Measurement of Area Questions in English

(A) Let the two sides of the right triangle be $a$ and $b$,and the hypotenuse be $c = 20\, cm$.
The semi-perimeter $s = \frac{a + b + c}{2} = 24\, cm$.
Therefore,$a + b + 20 = 2 \times 24 = 48$,which gives $a + b = 28\, cm$ $...(1)$.
By the Pythagorean theorem,$a^2 + b^2 = c^2 = 20^2 = 400$.
We know that $(a + b)^2 = a^2 + b^2 + 2ab$.
Substituting the values,$28^2 = 400 + 2ab$.
$784 = 400 + 2ab$,so $2ab = 384$,which means $ab = 192$.
Now,$(a - b)^2 = (a + b)^2 - 4ab = 28^2 - 4(192) = 784 - 768 = 16$.
Thus,$a - b = \sqrt{16} = 4\, cm$ $...(2)$.
Adding $(1)$ and $(2)$,$2a = 32$,so $a = 16\, cm$.
Subtracting $(2)$ from $(1)$,$2b = 24$,so $b = 12\, cm$.
Thus,the other two sides are $16\, cm$ and $12\, cm$.

(B) Let the sides of the triangle be $3x$,$4x$,and $5x$.
Given that the perimeter is $36 \text{ cm}$,we have:
$3x + 4x + 5x = 36$
$12x = 36$
$x = 3$
Therefore,the sides are:
$a = 3(3) = 9 \text{ cm}$
$b = 4(3) = 12 \text{ cm}$
$c = 5(3) = 15 \text{ cm}$
The semi-perimeter $s$ is given by:
$s = \frac{a+b+c}{2} = \frac{36}{2} = 18 \text{ cm}$
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{18(18-9)(18-12)(18-15)}$
$\text{Area} = \sqrt{18 \times 9 \times 6 \times 3}$
$\text{Area} = \sqrt{2916} = 54 \text{ cm}^2$
Alternatively,since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$,it is a right-angled triangle.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \times 12 = 54 \text{ cm}^2$.

(C) Let the sides of the triangle be $a = 50\, m$,$b = 78\, m$,and $c = 112\, m$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{50 + 78 + 112}{2} = \frac{240}{2} = 120\, m$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{120(120-50)(120-78)(120-112)}$
$\text{Area} = \sqrt{120 \times 70 \times 42 \times 8}$
$\text{Area} = \sqrt{2822400} = 1680\, m^2$.
The area of a triangle is also given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Here,the base is $112\, m$ and the height $h$ is the perpendicular length.
$\frac{1}{2} \times 112 \times h = 1680$
$56 \times h = 1680$
$h = \frac{1680}{56} = 30\, m$.

(A) Let the wall be represented by $AC$ and the ground by $AB$,where $AC = 12\, m$ and $AB = 5\, m$.
The ladder is represented by the hypotenuse $BC$ of the right-angled triangle $\triangle ABC$.
According to the Pythagorean theorem,$BC^2 = AC^2 + AB^2$.
$BC^2 = 12^2 + 5^2 = 144 + 25 = 169$.
Therefore,$BC = \sqrt{169} = 13\, m$.
Thus,the length of the ladder is $13\, m$.

(C) Let the ladder be $AC$ and $CE$,where $AC = CE = 65 \ cm$.
Let $AB = 63 \ cm$ be the height of the window on one side and $ED = 56 \ cm$ be the height of the point on the opposite side.
In right-angled triangle $\triangle ABC$,by Pythagoras theorem:
$BC^2 + AB^2 = AC^2$
$BC^2 + 63^2 = 65^2$
$BC^2 = 4225 - 3969 = 256$
$BC = \sqrt{256} = 16 \ cm$.
In right-angled triangle $\triangle CDE$,by Pythagoras theorem:
$CD^2 + ED^2 = CE^2$
$CD^2 + 56^2 = 65^2$
$CD^2 = 4225 - 3136 = 1089$
$CD = \sqrt{1089} = 33 \ cm$.
Therefore,the width of the street is $BD = BC + CD = 16 + 33 = 49 \ cm$.

(C) The area of a triangle with base $x$ and altitude $h$ is given by $\frac{1}{2} \times x \times h$.
The area of a square with side $x$ is given by $x^2$.
According to the problem,the area of the triangle is equal to the area of the square:
$\frac{1}{2} \times x \times h = x^2$
Multiplying both sides by $2$ and dividing by $x$ (assuming $x \neq 0$):
$h = 2x$
Therefore,the altitude of the triangle is $2x$.

