Area of a Triangle by Heron's Formula Questions in English

(A) Given,the perimeter of the triangle is $32 \, cm$. Let the sides be $a = 8 \, cm$ and $b = 11 \, cm$.
The third side $c$ is calculated as:
$c = 32 \, cm - (8 + 11) \, cm = 32 \, cm - 19 \, cm = 13 \, cm$.
The semi-perimeter $s$ is:
$s = \frac{\text{Perimeter}}{2} = \frac{32}{2} \, cm = 16 \, cm$.
Now,calculate $(s - a)$,$(s - b)$,and $(s - c)$:
$s - a = 16 - 8 = 8 \, cm$
$s - b = 16 - 11 = 5 \, cm$
$s - c = 16 - 13 = 3 \, cm$
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$\text{Area} = \sqrt{16 \times 8 \times 5 \times 3} \, cm^2$
$\text{Area} = \sqrt{16 \times 120} \, cm^2$
$\text{Area} = 4 \times \sqrt{4 \times 30} \, cm^2$
$\text{Area} = 4 \times 2 \sqrt{30} \, cm^2 = 8 \sqrt{30} \, cm^2$.

(B) To find the area of the park,we use Heron's formula.
First,calculate the semi-perimeter $s$:
$2s = 50\,m + 80\,m + 120\,m = 250\,m$
$s = 125\,m$
Now,calculate the differences:
$s - a = 125\,m - 120\,m = 5\,m$
$s - b = 125\,m - 80\,m = 45\,m$
$s - c = 125\,m - 50\,m = 75\,m$
Area of the park $= \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{125 \times 5 \times 45 \times 75}\,m^2$
$= \sqrt{2109375}\,m^2 = 375\sqrt{15}\,m^2$
Next,calculate the cost of fencing:
Perimeter of the park $= 250\,m$
Length of wire needed $= 250\,m - 3\,m = 247\,m$
Cost of fencing $= 247\,m \times Rs. 20/m = Rs. 4940$

(C) Let the sides of the triangle be $3x, 5x$ and $7x$ meters.
The perimeter of the triangle is given as $300\, m$.
Therefore,$3x + 5x + 7x = 300$.
$15x = 300$,which gives $x = 20$.
Thus,the sides of the triangle are $3 \times 20 = 60\, m$,$5 \times 20 = 100\, m$,and $7 \times 20 = 140\, m$.
Using Heron's formula,the semi-perimeter $s$ is:
$s = \frac{60 + 100 + 140}{2} = \frac{300}{2} = 150\, m$.
The area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{150(150 - 60)(150 - 100)(150 - 140)}\, m^2$.
Area $= \sqrt{150 \times 90 \times 50 \times 10}\, m^2$.
Area $= \sqrt{6750000}\, m^2$.
Area $= 1500\sqrt{3}\, m^2$.

(D) For an equilateral triangle with side $a$,the area is given by $\frac{\sqrt{3}}{4} a^2$.
Given that the perimeter of the equilateral triangle is $180 \, cm$.
Since all three sides are equal,$3a = 180 \, cm$,which gives $a = 60 \, cm$.
Using Heron's formula,the semi-perimeter $s = \frac{\text{Perimeter}}{2} = \frac{180}{2} = 90 \, cm$.
The area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$.
Substituting the values: $\text{Area} = \sqrt{90(90-60)(90-60)(90-60)} \, cm^2$.
$\text{Area} = \sqrt{90 \times 30 \times 30 \times 30} \, cm^2$.
$\text{Area} = \sqrt{3 \times 30 \times 30 \times 30 \times 30} \, cm^2$.
$\text{Area} = \sqrt{3 \times (30)^2 \times (30)^2} \, cm^2$.
$\text{Area} = 30 \times 30 \times \sqrt{3} \, cm^2 = 900 \sqrt{3} \, cm^2$.

(A) The sides of the triangular wall are $a = 122\, m, b = 120\, m, c = 22\, m$.
The semi-perimeter $s$ is given by:
$s = \frac{a + b + c}{2} = \frac{122 + 120 + 22}{2}\, m = \frac{264}{2}\, m = 132\, m$.
The area of the triangle using Heron's formula is $\sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{132(132 - 122)(132 - 120)(132 - 22)}\, m^2$
$= \sqrt{132 \times 10 \times 12 \times 110}\, m^2$
$= \sqrt{12 \times 11 \times 10 \times 12 \times 11 \times 10}\, m^2$
$= \sqrt{12^2 \times 11^2 \times 10^2}\, m^2 = 12 \times 11 \times 10\, m^2 = 1320\, m^2$.
Rent for $1$ year ($12$ months) per $m^2 = Rs. 5000$.
Rent for $3$ months per $m^2 = Rs. 5000 \times \frac{3}{12} = Rs. 1250$.
Total rent for $1320\, m^2$ for $3$ months $= 1320 \times 1250 = Rs. 16,50,000$.

(B) The sides of the triangular wall are $a = 15 \, m$,$b = 11 \, m$,and $c = 6 \, m$.
First,we calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{15 + 11 + 6}{2} \, m = \frac{32}{2} \, m = 16 \, m$.
Now,using Heron's formula,the area of the triangular surface is:
Area $= \sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{16(16 - 15)(16 - 11)(16 - 6)} \, m^2$
$= \sqrt{16 \times 1 \times 5 \times 10} \, m^2$
$= \sqrt{16 \times 50} \, m^2$
$= \sqrt{800} \, m^2$
$= \sqrt{400 \times 2} \, m^2 = 20 \sqrt{2} \, m^2$.
Thus,the required area painted in colour is $20 \sqrt{2} \, m^2$.

