 \text{ m}$.
$4$. The area of the room is $10.5 \times 3 = 31.5 \text{ m}^2$.
$5$. The area of one square tile is $1.5 \times 1.5 = 2.25 \text{ m}^2$.
$6$. The number of tiles needed = $\frac{\text{Area of room}}{\text{Area of one tile}} = \frac{31.5}{2.25} = 14$.

(C) Let the breadth of the rectangle be $x \, cm$.
Then,the length of the rectangle is $(x + 2) \, cm$.
The perimeter of a rectangle is given by the formula $P = 2 \times (\text{length} + \text{breadth})$.
Given $P = 48 \, cm$,we have $2(x + x + 2) = 48$.
Dividing by $2$,we get $2x + 2 = 24$.
Subtracting $2$ from both sides,$2x = 22$,which gives $x = 11 \, cm$.
So,the breadth is $11 \, cm$ and the length is $11 + 2 = 13 \, cm$.
The area of the rectangle is $\text{length} \times \text{breadth} = 13 \, cm \times 11 \, cm = 143 \, cm^2$.

(C) The total cost of levelling is given by the product of the area and the rate per square metre.
Total Cost = Area $\times$ Rate
$900 = \text{Area} \times 1.25$
Area = $\frac{900}{1.25} = 720 \ m^2$
Since the ground is rectangular,Area = $\text{length} \times \text{width}$.
$720 = 30 \times \text{width}$
Width = $\frac{720}{30} = 24 \ m$.

(C) Let the breadth of the rectangle be $x\, cm$. Then,the length is $2x\, cm$.
The initial area of the rectangle is $A_1 = \text{length} \times \text{breadth} = (2x)(x) = 2x^2\, cm^2$.
According to the problem,the new length is $(2x - 5)\, cm$ and the new breadth is $(x + 5)\, cm$.
The new area is $A_2 = (2x - 5)(x + 5) = 2x^2 + 10x - 5x - 25 = 2x^2 + 5x - 25$.
Given that the area increases by $75\, cm^2$,we have $A_2 = A_1 + 75$.
Substituting the expressions: $2x^2 + 5x - 25 = 2x^2 + 75$.
Subtracting $2x^2$ from both sides: $5x - 25 = 75$.
$5x = 100 \Rightarrow x = 20\, cm$.
The length of the rectangle is $2x = 2 \times 20 = 40\, cm$.

(C) The area of the floor is $4\, m \times 3\, m = 12\, m^2$.
Since $1\, m = 100\, cm$,then $1\, m^2 = 10,000\, cm^2$.
Therefore,the area of the floor in square centimeters is $12 \times 10,000 = 120,000\, cm^2$.
The area of one rectangular tile is $8\, cm \times 6\, cm = 48\, cm^2$.
The number of tiles required is the total area of the floor divided by the area of one tile.
Number of tiles $= \frac{120,000}{48} = 2,500$.

(A) Step $1$: Calculate the area of the square room.
Side of the square room $= 3\, m = 300\, cm$.
Area of the room $= 300\, cm \times 300\, cm = 90,000\, cm^2$.
Step $2$: Calculate the area of one marble slab.
Area of one slab $= 20\, cm \times 30\, cm = 600\, cm^2$.
Step $3$: Calculate the number of slabs required.
Number of slabs $= \frac{\text{Total Area of Room}}{\text{Area of one slab}} = \frac{90,000}{600} = 150$.
Therefore,$150$ marble slabs are required.

(B) Given,perimeter of the rectangle $= 200\, m$.
The formula for the perimeter of a rectangle is $P = 2(l + b)$,where $l$ is the length and $b$ is the breadth.
$2(l + 40) = 200$
$l + 40 = 100$
$l = 100 - 40 = 60\, m$.
The area of a rectangle is given by $A = l \times b$.
$A = 60\, m \times 40\, m = 2400\, m^2$.

(A) The formula for the area of a triangle is $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
First,convert the base from meters to centimeters: $1.5\, m = 1.5 \times 100\, cm = 150\, cm$.
Now,substitute the values into the formula:
$\text{Area} = \frac{1}{2} \times 150\, cm \times 75\, cm$.
$\text{Area} = 75\, cm \times 75\, cm = 5625\, cm^2$.

(C) The triangle is an equilateral triangle because all sides are equal ($8 \text{ cm}$ each).
The formula for the area of an equilateral triangle is $\text{Area} = \frac{\sqrt{3}}{4} \times \text{side}^2$.
Substituting the side length $a = 8 \text{ cm}$:
$\text{Area} = \frac{\sqrt{3}}{4} \times 8^2$
$\text{Area} = \frac{\sqrt{3}}{4} \times 64$
$\text{Area} = 16 \sqrt{3} \text{ cm}^2$.