(C) Let the base of the triangle be $3x$ and the height be $4x$.
The formula for the area of a triangle is $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Substituting the given values: $150 = \frac{1}{2} \times 3x \times 4x$.
$150 = 6x^{2}$.
$x^{2} = \frac{150}{6} = 25$.
$x = \sqrt{25} = 5$.
Therefore,the height is $4x = 4 \times 5 = 20 \, m$.

(C) The cost of cultivation is ₹ $495.72$ at a rate of ₹ $36.72$ per hectare.
Area of the field in hectares $= \frac{495.72}{36.72} = 13.5 \text{ hectares}$.
Since $1 \text{ hectare} = 10,000 \, m^2$,the area in square meters is $13.5 \times 10,000 = 135,000 \, m^2$.
Let the height of the triangular field be $h = x \, m$.
Then,the base $b = 3x \, m$.
The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = 135,000$.
$\frac{1}{2} \times 3x \times x = 135,000$.
$1.5x^2 = 135,000$.
$x^2 = \frac{135,000}{1.5} = 90,000$.
$x = \sqrt{90,000} = 300$.
Therefore,height $= 300 \, m$ and base $= 3 \times 300 = 900 \, m$.

(C) Let the original sides of the triangle be $a, b,$ and $c$.
The semi-perimeter $s$ is given by $s = \frac{1}{2}(a + b + c)$.
The area of the original triangle is $A = \sqrt{s(s - a)(s - b)(s - c)}$.
If the sides are doubled,the new sides are $2a, 2b,$ and $2c$.
The new semi-perimeter $S$ is $S = \frac{1}{2}(2a + 2b + 2c) = a + b + c = 2s$.
The area of the new triangle $A'$ is $\sqrt{S(S - 2a)(S - 2b)(S - 2c)}$.
Substituting $S = 2s$,we get $A' = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}$.
Factoring out $2$ from each term inside the square root,we get $A' = \sqrt{2s \cdot 2(s - a) \cdot 2(s - b) \cdot 2(s - c)} = \sqrt{16s(s - a)(s - b)(s - c)}$.
$A' = 4 \sqrt{s(s - a)(s - b)(s - c)} = 4A$.
Thus,the area becomes $4$ times the original area.

(A) Let the sides of the triangle be $a = 85 \, m$,$b = 154 \, m$,and $c$ be the third side.
Given perimeter $P = 324 \, m$.
So,$c = P - (a + b) = 324 - (85 + 154) = 324 - 239 = 85 \, m$.
The semi-perimeter $s = \frac{P}{2} = \frac{324}{2} = 162 \, m$.
Using Heron's formula,the area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{162(162-85)(162-154)(162-85)} = \sqrt{162 \times 77 \times 8 \times 77}$.
Area $= \sqrt{162 \times 8 \times 77^2} = \sqrt{1296 \times 5929} = 36 \times 77 = 2772 \, m^2$.
The cost of ploughing at the rate of ₹ $10$ per $m^2$ is $2772 \times 10 = ₹ 27720$.

(C) The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4} \times (\text{side})^2$.
Given that the side length is $2 \sqrt{3} \text{ cm}$.
Substituting the value into the formula:
Area $= \frac{\sqrt{3}}{4} \times (2 \sqrt{3})^2$
Area $= \frac{\sqrt{3}}{4} \times (4 \times 3)$
Area $= \frac{\sqrt{3}}{4} \times 12$
Area $= 3 \sqrt{3} \text{ cm}^2$.

(C) The formula for the height $(h)$ of an equilateral triangle with side length $(a)$ is given by:
$h = \frac{\sqrt{3}}{2} \times a$
Given that the height $h = 2 \sqrt{3} \text{ cm}$.
Substituting the value into the formula:
$2 \sqrt{3} = \frac{\sqrt{3}}{2} \times a$
Dividing both sides by $\sqrt{3}$:
$2 = \frac{a}{2}$
Multiplying both sides by $2$:
$a = 4 \text{ cm}$.
Therefore,the length of the side is $4 \text{ cm}$.