(C) Let the sides of the triangle be $a = 18 \, cm$,$b = 10 \, cm$,and $c$ be the third side.
Given that the perimeter $(2s) = 42 \, cm$.
Therefore,the semi-perimeter $s = \frac{42}{2} = 21 \, cm$.
The third side $c = 42 - (18 + 10) = 42 - 28 = 14 \, cm$.
Using Heron's formula,the area of a triangle $= \sqrt{s(s-a)(s-b)(s-c)}$.
Substituting the values: Area $= \sqrt{21(21-18)(21-10)(21-14)} \, cm^2$.
Area $= \sqrt{21 \times 3 \times 11 \times 7} \, cm^2$.
Area $= \sqrt{(3 \times 7) \times 3 \times 11 \times 7} \, cm^2 = \sqrt{3^2 \times 7^2 \times 11} \, cm^2$.
Area $= 3 \times 7 \sqrt{11} \, cm^2 = 21 \sqrt{11} \, cm^2$.

(D) Perimeter of the triangle $= 540 \, cm$.
Semi-perimeter $s = \frac{540}{2} = 270 \, cm$.
Let the sides be $12x, 17x,$ and $25x$.
$12x + 17x + 25x = 540 \Rightarrow 54x = 540 \Rightarrow x = 10$.
Thus,the sides are $a = 120 \, cm, b = 170 \, cm, c = 250 \, cm$.
Using Heron's formula: Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
$s-a = 270 - 120 = 150 \, cm$.
$s-b = 270 - 170 = 100 \, cm$.
$s-c = 270 - 250 = 20 \, cm$.
Area $= \sqrt{270 \times 150 \times 100 \times 20} = \sqrt{81,000,000} = 9,000 \, cm^2$.

(A) The equal sides of the triangle are $12 \, cm$ each.
Let the third side be $x \, cm$.
Since the perimeter is $30 \, cm$,we have:
$12 \, cm + 12 \, cm + x \, cm = 30 \, cm$
$24 \, cm + x = 30 \, cm$
$x = 30 - 24 = 6 \, cm$.
Now,the semi-perimeter $s = \frac{\text{Perimeter}}{2} = \frac{30}{2} = 15 \, cm$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{15(15-12)(15-12)(15-6)} \, cm^2$
$\text{Area} = \sqrt{15 \times 3 \times 3 \times 9} \, cm^2$
$\text{Area} = \sqrt{15 \times 9 \times 9} \, cm^2 = 9 \sqrt{15} \, cm^2$.
Thus,the area of the triangle is $9 \sqrt{15} \, cm^2$.

(A) For the triangular field with sides $a = 200 \, m, b = 240 \, m, c = 360 \, m$:
Semi-perimeter $s = \frac{200 + 240 + 360}{2} = 400 \, m$.
Area of the wheat field = $\sqrt{400(400-200)(400-240)(400-360)} = \sqrt{400 \times 200 \times 160 \times 40} = \sqrt{512000000} = 16000\sqrt{2} \, m^2 \approx 22627.4 \, m^2 \approx 2.26 \, \text{hectares}$.
For the second triangular field with sides $240 \, m, 320 \, m, 400 \, m$:
Semi-perimeter $s = \frac{240 + 320 + 400}{2} = 480 \, m$.
Area of the second field = $\sqrt{480(480-240)(480-320)(480-400)} = \sqrt{480 \times 240 \times 160 \times 80} = \sqrt{1474560000} = 38400 \, m^2 = 3.84 \, \text{hectares}$.
The median divides the triangle into two triangles of equal area.
Area for potatoes = Area for onions = $\frac{3.84}{2} = 1.92 \, \text{hectares}$ each.

(C) Since $AB = 9 \, m$ and $BC = 40 \, m$ with $\angle B = 90^{\circ}$,we use the Pythagorean theorem to find $AC$:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41 \, m$.
The first group cleans the area of the right-angled triangle $ABC$:
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \times 9 = 180 \, m^2$.
The second group cleans the area of triangle $ACD$ with sides $41 \, m, 15 \, m$ and $28 \, m$. Using Heron's formula,the semi-perimeter $s = \frac{41 + 15 + 28}{2} = \frac{84}{2} = 42 \, m$.
Area of $\Delta ACD = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{42(42 - 41)(42 - 15)(42 - 28)} = \sqrt{42 \times 1 \times 27 \times 14} = \sqrt{15876} = 126 \, m^2$.
The first group cleaned $180 \, m^2$ and the second group cleaned $126 \, m^2$. The first group cleaned more area by $(180 - 126) = 54 \, m^2$.
Total area cleaned by the students $= 180 + 126 = 306 \, m^2$.