(D) Let the lateral sides be $5x$ and the base be $4x$.
Since it is an isosceles triangle,the two equal sides are $5x$ and $5x$.
The perimeter is the sum of all sides: $5x + 5x + 4x = 14x$.
Given the perimeter is $14 \ cm$,we have $14x = 14$,which gives $x = 1$.
Thus,the sides of the triangle are $5 \ cm, 5 \ cm$,and $4 \ cm$.
The semi-perimeter $s$ is calculated as $s = \frac{5+5+4}{2} = 7 \ cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{7(7-5)(7-5)(7-4)} = \sqrt{7 \times 2 \times 2 \times 3}$.
Area $= \sqrt{7 \times 4 \times 3} = 2 \sqrt{21} \ cm^2$.

(C) The formula for the altitude $(h)$ of an equilateral triangle with side length $(a)$ is given by:
$h = \frac{\sqrt{3}}{2} \times a$
Given that the altitude $h = 6 \, cm$.
Substituting the value in the formula:
$6 = \frac{\sqrt{3}}{2} \times a$
$a = \frac{6 \times 2}{\sqrt{3}}$
$a = \frac{12}{\sqrt{3}}$
Rationalizing the denominator:
$a = \frac{12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{12 \sqrt{3}}{3} = 4 \sqrt{3} \, cm$
Therefore,the side of the equilateral triangle is $4 \sqrt{3} \, cm$.

(A) The height $(h)$ of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} \times a$,where $a$ is the side length.
Given $h = 9 \ cm$,we have $9 = \frac{\sqrt{3}}{2} \times a$.
Thus,$a = \frac{18}{\sqrt{3}} = 6\sqrt{3} \ cm$.
The area $(A)$ of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} \times a^2$.
Substituting the value of $a$: $A = \frac{\sqrt{3}}{4} \times (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times (36 \times 3) = \frac{\sqrt{3}}{4} \times 108 = 27\sqrt{3} \ cm^2$.

(A) For a square,the diagonal is given by $d = a \sqrt{2}$,where $a$ is the side length.
Given $d = 12 \sqrt{2} \text{ cm}$,so $a = 12 \text{ cm}$.
The perimeter of the square is $P = 4a = 4 \times 12 = 48 \text{ cm}$.
Since the perimeter of the equilateral triangle is equal to the perimeter of the square,the perimeter of the triangle is $48 \text{ cm}$.
Let the side of the equilateral triangle be $s$. Then $3s = 48 \text{ cm}$,which gives $s = 16 \text{ cm}$.
The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} s^2$.
Area $= \frac{\sqrt{3}}{4} \times (16)^2 = \frac{\sqrt{3}}{4} \times 256 = 64 \sqrt{3} \text{ cm}^2$.

(D) Let the width of the rectangular lawn be $w$ meters.
Then,the length $l = 2w$ meters.
The area of the lawn is given as $\frac{2}{3}$ hectares.
Since $1 \text{ hectare} = 10,000 \, m^2$,the area is $\frac{2}{3} \times 10,000 = \frac{20,000}{3} \, m^2$.
The area of a rectangle is $\text{length} \times \text{width} = l \times w$.
Substituting the values: $(2w) \times w = \frac{20,000}{3}$.
$2w^2 = \frac{20,000}{3} \Rightarrow w^2 = \frac{10,000}{3}$.
$w = \sqrt{\frac{10,000}{3}} = \frac{100}{\sqrt{3}} \, m$.
The length $l = 2w = 2 \times \frac{100}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, m$.

(D) The shortest route between $X$ and $Z$ is the hypotenuse of the right-angled triangle $\triangle XYZ$.
By Pythagoras theorem,$XZ^2 = XY^2 + YZ^2$.
Given $XY = 15 \ m$ and $YZ = 20 \ m$.
$XZ = \sqrt{15^2 + 20^2} \ m$.
$XZ = \sqrt{225 + 400} \ m$.
$XZ = \sqrt{625} \ m$.
$XZ = 25 \ m$.

(A) Area of the field $= \frac{\text{Total cost}}{\text{Rate}} = \frac{675}{50} = 13.5 \text{ hectares}$.
Since $1 \text{ hectare} = 10,000 \, m^2$,the area is $13.5 \times 10,000 = 135,000 \, m^2$.
Let the altitude be $x \, m$. Then the base is $3x \, m$.
The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{altitude}$.
So,$\frac{1}{2} \times 3x \times x = 135,000$.
$\frac{3x^2}{2} = 135,000$.
$3x^2 = 270,000$.
$x^2 = 90,000$.
$x = 300 \, m$ (Altitude).
Base $= 3x = 3 \times 300 = 900 \, m$.