(B) The perimeter of an equilateral triangle is given by $3 \times \text{side}$.
Given,$3 \times \text{side} = 12\, m$.
Therefore,$\text{side} = \frac{12}{3} = 4\, m$.
The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4} \times (\text{side})^2$.
Substituting the value of the side: $\text{Area} = \frac{\sqrt{3}}{4} \times (4)^2$.
$\text{Area} = \frac{\sqrt{3}}{4} \times 16 = 4\sqrt{3}\, m^2$.

(A) The perimeter of an equilateral triangle is given by $3 \times \text{side} = 24 \, cm$.
Therefore,the side length is $\text{side} = \frac{24}{3} = 8 \, cm$.
The formula for the height $(h)$ of an equilateral triangle is $h = \frac{\sqrt{3}}{2} \times \text{side}$.
Substituting the value of the side,we get $h = \frac{\sqrt{3}}{2} \times 8 = 4 \sqrt{3} \, cm$.

(B) Let the two legs of the right-angled triangle be $a$ and $b$,and the hypotenuse be $c = 39\, cm$.
Given the perimeter $P = a + b + c = 90\, cm$.
Therefore,$a + b + 39 = 90$,which implies $a + b = 51\, cm$.
From the Pythagorean theorem,$a^2 + b^2 = c^2 = 39^2 = 1521$.
We know that $(a + b)^2 = a^2 + b^2 + 2ab$.
Substituting the values: $51^2 = 1521 + 2ab$.
$2601 = 1521 + 2ab \Rightarrow 2ab = 1080 \Rightarrow ab = 540$.
Now,we have $a + b = 51$ and $ab = 540$. These are roots of the quadratic equation $t^2 - 51t + 540 = 0$.
Solving for $t$: $t = \frac{51 \pm \sqrt{51^2 - 4(540)}}{2} = \frac{51 \pm \sqrt{2601 - 2160}}{2} = \frac{51 \pm \sqrt{441}}{2} = \frac{51 \pm 21}{2}$.
$t_1 = \frac{72}{2} = 36$ and $t_2 = \frac{30}{2} = 15$.
Thus,the other two sides are $36\, cm$ and $15\, cm$.

(B) Let the base of the isosceles triangle be $b$ and each equal side be $a$.
Given that $a = \frac{5}{8} b$.
The perimeter of the triangle is $2a + b = 306\, m$.
Substituting $a$ in the perimeter equation: $2(\frac{5}{8} b) + b = 306$.
$\frac{5}{4} b + b = 306 \Rightarrow \frac{9}{4} b = 306$.
$b = \frac{306 \times 4}{9} = 34 \times 4 = 136\, m$.
Now,$a = \frac{5}{8} \times 136 = 5 \times 17 = 85\, m$.
The area of an isosceles triangle is given by $\frac{b}{4} \sqrt{4a^2 - b^2}$.
Area $= \frac{136}{4} \sqrt{4(85)^2 - (136)^2}$.
Area $= 34 \sqrt{4(7225) - 18496} = 34 \sqrt{28900 - 18496}$.
Area $= 34 \sqrt{10404} = 34 \times 102 = 3468\, m^2$.

(C) In an isosceles right-angled triangle,let the two equal sides be $a$.
By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{a^2 + a^2} = a\sqrt{2}$.
Given that the hypotenuse $h = 8\, cm$,we have $a\sqrt{2} = 8$.
Thus,$a = \frac{8}{\sqrt{2}} = 4\sqrt{2}\, cm$.
The area of a right-angled triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Since the base and height are both $a$,$\text{Area} = \frac{1}{2} \times a^2$.
Substituting the value of $a$,$\text{Area} = \frac{1}{2} \times (4\sqrt{2})^2 = \frac{1}{2} \times (16 \times 2) = \frac{1}{2} \times 32 = 16\, cm^{2}$.

(B) Let the lateral side be $5x$ and the base be $4x$.
Since it is an isosceles triangle,the two lateral sides are equal.
Perimeter $= 5x + 5x + 4x = 14x$.
Given perimeter $= 14 \ cm$,so $14x = 14 \Rightarrow x = 1$.
Thus,the sides are $5 \ cm, 5 \ cm$,and $4 \ cm$.
Using the formula for the area of an isosceles triangle: $\text{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}$,where $a = 5$ and $b = 4$.
$\text{Area} = \frac{4}{4} \sqrt{4(5)^2 - (4)^2} = \sqrt{100 - 16} = \sqrt{84} = 2\sqrt{21} \ cm^2$.