(A) The quadrilateral $ABCD$ is divided into two triangles,$\Delta ABC$ and $\Delta ACD$,by the diagonal $AC$.
For $\Delta ABC$:
The sides are $a = 3 \, cm, b = 4 \, cm, c = 5 \, cm$.
Semi-perimeter $s_1 = \frac{3 + 4 + 5}{2} = 6 \, cm$.
Area of $\Delta ABC = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6 \, cm^2$.
For $\Delta ACD$:
The sides are $a = 5 \, cm, b = 4 \, cm, c = 5 \, cm$.
Semi-perimeter $s_2 = \frac{5 + 4 + 5}{2} = 7 \, cm$.
Area of $\Delta ACD = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} = \sqrt{7 \times 2 \times 3 \times 2} = \sqrt{84} = 2\sqrt{21} \, cm^2$.
Using $\sqrt{21} \approx 4.58$,Area $\approx 2 \times 4.58 = 9.16 \, cm^2$.
Total area of quadrilateral $ABCD = \text{Area}(\Delta ABC) + \text{Area}(\Delta ACD) = 6 + 9.16 = 15.16 \, cm^2 \approx 15.2 \, cm^2$.

(B) Area of surface $I$:
It is an isosceles triangle with sides $a = 5\, cm, b = 5\, cm, c = 1\, cm$.
Semi-perimeter $s = \frac{5+5+1}{2} = 5.5\, cm$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5.5(5.5-5)(5.5-5)(5.5-1)} = \sqrt{5.5 \times 0.5 \times 0.5 \times 4.5} = \sqrt{6.1875} \approx 2.48\, cm^{2}$.
Area of surface $II$:
It is a rectangle with length $6.5\, cm$ and breadth $1\, cm$.
Area $= 6.5 \times 1 = 6.5\, cm^{2}$.
Area of surface $III$:
It is a trapezium with parallel sides $1\, cm$ and $2\, cm$. The height $h$ is calculated using Pythagoras theorem: $h = \sqrt{1^{2} - (0.5)^{2}} = \sqrt{0.75} \approx 0.866\, cm$.
Area $= \frac{1}{2} \times (1 + 2) \times 0.866 = 1.5 \times 0.866 \approx 1.3\, cm^{2}$.
Area of surface $IV$ and $V$:
These are two congruent right-angled triangles with base $6\, cm$ and height $1.5\, cm$.
Area of $IV = \frac{1}{2} \times 6 \times 1.5 = 4.5\, cm^{2}$.
Area of $V = 4.5\, cm^{2}$.
Total area $= 2.48 + 6.5 + 1.3 + 4.5 + 4.5 = 19.28\, cm^{2} \approx 19.3\, cm^{2}$.

(D) Here,each side of the rhombus $= 30\, m$.
One of the diagonals $= 48\, m$.
Since a diagonal divides the rhombus into two congruent triangles,the sides of the first triangle are $a = 30\, m, b = 30\, m, c = 48\, m$.
The semi-perimeter $s = \frac{a + b + c}{2} = \frac{30 + 30 + 48}{2}\, m = \frac{108}{2} = 54\, m$.
Area of triangle $I = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{54(54 - 30)(54 - 30)(54 - 48)}\, m^{2}$.
$= \sqrt{54 \times 24 \times 24 \times 6}\, m^{2} = \sqrt{(9 \times 6) \times 24 \times 24 \times 6}\, m^{2} = \sqrt{9 \times 36 \times 24^{2}}\, m^{2} = 3 \times 6 \times 24\, m^{2} = 432\, m^{2}$.
Area of the second triangle is also $432\, m^{2}$.
Total area of the rhombus $= 432\, m^{2} + 432\, m^{2} = 864\, m^{2}$.
Thus,the area of grass for $18$ cows $= 864\, m^{2}$.
Area of grass for $1$ cow $= \frac{864}{18}\, m^{2} = 48\, m^{2}$.

(N/A) The sides of each triangular piece are $a = 20\, cm$,$b = 50\, cm$,and $c = 50\, cm$.
The semi-perimeter $s$ is calculated as:
$s = \frac{a + b + c}{2} = \frac{20 + 50 + 50}{2}\, cm = \frac{120}{2}\, cm = 60\, cm$.
Using Heron's formula,the area of each triangular piece is:
Area $= \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{60(60 - 20)(60 - 50)(60 - 50)}\, cm^2$
$= \sqrt{60 \times 40 \times 10 \times 10}\, cm^2 = \sqrt{240000}\, cm^2 = 200\sqrt{6}\, cm^2$.
Since there are $10$ triangular pieces in total,there are $5$ pieces of each colour.
Area of cloth required for each colour $= 5 \times 200\sqrt{6}\, cm^2 = 1000\sqrt{6}\, cm^2$.

(N/A) Area of triangle $I$:
Diagonal $= 32\, cm$.
Since the diagonals of a square bisect each other at $90^\circ$,the height of triangle $I$ (which is half the diagonal) $= \frac{1}{2} \times 32\, cm = 16\, cm$.
The base of the triangle is the other diagonal of the square,which is also $32\, cm$.
Area of triangle $I = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 32\, cm \times 16\, cm = 256\, cm^2$.
Area of triangle $II$:
Since the diagonal of a square divides it into two congruent triangles,the area of triangle $II$ is equal to the area of triangle $I$.
Area of triangle $II = 256\, cm^2$.
Area of triangle $III$:
The triangle at the base has sides $a = 8\, cm, b = 6\, cm, c = 6\, cm$.
Semi-perimeter $s = \frac{a + b + c}{2} = \frac{8 + 6 + 6}{2} = 10\, cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{10(10-8)(10-6)(10-6)} = \sqrt{10 \times 2 \times 4 \times 4} = \sqrt{320} = 8\sqrt{5}\, cm^2$.
Taking $\sqrt{5} \approx 2.24$,Area $\approx 8 \times 2.24 = 17.92\, cm^2$.
Thus,the area of paper used for each shade is:
Shade $I = 256\, cm^2$,Shade $II = 256\, cm^2$,Shade $III = 17.92\, cm^2$.