(A) Let the two sides of the right triangle be $a$ and $b$. The hypotenuse is $5 \, cm$.
Perimeter $= a + b + 5 = 12 \, cm \implies a + b = 7 \, cm$.
By Pythagoras theorem,$a^2 + b^2 = 5^2 = 25$.
We know that $(a + b)^2 = a^2 + b^2 + 2ab$.
Substituting the values: $(7)^2 = 25 + 2ab$.
$49 = 25 + 2ab \implies 2ab = 24 \implies ab = 12$.
Now,$(a - b)^2 = (a + b)^2 - 4ab = (7)^2 - 4(12) = 49 - 48 = 1$.
So,$a - b = 1$ (assuming $a > b$).
Solving $a + b = 7$ and $a - b = 1$,we get $2a = 8 \implies a = 4 \, cm$ and $b = 3 \, cm$.
Area of the triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times b = \frac{1}{2} \times 4 \times 3 = 6 \, cm^2$.

(B) Radius of outer circle $(R) = 20 \, cm$
Radius of inner circle $(r) = 15 \, cm$
$\therefore$ Area of ring $= (\text{Area of outer circle} - \text{Area of inner circle})$
$= \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \frac{22}{7}(20^2 - 15^2)$
$= \frac{22}{7}(20 + 15)(20 - 15) = \frac{22}{7}(35)(5) = 22 \times 5 \times 5 = 550 \, cm^2$

(A) The minute hand of a clock completes a full circle $(360^{\circ})$ in $60 \, min$.
Therefore,the angle covered by the minute hand in $30 \, min$ is $\theta = \frac{360^{\circ}}{60} \times 30 = 180^{\circ}$.
The length of the minute hand is the radius of the circle,so $r = 14 \, cm$.
The area covered by the minute hand is the area of the sector formed by the angle $\theta$:
$\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^2$
$\text{Area} = \frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14$
$\text{Area} = \frac{1}{2} \times 22 \times 2 \times 14$
$\text{Area} = 11 \times 28 = 308 \, cm^2$.

(D) Let the radius of the inner circle be $r$ and the radius of the outer circle be $R$.
Given that the sum of their areas is $116 \, \pi \, cm^2$,we have $\pi R^2 + \pi r^2 = 116 \, \pi$,which simplifies to $R^2 + r^2 = 116$.
Since the circles touch internally,the distance between their centers is $R - r = 6 \, cm$.
We know the identity $(R + r)^2 + (R - r)^2 = 2(R^2 + r^2)$.
Substituting the known values: $(R + r)^2 + (6)^2 = 2(116)$.
$(R + r)^2 + 36 = 232$.
$(R + r)^2 = 232 - 36 = 196$.
Taking the square root,$R + r = 14$.
Now we have a system of linear equations: $R + r = 14$ and $R - r = 6$.
Adding the two equations: $2R = 20 \Rightarrow R = 10 \, cm$.
Subtracting the equations: $2r = 8 \Rightarrow r = 4 \, cm$.
Thus,the radii are $10 \, cm$ and $4 \, cm$.

(D) The side of the square is $a = 21 \, m$.
Area of the square $= a^{2} = 21 \times 21 = 441 \, m^{2}$.
There are four semi-circles on the four sides of the square. The diameter of each semi-circle is equal to the side of the square,which is $21 \, m$.
Therefore,the radius of each semi-circle is $r = \frac{21}{2} = 10.5 \, m$.
Area of four semi-circles $= 4 \times (\frac{1}{2} \times \pi \times r^{2}) = 2 \times \pi \times r^{2}$.
Area $= 2 \times \frac{22}{7} \times 10.5 \times 10.5 = 2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 22 \times 3 \times 10.5 = 693 \, m^{2}$.
Total area of the bed $= 441 + 693 = 1134 \, m^{2}$.
Given that each rose plant needs $6 \, m^{2}$ of space.
Number of plants $= \frac{\text{Total Area}}{\text{Area per plant}} = \frac{1134}{6} = 189$.

(D) In an equilateral triangle,the median is the same as the altitude.
Let the side of the equilateral triangle be '$a$'.
The altitude (median) '$x$' is given by the formula: $x = \frac{\sqrt{3}}{2} a$.
From this,we can express the side '$a$' in terms of '$x$': $a = \frac{2x}{\sqrt{3}}$.
The area of an equilateral triangle is given by: $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Substituting the value of '$a$': $\text{Area} = \frac{\sqrt{3}}{4} \left( \frac{2x}{\sqrt{3}} \right)^2$.
$\text{Area} = \frac{\sqrt{3}}{4} \times \frac{4x^2}{3}$.
$\text{Area} = \frac{\sqrt{3} x^2}{3} = \frac{x^2}{\sqrt{3}}$.