(C) Let the original sides of the triangle be $a, b,$ and $c$. The original area $A$ is given by Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{a+b+c}{2}$.
When each side is increased by $200 \%$,the new sides become $a' = a + 2a = 3a$,$b' = b + 2b = 3b$,and $c' = c + 2c = 3c$.
The new semi-perimeter $s'$ is $s' = \frac{3a+3b+3c}{2} = 3 \left( \frac{a+b+c}{2} \right) = 3s$.
The new area $A'$ is $A' = \sqrt{s'(s'-a')(s'-b')(s'-c')} = \sqrt{3s(3s-3a)(3s-3b)(3s-3c)}$.
Factoring out $3$ from each term inside the square root: $A' = \sqrt{3s \cdot 3(s-a) \cdot 3(s-b) \cdot 3(s-c)} = \sqrt{81 \cdot s(s-a)(s-b)(s-c)} = 9 \sqrt{s(s-a)(s-b)(s-c)} = 9A$.
The increase in area is $A' - A = 9A - A = 8A$.
The percentage increase is $\frac{8A}{A} \times 100 \% = 800 \%$.

(C) Let the length of each equal side of the right-angled isosceles triangle be $x \text{ m}$.
According to the Pythagorean theorem,$x^2 + x^2 = (50 \sqrt{2})^2$.
$2x^2 = 2500 \times 2 = 5000$.
$x^2 = 2500$,which gives $x = 50 \text{ m}$.
The perimeter of the triangle is the sum of all sides: $P = x + x + 50 \sqrt{2} = 50 + 50 + 50 \times 1.414 = 100 + 70.7 = 170.7 \text{ m}$.
The total cost of fencing is $170.7 \times 3 = ₹512.1$.
Since $512.1 > 500$,the correct option is more than ₹$500$.

(D) The area of an isosceles triangle with base $b$ and equal sides $a$ is given by the formula: $\text{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}$.
Given $b = 16 \, cm$ and $a = 10 \, cm$,we have:
$\text{Area} = \frac{16}{4} \sqrt{4(10)^2 - (16)^2} = 4 \sqrt{400 - 256} = 4 \sqrt{144} = 4 \times 12 = 48 \, cm^2$.
Let the side of the equilateral triangle be $s$. The area of an equilateral triangle is $\frac{\sqrt{3}}{4} s^2$.
Equating the areas: $\frac{\sqrt{3}}{4} s^2 = 48$.
$s^2 = \frac{48 \times 4}{\sqrt{3}} = \frac{192}{1.732} \approx 110.85$.
$s = \sqrt{110.85} \approx 10.53 \, cm$.
Since $10.53$ is not exactly $10.5$,the correct option is $D$ (None of these).

(C) Let the two equal sides of the right-angled isosceles triangle be $x \, m$.
By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{x^2 + x^2} = x\sqrt{2}$.
The perimeter $P$ is the sum of all sides: $P = x + x + x\sqrt{2} = 2x + x\sqrt{2} = x(2 + \sqrt{2})$.
We can rewrite this as $P = x\sqrt{2}(\sqrt{2} + 1) = h(\sqrt{2} + 1)$.
Given the perimeter $P = 4\sqrt{2} + 4 = 4(\sqrt{2} + 1) \, m$.
Equating the two expressions: $h(\sqrt{2} + 1) = 4(\sqrt{2} + 1)$.
Therefore,the hypotenuse $h = 4 \, m$.

(A) Let the sides of the parallelogram be $a = 60 \, m$ and $b = 40 \, m$,and the diagonal be $d = 80 \, m$.
The area of a parallelogram can be calculated as twice the area of the triangle formed by two sides and the diagonal.
First,calculate the semi-perimeter $(s)$ of the triangle:
$s = \frac{a + b + d}{2} = \frac{60 + 40 + 80}{2} = \frac{180}{2} = 90 \, m$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-d)}$
$= \sqrt{90(90-60)(90-40)(90-80)}$
$= \sqrt{90 \times 30 \times 50 \times 10}$
$= \sqrt{1350000} = 300 \sqrt{15} \, m^2$.
The area of the parallelogram is twice the area of the triangle:
$\text{Area of parallelogram} = 2 \times 300 \sqrt{15} = 600 \sqrt{15} \, m^2$.

(C) The area of a parallelogram is calculated using the formula: $\text{Area} = \text{base} \times \text{height}$.
Here, the base is the side of the parallelogram, which is $14\, cm$.
The distance from the opposite side is the height (altitude) of the parallelogram, which is $16\, cm$.
Therefore, $\text{Area} = 14\, cm \times 16\, cm = 224\, cm^2$.