(C) There are $16$ equal triangular tiles.
Sides of the triangle are $a = 9\, cm, b = 28\, cm, c = 35\, cm$.
Semi-perimeter $s = \frac{a + b + c}{2} = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36\, cm$.
Area of one triangle using Heron's formula $= \sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{36(36 - 9)(36 - 28)(36 - 35)} = \sqrt{36 \times 27 \times 8 \times 1} = \sqrt{7776} = 36\sqrt{6}\, cm^2$.
Using $\sqrt{6} \approx 2.449$,Area $\approx 36 \times 2.449 = 88.164\, cm^2$.
Total area of $16$ tiles $= 16 \times 88.164 = 1410.624\, cm^2$.
Cost of polishing at $50p$ (or $Rs. 0.50$) per $cm^2$ $= 1410.624 \times 0.50 = Rs. 705.312$.
Rounding to the nearest value provided in the options,the cost is $Rs. 705.60$.

(D) The given field is in the form of a trapezium $ABCD$ such that parallel sides are $AB = 10\, m$ and $DC = 25\, m$.
Non-parallel sides are $AD = 13\, m$ and $BC = 14\, m$.
Draw $BE \parallel AD$,such that $E$ lies on $DC$. Then $ABED$ is a parallelogram.
$DE = AB = 10\, m$,so $EC = DC - DE = 25\, m - 10\, m = 15\, m$.
In $\Delta BCE$,the sides are $a = 13\, m$,$b = 14\, m$,and $c = 15\, m$.
Semi-perimeter $s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\, m$.
Area of $\Delta BCE = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84\, m^{2}$.
Let $h$ be the height of $\Delta BCE$ corresponding to base $EC = 15\, m$.
Area of $\Delta BCE = \frac{1}{2} \times \text{base} \times \text{height} = 84\, m^{2}$.
$\frac{1}{2} \times 15 \times h = 84 \Rightarrow h = \frac{168}{15} = 11.2\, m$.
Area of parallelogram $ABED = \text{base} \times \text{height} = AB \times h = 10 \times 11.2 = 112\, m^{2}$.
Total area of the field = Area of $\Delta BCE +$ Area of parallelogram $ABED = 84 + 112 = 196\, m^{2}$.

(A) Let $ABCD$ be the rhombus-shaped field.
Perimeter of the rhombus $= 400\, m$.
Since all sides of a rhombus are equal,each side $= 400\, m / 4 = 100\, m$.
Thus,$AB = BC = CD = DA = 100\, m$.
Let the diagonal $BD = 160\, m$.
The diagonal divides the rhombus into two congruent triangles,$\triangle ABD$ and $\triangle BCD$.
For $\triangle ABD$,the sides are $a = 100\, m$,$b = 100\, m$,and $c = 160\, m$.
The semi-perimeter $s$ is given by:
$s = (100 + 100 + 160) / 2 = 360 / 2 = 180\, m$.
Using Heron's formula,the area of $\triangle ABD$ is:
Area $= \sqrt{s(s-a)(s-b)(s-c)}$
Area $= \sqrt{180(180-100)(180-100)(180-160)}$
Area $= \sqrt{180 \times 80 \times 80 \times 20}$
Area $= \sqrt{180 \times 20 \times 80^2}$
Area $= \sqrt{3600 \times 6400}$
Area $= 60 \times 80 = 4800\, m^2$.
Since the land is divided into two equal parts,each of them will get an area of $4800\, m^2$.

(A) Let the right triangle be $\triangle ABC$ where $\angle B = 90^{\circ}$.
Given: Base $BC = 8 \, cm$ and hypotenuse $AC = 10 \, cm$.
According to the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Substituting the values: $10^2 = AB^2 + 8^2$.
$100 = AB^2 + 64$.
$AB^2 = 100 - 64 = 36$.
$AB = \sqrt{36} = 6 \, cm$.
The area of a right triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} \times 8 \times 6 = 24 \, cm^2$.

(C) The perimeter of an equilateral triangle is given by $P = 3a$,where $a$ is the side length.
Given $3a = 60 \, m$,we find the side length $a = 60 / 3 = 20 \, m$.
The area of an equilateral triangle is calculated using the formula $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Substituting $a = 20 \, m$ into the formula:
$\text{Area} = \frac{\sqrt{3}}{4} \times (20)^2 = \frac{\sqrt{3}}{4} \times 400 = 100 \sqrt{3} \, m^2$.

(D) Given,the three sides of the triangle are $a = 56 \ cm$,$b = 60 \ cm$,and $c = 52 \ cm$.
First,calculate the semi-perimeter $(s)$ of the triangle:
$s = \frac{a + b + c}{2} = \frac{56 + 60 + 52}{2} = \frac{168}{2} = 84 \ cm$.
Using Heron's formula,the area of the triangle is given by $\sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{84(84 - 56)(84 - 60)(84 - 52)}$
Area $= \sqrt{84 \times 28 \times 24 \times 32}$
Area $= \sqrt{(4 \times 3 \times 7) \times (4 \times 7) \times (8 \times 3) \times (8 \times 4)}$
Area $= \sqrt{4^4 \times 7^2 \times 3^2 \times 8^2} = 4^2 \times 7 \times 3 \times 8 = 16 \times 168 = 1344 \ cm^2$.
Thus,the area of the triangle is $1344 \ cm^2$.