(B) The circumference of the circle is given by the formula $C = 2 \pi r$.
Given $r = 42 \ cm$,the circumference is $C = 2 \times \frac{22}{7} \times 42 = 2 \times 22 \times 6 = 264 \ cm$.
Since the wire is bent into a square,the perimeter of the square is equal to the circumference of the circle.
Perimeter of the square $= 4 \times \text{side} = 264 \ cm$.
Therefore,the side of the square is $\frac{264}{4} = 66 \ cm$.

(A) The circumference of the wheel is given by the formula $C = \pi \times d$,where $d$ is the diameter.
Given $d = 105\, cm$,the circumference $C = \frac{22}{7} \times 105 = 22 \times 15 = 330\, cm$.
This means the wheel covers $330\, cm$ in one complete revolution.
The total distance to be covered is $330\, m$. Converting this to centimeters,we get $330\, m = 330 \times 100 = 33000\, cm$.
The number of revolutions is calculated by dividing the total distance by the circumference of the wheel:
$\text{Number of revolutions} = \frac{\text{Total distance}}{\text{Circumference}} = \frac{33000\, cm}{330\, cm} = 100$.
Therefore,the wheel will rotate $100$ times.

(A) Let the original length be $L$ and the original width be $W$. The original area is $A = L \times W$.
The new length $L' = L + 0.60L = 1.6L = \frac{8}{5}L$.
Let the new width be $W'$. To maintain the same area,$L' \times W' = L \times W$.
Substituting $L'$,we get $\frac{8}{5}L \times W' = L \times W$,which simplifies to $W' = \frac{5}{8}W$.
The decrease in width is $W - W' = W - \frac{5}{8}W = \frac{3}{8}W$.
The percentage decrease is $\frac{\text{Decrease}}{\text{Original Width}} \times 100 = \frac{\frac{3}{8}W}{W} \times 100 = \frac{3}{8} \times 100 = 37.5 \% = 37 \frac{1}{2} \%$.

(C) Let the original length be $x$ and the original breadth be $y$.
Original Area $= x \times y = xy$.
New length $= x + 50\% \text{ of } x = x + 0.5x = 1.5x = \frac{3x}{2}$.
New breadth $= y + 20\% \text{ of } y = y + 0.2y = 1.2y = \frac{6y}{5}$.
New Area $= \text{New length} \times \text{New breadth} = \frac{3x}{2} \times \frac{6y}{5} = \frac{18xy}{10} = \frac{9}{5}xy$.
Therefore,the new area is $\frac{9}{5}$ times the original area.

(B) When two cubes of edge $10 \, cm$ are joined end-to-end,they form a cuboid.
Length of the resulting cuboid $(l) = 10 \, cm + 10 \, cm = 20 \, cm$.
Breadth of the resulting cuboid $(b) = 10 \, cm$.
Height of the resulting cuboid $(h) = 10 \, cm$.
The surface area of a cuboid is given by the formula: $2(lb + bh + hl)$.
Substituting the values: $2(20 \times 10 + 10 \times 10 + 10 \times 20) \, cm^2$.
$= 2(200 + 100 + 200) \, cm^2$.
$= 2(500) \, cm^2 = 1000 \, cm^2$.

(B) Let the dimensions of the cuboid be $l, b,$ and $h.$
The volume of the cuboid is given by $V = l \times b \times h.$
The areas of the three adjacent faces are given as:
$x = l \times b$
$y = b \times h$
$z = h \times l$
Multiplying these three areas together,we get:
$x \times y \times z = (l \times b) \times (b \times h) \times (h \times l)$
$xyz = l^{2} \times b^{2} \times h^{2}$
$xyz = (l \times b \times h)^{2}$
Since $V = l \times b \times h$,substituting $V$ into the equation gives:
$xyz = V^{2}$
Therefore,$V^{2} = xyz$.

(B) Given: Volume of cylinder $V = 448 \pi \, cm^3$ and height $h = 7 \, cm$.
Let the radius be $r$.
The formula for volume is $V = \pi r^2 h$.
Substituting the values: $448 \pi = \pi r^2 (7)$.
$r^2 = \frac{448}{7} = 64$,so $r = 8 \, cm$.
Lateral Surface Area $(LSA)$ $= 2 \pi r h = 2 \times \frac{22}{7} \times 8 \times 7 = 352 \, cm^2$.
Total Surface Area $(TSA)$ $= 2 \pi r h + 2 \pi r^2 = 2 \pi r (h + r)$.
$TSA$ $= 2 \times \frac{22}{7} \times 8 \times (7 + 8) = 2 \times \frac{22}{7} \times 8 \times 15 = \frac{5280}{7} \approx 754.286 \, cm^2$.