(B) Let the sides of the parallelogram be $a = 65 \, m$,$b = 119 \, m$,and the diagonal be $d = 156 \, m$.
The area of the parallelogram is twice the area of the triangle formed by these three sides.
First,calculate the semi-perimeter $s$ of the triangle:
$s = \frac{a + b + d}{2} = \frac{65 + 119 + 156}{2} = \frac{340}{2} = 170 \, m$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-d)}$
$= \sqrt{170(170-65)(170-119)(170-156)}$
$= \sqrt{170 \times 105 \times 51 \times 14}$
$= \sqrt{12751200} = 3570 \, m^2$.
Area of the parallelogram $= 2 \times 3570 = 7140 \, m^2$.
Cost of gravelling $= \text{Area} \times \text{Rate} = 7140 \times 10 = ₹ 71400$.

(A) The area of a parallelogram is calculated by the formula: $\text{Area} = \text{base} \times \text{corresponding altitude}$.
Given,$\text{base} = 10\, m$ and $\text{altitude} = 7\, m$.
Therefore,$\text{Area} = 10\, m \times 7\, m = 70\, m^2$.

(B) The area of a parallelogram is given by the product of a base and its corresponding height (distance between parallel sides).
Let the sides be $a = 8\, m$ and $b = 5\, m$.
The distance between the longer sides $(8\, m)$ is given as $h_1 = 4\, m$.
Therefore,the area of the parallelogram $= \text{base} \times \text{height} = 8\, m \times 4\, m = 32\, m^2$.
Let the distance between the shorter sides $(5\, m)$ be $h_2$.
Since the area remains the same,we have $5\, m \times h_2 = 32\, m^2$.
$h_2 = \frac{32}{5} = 6.4\, m$.
Thus,the distance between the shorter sides is $6.4\, m$.

(C) The area of a quadrilateral can be calculated using the formula: $\text{Area} = \frac{1}{2} \times d \times (p_{1} + p_{2})$,where $d$ is the length of the diagonal and $p_{1}, p_{2}$ are the lengths of the perpendiculars from the opposite vertices to the diagonal.
Given: $\text{Area} = 420 \, m^{2}$,$p_{1} = 18 \, m$,and $p_{2} = 12 \, m$.
Substituting the values into the formula:
$420 = \frac{1}{2} \times d \times (18 + 12)$
$420 = \frac{1}{2} \times d \times 30$
$420 = 15 \times d$
$d = \frac{420}{15}$
$d = 28 \, m$.
Thus,the length of the diagonal is $28 \, m$.

(A) Let the base of the parallelogram be $x \, cm$.
Given that the altitude is twice the corresponding base,so the altitude is $2x \, cm$.
The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{altitude}$.
Substituting the given values: $72 = x \times 2x$.
$72 = 2x^{2}$.
Dividing both sides by $2$,we get $x^{2} = 36$.
Taking the square root of both sides,$x = 6$.
Therefore,the length of the base is $6 \, cm$.

(B) The formula for the area of a parallelogram is given by: $\text{Area} = \text{base} \times \text{height}$.
Given that the area is $240 \, cm^{2}$ and the height is $12 \, cm$.
Let the base be $x \, cm$.
Substituting the values into the formula: $240 = x \times 12$.
To find $x$,divide both sides by $12$: $x = \frac{240}{12} = 20 \, cm$.
Therefore,the base of the parallelogram is $20 \, cm$.

(A) The area of quadrilateral $ABCD$ is the sum of the areas of $\Delta ABC$ and $\Delta ADC$.
For $\Delta ABC$,the sides are $a = 20\, m, b = 13\, m, c = 21\, m$.
Semi-perimeter $s_1 = \frac{20 + 13 + 21}{2} = \frac{54}{2} = 27\, m$.
Area of $\Delta ABC = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{27(27 - 20)(27 - 13)(27 - 21)} = \sqrt{27 \times 7 \times 14 \times 6} = \sqrt{15876} = 126\, m^2$.
For $\Delta ADC$,the sides are $a = 10\, m, b = 17\, m, c = 21\, m$.
Semi-perimeter $s_2 = \frac{10 + 17 + 21}{2} = \frac{48}{2} = 24\, m$.
Area of $\Delta ADC = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{24(24 - 10)(24 - 17)(24 - 21)} = \sqrt{24 \times 14 \times 7 \times 3} = \sqrt{7056} = 84\, m^2$.
Total area of quadrilateral $ABCD = 126\, m^2 + 84\, m^2 = 210\, m^2$.