(A) The area of an equilateral triangle is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} \times (\text{side})^2$.
Given the side length is $2\sqrt{3} \, cm$.
Substituting the value into the formula:
$\text{Area} = \frac{\sqrt{3}}{4} \times (2\sqrt{3})^2$
$= \frac{\sqrt{3}}{4} \times (4 \times 3)$
$= \frac{\sqrt{3}}{4} \times 12$
$= 3\sqrt{3} \, cm^2$.
Using the value $\sqrt{3} \approx 1.732$:
$\text{Area} = 3 \times 1.732 = 5.196 \, cm^2$.

(B) The area of an equilateral triangle is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} \times (\text{side})^2$.
Given that the area is $9 \sqrt{3} \text{ cm}^2$,we set up the equation:
$9 \sqrt{3} = \frac{\sqrt{3}}{4} \times (\text{side})^2$.
Dividing both sides by $\sqrt{3}$,we get:
$9 = \frac{1}{4} \times (\text{side})^2$.
Multiplying both sides by $4$,we get:
$(\text{side})^2 = 36$.
Taking the square root of both sides,we find:
$\text{side} = \sqrt{36} = 6 \text{ cm}$.

(C) The area of an equilateral triangle is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} \times (\text{side})^2$.
Given that the area is $16 \sqrt{3} \, cm^{2}$,we have:
$16 \sqrt{3} = \frac{\sqrt{3}}{4} \times (\text{side})^2$.
Dividing both sides by $\sqrt{3}$:
$16 = \frac{1}{4} \times (\text{side})^2$.
Multiplying by $4$:
$(\text{side})^2 = 16 \times 4 = 64$.
Taking the square root:
$\text{side} = \sqrt{64} = 8 \, cm$.
The perimeter of an equilateral triangle is $3 \times \text{side}$:
$\text{Perimeter} = 3 \times 8 = 24 \, cm$.

(D) The sides of the triangle are $a = 35 \text{ cm}$,$b = 54 \text{ cm}$,and $c = 61 \text{ cm}$.
First,calculate the semi-perimeter $s$:
$s = \frac{35 + 54 + 61}{2} = \frac{150}{2} = 75 \text{ cm}$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{75(75-35)(75-54)(75-61)}$
$= \sqrt{75 \times 40 \times 21 \times 14} = \sqrt{(25 \times 3) \times (8 \times 5) \times (3 \times 7) \times (2 \times 7)}$
$= \sqrt{25 \times 3^2 \times 7^2 \times 16 \times 5} = 5 \times 3 \times 7 \times 4 \sqrt{5} = 420 \sqrt{5} \text{ cm}^2$.
The longest altitude corresponds to the smallest side of the triangle.
Since the smallest side is $35 \text{ cm}$,the longest altitude $h$ is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 420 \sqrt{5} = \frac{1}{2} \times 35 \times h$.
$h = \frac{420 \sqrt{5} \times 2}{35} = 12 \times 2 \sqrt{5} = 24 \sqrt{5} \text{ cm}$.

(A) The sides of the isosceles triangle are $a = 4 \, cm$,$b = 4 \, cm$,and $c = 2 \, cm$.
The semi-perimeter $s$ is calculated as:
$s = \frac{a + b + c}{2} = \frac{4 + 4 + 2}{2} = \frac{10}{2} = 5 \, cm$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
Substituting the values:
$\text{Area} = \sqrt{5(5 - 4)(5 - 4)(5 - 2)}$
$\text{Area} = \sqrt{5 \times 1 \times 1 \times 3}$
$\text{Area} = \sqrt{15} \, cm^2$.

(B) The sides of the triangle are $a = 6 \, cm$,$b = 8 \, cm$,and $c = 10 \, cm$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{6 + 8 + 10}{2} = \frac{24}{2} = 12 \, cm$.
Using Heron's formula,the area of the triangle is:
$Area = \sqrt{s(s - a)(s - b)(s - c)}$
$Area = \sqrt{12(12 - 6)(12 - 8)(12 - 10)}$
$Area = \sqrt{12 \times 6 \times 4 \times 2}$
$Area = \sqrt{576} = 24 \, cm^2$.
The cost of painting is $9$ paise per $cm^2$,which is $Rs \, 0.09$ per $cm^2$.
Total cost $= 24 \times 0.09 = Rs \, 2.16$.

(B) The statement is False.
According to Heron's formula, the area of a triangle with sides $a, b, c$ is given by $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semi-perimeter of the triangle.
The semi-perimeter $s$ is defined as $s = \frac{a+b+c}{2}$, which is equal to half of the perimeter of the triangle.
In the given statement, $s$ is incorrectly defined as the perimeter of the triangle instead of the semi-perimeter.

(TRUE) Let the equal sides of the isosceles triangle be $a$ and the base be $b = 5 \, cm$.
The perimeter of the triangle is given as $11 \, cm$.
So,$a + a + 5 = 11$.
$2a = 11 - 5 = 6$.
$a = 3 \, cm$.
The area of an isosceles triangle is given by the formula $\text{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}$.
Substituting the values: $\text{Area} = \frac{5}{4} \sqrt{4(3)^2 - 5^2} = \frac{5}{4} \sqrt{4(9) - 25} = \frac{5}{4} \sqrt{36 - 25} = \frac{5}{4} \sqrt{11} \, cm^2$.
Since the calculated area matches the given value,the statement is True.