(D) Given: Radius $(r) = 5 \text{ cm}$,Height $(h) = 12 \text{ cm}$.
First,calculate the slant height $(l)$ using the formula $l = \sqrt{h^2 + r^2}$.
$l = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}$.
The lateral (curved) surface area of a cone is given by the formula $A = \pi r l$.
Using $\pi \approx \frac{22}{7}$:
$A = \frac{22}{7} \times 5 \times 13 = \frac{1430}{7} \approx 204.285 \text{ cm}^2$.
Rounding to one decimal place,we get $204.3 \text{ cm}^2$.

(A) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^{3}$.
Given $V = 38808 \, cm^{3}$.
$\frac{4}{3} \times \frac{22}{7} \times r^{3} = 38808$.
$r^{3} = 38808 \times \frac{3 \times 7}{4 \times 22} = 9261$.
Taking the cube root,$r = \sqrt[3]{9261} = 21 \, cm$.
The surface area of a sphere is given by $A = 4 \pi r^{2}$.
$A = 4 \times \frac{22}{7} \times 21 \times 21 = 4 \times 22 \times 3 \times 21 = 5544 \, cm^{2}$.
Thus,the radius is $21 \, cm$ and the surface area is $5544 \, cm^{2}$.

(A) Let the volumes of the two hemispheres be $V_1$ and $V_2$,and their radii be $r_1$ and $r_2$ respectively.
The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$.
Given the ratio of volumes: $V_1 : V_2 = 8 : 27$.
Substituting the formula: $\frac{\frac{2}{3} \pi r_1^3}{\frac{2}{3} \pi r_2^3} = \frac{8}{27}$.
Simplifying the expression: $\frac{r_1^3}{r_2^3} = \frac{8}{27}$.
Taking the cube root on both sides: $\frac{r_1}{r_2} = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}$.
Therefore,the ratio of their radii is $2:3$.

(A) The volume of the material remains constant when a sphere is reshaped into a wire.
Radius of the sphere $R = \frac{18}{2} = 9\, cm$.
Diameter of the wire $= 40\, mm = 4\, cm$,so the radius of the wire $r = \frac{4}{2} = 2\, cm$.
Let the length of the wire be $h\, cm$.
Volume of the sphere $= \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (9)^3 = \frac{4}{3} \pi (729) = 972\pi\, cm^3$.
Volume of the cylindrical wire $= \pi r^2 h = \pi (2)^2 h = 4\pi h\, cm^3$.
Equating the volumes: $972\pi = 4\pi h$.
$h = \frac{972}{4} = 243\, cm$.
Thus,the length of the wire is $243\, cm$.

(A) Let the initial edge of the cube be $x \text{ cm}$.
The initial surface area of the cube is $S_1 = 6x^2$.
The new edge after a $50 \%$ increase is $x + 0.5x = 1.5x = \frac{3x}{2} \text{ cm}$.
The new surface area is $S_2 = 6 \left( \frac{3x}{2} \right)^2 = 6 \left( \frac{9x^2}{4} \right) = \frac{54x^2}{4} = 13.5x^2$.
The increase in surface area is $S_2 - S_1 = 13.5x^2 - 6x^2 = 7.5x^2$.
The percentage increase is $\left( \frac{\text{Increase}}{\text{Initial Area}} \right) \times 100 = \left( \frac{7.5x^2}{6x^2} \right) \times 100 = 1.25 \times 100 = 125 \%$.

(C) First,convert all dimensions to meters $(m)$:
Wall dimensions: $10\, m$,$4\, dm = 0.4\, m$,$5\, m$.
Volume of the wall $= 10\, m \times 0.4\, m \times 5\, m = 20\, m^3$.
Volume occupied by mortar $= \frac{1}{10} \times 20\, m^3 = 2\, m^3$.
Volume to be occupied by bricks $= 20\, m^3 - 2\, m^3 = 18\, m^3$.
Volume of one brick $= 0.25\, m \times 0.15\, m \times 0.08\, m = 0.003\, m^3 = \frac{3}{1000}\, m^3$.
Number of bricks required $= \frac{\text{Volume of bricks}}{\text{Volume of one brick}} = \frac{18}{3/1000} = \frac{18 \times 1000}{3} = 6000$.