(A) Let the parallelogram be $ABCD$ with diagonals $AC = 72 \, cm$ and $BD = 30 \, cm$.
In a parallelogram,the diagonals bisect each other at point $O$.
Therefore,$OA = OC = \frac{72}{2} = 36 \, cm$ and $OB = OD = \frac{30}{2} = 15 \, cm$.
Since the diagonals of a rhombus bisect each other at right angles,if we assume this is a rhombus (or calculate the side length using the property of diagonals),the side length $s$ is given by $s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}$.
$s = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \, cm$.
Since all four sides of a rhombus are equal,the perimeter is $4 \times 39 = 156 \, cm$.
Note: $A$ parallelogram with diagonals bisecting at right angles is a rhombus. If it is not a rhombus,the perimeter cannot be uniquely determined. Given the options,it is implied to be a rhombus.

(C) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{altitude}$.
Given,$\text{base} = (x+4)$,$\text{altitude} = (x-3)$,and $\text{Area} = (x^{2}-4)$.
Substituting these values into the formula:
$(x^{2}-4) = (x+4)(x-3)$
$x^{2}-4 = x^{2} - 3x + 4x - 12$
$x^{2}-4 = x^{2} + x - 12$
Subtracting $x^{2}$ from both sides:
$-4 = x - 12$
$x = 12 - 4 = 8$.
Now,substitute $x = 8$ into the expression for the area:
$\text{Area} = x^{2}-4 = (8)^{2}-4 = 64 - 4 = 60$ sq units.

(B) For any parallelogram with adjacent sides $a$ and $b$ and diagonals $d_1$ and $d_2$,the parallelogram law states: $d_1^2 + d_2^2 = 2(a^2 + b^2)$.
Given $a = 12\, cm$,$b = 14\, cm$,and $d_1 = 16\, cm$.
Substituting the values into the formula:
$16^2 + d_2^2 = 2(12^2 + 14^2)$
$256 + d_2^2 = 2(144 + 196)$
$256 + d_2^2 = 2(340)$
$256 + d_2^2 = 680$
$d_2^2 = 680 - 256 = 424$
$d_2 = \sqrt{424} \approx 20.59\, cm \approx 20.6\, cm$.

(B) The area of a circle is given by the formula $A = \pi r^{2}$.
Given $A = 154 \, m^{2}$ and taking $\pi = \frac{22}{7}$:
$\frac{22}{7} \times r^{2} = 154$
$r^{2} = 154 \times \frac{7}{22}$
$r^{2} = 7 \times 7 = 49$
$r = \sqrt{49} = 7 \, m$.
The perimeter (circumference) of a circle is given by $P = 2 \pi r$.
$P = 2 \times \frac{22}{7} \times 7 = 44 \, m$.

(A) Let the side of the square be $x$ and the radius of the circle be $r$.
According to the problem,the perimeter of the circle is equal to the perimeter of the square:
$2 \pi r = 4x$
$x = \frac{\pi r}{2}$
Now,we compare the area of the circle $(A_c = \pi r^2)$ to the area of the square $(A_s = x^2)$:
$\frac{A_c}{A_s} = \frac{\pi r^2}{x^2} = \frac{\pi r^2}{(\frac{\pi r}{2})^2} = \frac{\pi r^2}{\frac{\pi^2 r^2}{4}}$
$\frac{A_c}{A_s} = \frac{4}{\pi} = \frac{4}{22/7} = \frac{4 \times 7}{22} = \frac{28}{22} = \frac{14}{11}$
Thus,the ratio of their areas is $14: 11$.

(D) Let the breadth of the rectangle be $x \, m$.
Then,the length of the rectangle is $3x \, m$.
The perimeter of a rectangle is given by the formula $P = 2(l + b)$.
Given that the perimeter is $96 \, m$,we have $2(3x + x) = 96$.
$2(4x) = 96 \Rightarrow 8x = 96$.
$x = 12 \, m$.
So,the breadth is $12 \, m$ and the length is $3 \times 12 = 36 \, m$.
The area of the rectangle is $\text{length} \times \text{breadth} = 36 \times 12 = 432 \, m^2$.