(B) The formula for the area of an equilateral triangle is $\text{Area} = \frac{\sqrt{3}}{4} \times (\text{side})^2$.
Given that the side of the triangle is $8 \text{ cm}$,we substitute this value into the formula:
$\text{Area} = \frac{\sqrt{3}}{4} \times (8)^2$
$\text{Area} = \frac{\sqrt{3}}{4} \times 64$
$\text{Area} = 16 \sqrt{3} \text{ cm}^2$.
Since the calculated area $16 \sqrt{3} \text{ cm}^2$ is not equal to the given area $20 \sqrt{3} \text{ cm}^2$,the statement is False.

(TRUE) The sides of the triangle are $a = 51 \, m$,$b = 37 \, m$,and $c = 20 \, m$.
The semi-perimeter $s$ is calculated as:
$s = \frac{a + b + c}{2} = \frac{51 + 37 + 20}{2} = \frac{108}{2} = 54 \, m$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{54(54-51)(54-37)(54-20)}$
$= \sqrt{54 \times 3 \times 17 \times 34}$
$= \sqrt{(2 \times 3^3) \times 3 \times 17 \times (2 \times 17)}$
$= \sqrt{2^2 \times 3^4 \times 17^2} = 2 \times 3^2 \times 17 = 2 \times 9 \times 17 = 306 \, m^2$.
The cost of levelling the ground is $\text{Area} \times \text{Rate} = 306 \, m^2 \times Rs \, 3/m^2 = Rs \, 918$.
Since the calculated cost matches the given value,the statement is True.

(A) Let the sides of the triangle be $a = 11 \, cm$,$b = 12 \, cm$,and $c = 13 \, cm$.
The semi-perimeter $s$ is calculated as:
$s = \frac{a + b + c}{2} = \frac{11 + 12 + 13}{2} = \frac{36}{2} = 18 \, cm$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-11)(18-12)(18-13)}$
$= \sqrt{18 \times 7 \times 6 \times 5} = \sqrt{3780} = \sqrt{36 \times 105} = 6\sqrt{105} \, cm^2$.
We know that $\text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude}$.
Given the base is $12 \, cm$,let the altitude be $h$:
$6\sqrt{105} = \frac{1}{2} \times 12 \times h$
$6\sqrt{105} = 6h$
$h = \sqrt{105} \, cm$.
Since $\sqrt{100} = 10$ and $\sqrt{121} = 11$,$\sqrt{105} \approx 10.2469 \, cm$,which is approximately $10.25 \, cm$.
Thus,the given statement is True.

(A) Let the sides of the triangular field be $a = 41\, m$,$b = 40\, m$,and $c = 9\, m$.
The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{41 + 40 + 9}{2} = \frac{90}{2} = 45\, m$.
Using Heron's formula,the area of the triangular field is $\sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{45(45 - 41)(45 - 40)(45 - 9)} = \sqrt{45 \times 4 \times 5 \times 36}$.
Area $= \sqrt{9 \times 5 \times 4 \times 5 \times 36} = \sqrt{32400} = 180\, m^2$.
Each rose bed needs $900\, cm^2$ space. Converting this to square meters: $900\, cm^2 = \frac{900}{10000}\, m^2 = 0.09\, m^2$.
The number of rose beds $= \frac{\text{Total Area}}{\text{Area per bed}} = \frac{180}{0.09} = 2000$.

(C) The sides of the triangular field are $a = 50 \, m$,$b = 65 \, m$,and $c = 65 \, m$.
First,we calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{50 + 65 + 65}{2} = \frac{180}{2} = 90 \, m$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{90(90-50)(90-65)(90-65)}$
$= \sqrt{90 \times 40 \times 25 \times 25}$
$= \sqrt{3600 \times 625} = 60 \times 25 = 1500 \, m^2$.
The cost of laying grass is given by the product of the area and the rate:
$\text{Cost} = 1500 \, m^2 \times 7 \, Rs/m^2 = 10,500 \, Rs$.

(D) The sides of the triangular side walls of the flyover are $a = 13 \, m$,$b = 14 \, m$,and $c = 15 \, m$.
First,we calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \, m$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$
$= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{(3 \times 7) \times (2^3) \times 7 \times (2 \times 3)}$
$= \sqrt{7^2 \times 3^2 \times 2^4} = 7 \times 3 \times 2^2 = 84 \, m^2$.
The annual rent is $Rs \, 2000$ per $m^2$.
The rent for $1 \, m^2$ for $6$ months (half a year) is $Rs \, \frac{2000}{2} = Rs \, 1000$.
Therefore,the total rent for $84 \, m^2$ for $6$ months is $84 \times 1000 = Rs \, 84,000$.