(D) Since the cow is tied at the corner of a rectangular field,the area it can graze forms a sector of a circle with a radius $r = 14\, m$ and a central angle of $90^{\circ}$ (as the corner of a rectangle is $90^{\circ}$).
The area of the sector is given by the formula: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^2$.
Substituting the values: $\text{Area} = \frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14$.
$\text{Area} = \frac{1}{4} \times \frac{22}{7} \times 196$.
$\text{Area} = \frac{1}{4} \times 22 \times 28$.
$\text{Area} = 154\, m^2$.

(C) The speed of the scooter is $66\, km/h$.
To find the distance covered in one minute,convert the speed to $cm/min$:
Distance per minute $= \frac{66 \times 1000 \times 100}{60} = 110000\, cm/min$.
The distance covered in one revolution is equal to the circumference of the wheel:
Circumference $= \pi \times d = \frac{22}{7} \times 70 = 220\, cm$.
The number of revolutions per minute is the total distance covered in one minute divided by the distance covered in one revolution:
Number of revolutions $= \frac{110000}{220} = 500$.

(A) The area of a circle is given by the formula $A = \pi r^2$,where $r$ is the radius.
For the first park,the diameter is $16 \ m$,so the radius $r_1 = 16 / 2 = 8 \ m$.
The area of the first park is $A_1 = \pi (8)^2 = 64\pi \ m^2$.
For the second park,the diameter is $12 \ m$,so the radius $r_2 = 12 / 2 = 6 \ m$.
The area of the second park is $A_2 = \pi (6)^2 = 36\pi \ m^2$.
The total area of the two parks is $A_{total} = 64\pi + 36\pi = 100\pi \ m^2$.
Let the radius of the new park be $R$. Since the new park occupies the same space,its area must be equal to the total area of the two small parks:
$\pi R^2 = 100\pi$
$R^2 = 100$
$R = \sqrt{100} = 10 \ m$.
Therefore,the radius of the new park is $10 \ m$.

(C) The length of the rectangular park is $L = 65\, m$ and the width is $W = 50\, m$.
The width of the paths is $w = 2\, m$.
The area of the path parallel to the length is $L \times w = 65 \times 2 = 130\, m^2$.
The area of the path parallel to the width is $W \times w = 50 \times 2 = 100\, m^2$.
Since the two paths intersect at the center,the intersection area is a square of side $w = 2\, m$,which is $w^2 = 2 \times 2 = 4\, m^2$.
The total area of the paths is $(130 + 100 - 4) = 226\, m^2$.
The cost of construction is $226 \times 17.25 = ₹ 3898.50$.

(D) The formula for the area of a trapezium is $\text{Area} = \frac{1}{2} \times (a + b) \times h$,where $a$ and $b$ are the lengths of the parallel sides and $h$ is the height (distance between them).
Given: $\text{Area} = 2500 \, m^2$,$a = 75 \, m$,and $h = 40 \, m$.
Substituting the values into the formula:
$2500 = \frac{1}{2} \times (75 + b) \times 40$
$2500 = (75 + b) \times 20$
Divide both sides by $20$:
$125 = 75 + b$
$b = 125 - 75$
$b = 50 \, m$.
Therefore,the length of the other parallel side is $50 \, m$.

(C) Let the breadth of the rectangle be $x \, cm$.
Then,the length of the rectangle is $(x + 3) \, cm$.
According to the problem,the numerical value of the area is equal to the numerical value of the perimeter.
Area $= \text{Length} \times \text{Breadth} = x(x + 3) \, cm^2$.
Perimeter $= 2(\text{Length} + \text{Breadth}) = 2(x + 3 + x) = 2(2x + 3) = (4x + 6) \, cm$.
Equating the two: $x(x + 3) = 4x + 6$.
$x^2 + 3x = 4x + 6$.
$x^2 - x - 6 = 0$.
Factoring the quadratic equation: $(x - 3)(x + 2) = 0$.
This gives $x = 3$ or $x = -2$.
Since the breadth cannot be negative,the breadth is $3 \, cm$.