(D) Let $ABC$ be an equilateral triangle with side length $a$. Let $O$ be an interior point,and let the perpendiculars from $O$ to the sides $AB$,$BC$,and $AC$ be $h_1 = 14 \, cm$,$h_2 = 10 \, cm$,and $h_3 = 6 \, cm$ respectively.
The area of $\Delta ABC$ is the sum of the areas of $\Delta OAB$,$\Delta OBC$,and $\Delta OAC$.
Area of $\Delta OAB = \frac{1}{2} \times a \times 14 = 7a \, cm^2$.
Area of $\Delta OBC = \frac{1}{2} \times a \times 10 = 5a \, cm^2$.
Area of $\Delta OAC = \frac{1}{2} \times a \times 6 = 3a \, cm^2$.
Total Area $= 7a + 5a + 3a = 15a \, cm^2$.
We know the area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} a^2$.
Equating the two expressions for the area:
$\frac{\sqrt{3}}{4} a^2 = 15a$
Since $a \neq 0$,we divide by $a$:
$\frac{\sqrt{3}}{4} a = 15$
$a = \frac{15 \times 4}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \, cm$.
Now,substitute $a = 20\sqrt{3}$ into the area formula:
Area $= 15 \times (20\sqrt{3}) = 300\sqrt{3} \, cm^2$.
Thus,the area of the equilateral triangle is $300\sqrt{3} \, cm^2$.

(D) Let the equal sides of the isosceles triangle be $3x$ and $3x$,and the base be $2x$.
The perimeter of the triangle is the sum of its sides: $3x + 3x + 2x = 8x$.
Given that the perimeter is $32 \, cm$,we have $8x = 32$,which gives $x = 4$.
Thus,the sides of the triangle are $12 \, cm, 12 \, cm$,and $8 \, cm$.
The semi-perimeter $s$ is given by $s = \frac{12 + 12 + 8}{2} = \frac{32}{2} = 16 \, cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{16(16-12)(16-12)(16-8)} = \sqrt{16 \times 4 \times 4 \times 8}$.
Area $= \sqrt{16 \times 16 \times 8} = 16 \sqrt{8} = 16 \times 2\sqrt{2} = 32\sqrt{2} \, cm^2$.

(N/A) Let the field be $ABCD.$
Area of the parallelogram $ABCD = 2 \times (\text{Area of } \Delta ABC) \quad ...(1)$
Now,the sides of $\Delta ABC$ are $a = 40 \, m, b = 60 \, m$ and $c = 80 \, m$.
Semi-perimeter of $\Delta ABC$,$s = \frac{a+b+c}{2} = \frac{40+60+80}{2} = \frac{180}{2} = 90 \, m$.
Using Heron's Formula,Area of $\Delta ABC = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{90(90-40)(90-60)(90-80)}$
$= \sqrt{90 \times 50 \times 30 \times 10}$
$= \sqrt{1350000} = 300\sqrt{15} \, m^2 \approx 1161.895 \, m^2$.
From equation $(1)$,
Area of parallelogram $ABCD = 2 \times 300\sqrt{15} = 600\sqrt{15} \, m^2 \approx 2323.79 \, m^2$.

(A) Let the sides of the triangle be $6x, 7x,$ and $8x$ meters.
Given that the perimeter is $420\, m$,we have:
$6x + 7x + 8x = 420$
$21x = 420$
$x = 20$
Thus,the sides are $a = 120\, m, b = 140\, m, c = 160\, m$.
The semi-perimeter $s$ is given by $s = \frac{420}{2} = 210\, m$.
Using Heron's formula,the area is $\sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{210(210-120)(210-140)(210-160)}$
Area $= \sqrt{210 \times 90 \times 70 \times 50}$
Area $= \sqrt{66150000} \approx 8133.265\, m^2$.

(B) Let $a, b, c$ be the sides of the given triangle and $s$ be its semi-perimeter.
Then,$s = \frac{a+b+c}{2}$,which implies $2s = a+b+c$ $...(1)$.
The area of the given triangle is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
According to the problem,the sides of the new triangle are $2a, 2b$,and $2c$.
Let $S$ be the semi-perimeter of the new triangle.
$S = \frac{2a+2b+2c}{2} = a+b+c = 2s$ $...(2)$.
The area of the new triangle is $\Delta' = \sqrt{S(S-2a)(S-2b)(S-2c)}$.
Substituting $S = 2s$ into the formula:
$\Delta' = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$
$\Delta' = \sqrt{2s \cdot 2(s-a) \cdot 2(s-b) \cdot 2(s-c)}$
$\Delta' = \sqrt{16s(s-a)(s-b)(s-c)}$
$\Delta' = 4\sqrt{s(s-a)(s-b)(s-c)} = 4\Delta$.
Thus,the ratio of the area of the new triangle to the given triangle is $\frac{\Delta'}{\Delta} = \frac{4\Delta}{\Delta} = 4:1$.

(C) The kite consists of a square $ABCD$ and an isosceles triangle at the bottom.
Given that the diagonal of the square is $44 \, cm$.
The area of a square can be calculated using the diagonal $d$ as $\text{Area} = \frac{1}{2} d^2$.
$\text{Area of square } ABCD = \frac{1}{2} \times 44 \times 44 = 968 \, cm^2$.
The square is divided into four equal triangles by its diagonals. Each triangle has an area of $\frac{968}{4} = 242 \, cm^2$.
From the figure:
- Red shade (part $IV$) $= 242 \, cm^2$.
- Yellow shade (parts $I$ and $II$) $= 242 + 242 = 484 \, cm^2$.
- Green shade (part $III$ and the bottom triangle).
For the bottom triangle with sides $20 \, cm, 20 \, cm, 14 \, cm$:
Semi-perimeter $s = \frac{20+20+14}{2} = 27 \, cm$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-20)(27-20)(27-14)} = \sqrt{27 \times 7 \times 7 \times 13} = 21 \sqrt{39} \approx 21 \times 6.245 = 131.15 \, cm^2$.
Total Green shade $= 242 + 131.15 = 373.15 \, cm^2$.