(C) Let the sides of the two squares be $x$ and $y$ respectively.
The area of a square is given by the formula $\text{Area} = \text{side}^2$.
Given the ratio of the areas is $\frac{x^2}{y^2} = \frac{9}{1}$.
Taking the square root of both sides,we get $\frac{x}{y} = \sqrt{\frac{9}{1}} = \frac{3}{1}$.
The perimeter of a square is given by the formula $\text{Perimeter} = 4 \times \text{side}$.
The ratio of their perimeters is $\frac{4x}{4y} = \frac{x}{y}$.
Substituting the value of $\frac{x}{y}$,the ratio of the perimeters is $3: 1$.

(D) Let $x$ (in metres) be the width of the path.
The side of the inner square is $30\, m$.
The side of the outer square is $(30 + 2x)\, m$.
The area of the path is given by the difference between the area of the outer square and the area of the inner square.
Area of path $= (30 + 2x)^2 - 30^2 = 256$.
Expanding the equation: $(900 + 120x + 4x^2) - 900 = 256$.
$4x^2 + 120x = 256$.
Dividing by $4$: $x^2 + 30x = 64$.
$x^2 + 30x - 64 = 0$.
Factoring the quadratic equation: $(x + 32)(x - 2) = 0$.
This gives $x = -32$ or $x = 2$.
Since the width cannot be negative,we have $x = 2\, m$.

(C) The area grazed by the calf is the area of a circle with the rope length as the radius.
Initial radius $(r_1)$ = $12 \ m$.
Final radius $(r_2)$ = $23 \ m$.
Additional area grazed = Area with radius $r_2$ - Area with radius $r_1$.
Additional area = $\pi r_2^2 - \pi r_1^2 = \pi(r_2^2 - r_1^2)$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
Additional area = $\pi(23 - 12)(23 + 12) = \pi(11)(35)$.
Taking $\pi = \frac{22}{7}$:
Additional area = $\frac{22}{7} \times 11 \times 35 = 22 \times 11 \times 5 = 1210 \ m^2$.

(B) When four circles of radius $r = 7\, cm$ are placed such that each touches two others,their centers form a square with side length $s = 2r = 14\, cm$.
The area of this square is $s^2 = 14 \times 14 = 196\, cm^2$.
The space enclosed by the four pieces is the area of the square minus the area of the four quadrants of the circles lying inside the square.
Each quadrant has an angle of $90^\circ$. The total area of the four quadrants is $4 \times (\frac{90}{360} \times \pi r^2) = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154\, cm^2$.
Therefore,the required enclosed area is $196 - 154 = 42\, cm^2$.

(C) The cost of cultivation is given as ₹ $6000$ at a rate of $25$ $paise$ per square metre.
Since $100$ $paise = ₹ 1,$ the rate is $0.25$ ₹ per $m^2$.
Area of the field $= \frac{\text{Total Cost}}{\text{Rate}} = \frac{6000}{0.25} = 24000 \, m^2$.
Let the length be $5x$ and the breadth be $3x$.
The area of a rectangle is $\text{length} \times \text{breadth}$.
So,$5x \times 3x = 24000$.
$15x^2 = 24000$.
$x^2 = \frac{24000}{15} = 1600$.
$x = \sqrt{1600} = 40$.
Therefore,the length $= 5 \times 40 = 200 \, m$ and the breadth $= 3 \times 40 = 120 \, m$.

(C) The total cost of carpeting is ₹ $105$ and the rate is ₹ $3.50$ per $m^2$.
Area of the carpet = $\frac{\text{Total Cost}}{\text{Rate}} = \frac{105}{3.50} = 30 \ m^2$.
Since the area of the carpet is equal to the area of the room,the area of the room is $30 \ m^2$.
Given that the width of the room is $5 \ m$.
Area of the room = $\text{Length} \times \text{Width}$.
$30 = \text{Length} \times 5$.
Length = $\frac{30}{5} = 6 \ m$.

(A) Area of the rectangular field $= \frac{28000}{3500} = 8$ hectares.
Since $1$ hectare $= 10000 \text{ m}^2$,the area $= 8 \times 10000 = 80000 \text{ m}^2$.
Let the breadth be $x$ meters and the length be $2x$ meters.
Area $= \text{Length} \times \text{Breadth} = 2x \times x = 2x^2$.
$2x^2 = 80000 \implies x^2 = 40000 \implies x = 200 \text{ m}$.
Breadth $= 200 \text{ m}$,Length $= 400 \text{ m}$.
Perimeter $= 2(\text{Length} + \text{Breadth}) = 2(400 + 200) = 2(600) = 1200 \text{ m}$.
Cost of fencing $= 1200 \times 5 = ₹ 6000$.