(D) Let the smaller side of the triangle be $x \, cm$. Therefore,the second side is $(x + 4) \, cm$ and the third side is $(2x - 6) \, cm$.
The perimeter of the triangle is the sum of its sides:
$x + (x + 4) + (2x - 6) = 50$
$4x - 2 = 50$
$4x = 52$
$x = 13 \, cm$.
The three sides are $13 \, cm$,$17 \, cm$,and $20 \, cm$.
The semi-perimeter $s = \frac{13 + 17 + 20}{2} = \frac{50}{2} = 25 \, cm$.
Using Heron's formula,the area of the triangle is $\sqrt{s(s - a)(s - b)(s - c)}$:
Area $= \sqrt{25(25 - 13)(25 - 17)(25 - 20)}$
$= \sqrt{25 \times 12 \times 8 \times 5}$
$= \sqrt{25 \times (4 \times 3) \times (4 \times 2) \times 5}$
$= 5 \times 4 \times \sqrt{3 \times 2 \times 5}$
$= 20 \sqrt{30} \, cm^2$.

(B) The sides of $\Delta ABC$ are $a = 6.5\, cm$,$b = 7\, cm$,and $c = 7.5\, cm$.
The semi-perimeter $s$ of $\Delta ABC$ is:
$s = \frac{6.5 + 7 + 7.5}{2} = \frac{21}{2} = 10.5\, cm$.
Using Heron's formula,the area of $\Delta ABC$ is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{10.5(10.5 - 6.5)(10.5 - 7)(10.5 - 7.5)}$
$= \sqrt{10.5 \times 4 \times 3.5 \times 3}$
$= \sqrt{441} = 21\, cm^2$.
Since the parallelogram $DBCE$ is constructed on the same base $BC$ and has the same area as $\Delta ABC$:
$\text{Area of parallelogram } DBCE = \text{Area of } \Delta ABC = 21\, cm^2$.
The area of a parallelogram is given by $\text{base} \times \text{height}$.
$BC \times DF = 21\, cm^2$
$7\, cm \times DF = 21\, cm^2$
$DF = \frac{21}{7} = 3\, cm$.
Thus,the height $DF$ of the parallelogram is $3\, cm$.

(N/A) Given,the dimensions of the rectangular tile are $50\, cm \times 70\, cm$.
Area of the rectangular tile $= 50\, cm \times 70\, cm = 3500\, cm^2$.
The sides of each triangle are $a = 25\, cm, b = 17\, cm$,and $c = 26\, cm$.
Now,the semi-perimeter $s$ is given by $s = \frac{a + b + c}{2}$.
$s = \frac{25 + 17 + 26}{2} = \frac{68}{2} = 34\, cm$.
Using Heron's formula,the area of one triangle is $\sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{34(34 - 25)(34 - 17)(34 - 26)} = \sqrt{34 \times 9 \times 17 \times 8}$.
Area $= \sqrt{(17 \times 2) \times 3^2 \times 17 \times (2^3)} = \sqrt{17^2 \times 2^4 \times 3^2} = 17 \times 4 \times 3 = 204\, cm^2$.
Total area of $8$ triangles $= 8 \times 204\, cm^2 = 1632\, cm^2$.
Thus,the total area of the design is $1632\, cm^2$.
Remaining area of the tile $=$ Area of the rectangle $-$ Area of the design.
Remaining area $= 3500\, cm^2 - 1632\, cm^2 = 1868\, cm^2$.
Hence,the total area of the design is $1632\, cm^2$ and the remaining area of the tile is $1868\, cm^2$.

(N/A) $(1)$ The area of a rhombus is given by $\frac{1}{2} \times \text{product of its diagonals}$.
$(2)$ The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{corresponding altitude}$.

(B) Let the sides of the given triangle be $a = 12 \, cm$,$b = 17 \, cm$,and $c = 25 \, cm$.
First,calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{12 + 17 + 25}{2} = \frac{54}{2} = 27 \, cm$.
Now,calculate the differences:
$s - a = 27 - 12 = 15 \, cm$
$s - b = 27 - 17 = 10 \, cm$
$s - c = 27 - 25 = 2 \, cm$
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$\text{Area} = \sqrt{27 \times 15 \times 10 \times 2} \, cm^2$
$\text{Area} = \sqrt{(9 \times 3) \times (3 \times 5) \times (5 \times 2) \times 2} \, cm^2$
$\text{Area} = \sqrt{9 \times 9 \times 25 \times 4} \, cm^2$
$\text{Area} = 3 \times 3 \times 5 \times 2 = 90 \, cm^2$.

(C) For a triangular field,the side lengths are $a = 51 \,m$,$b = 52 \,m$,and $c = 53 \,m$.
First,calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{51 + 52 + 53}{2} = \frac{156}{2} = 78 \,m$.
Now,calculate the differences:
$s - a = 78 - 51 = 27 \,m$
$s - b = 78 - 52 = 26 \,m$
$s - c = 78 - 53 = 25 \,m$
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{78 \times 27 \times 26 \times 25}$
$= \sqrt{(13 \times 6) \times (9 \times 3) \times (13 \times 2) \times 25}$
$= \sqrt{13^2 \times 6 \times 27 \times 25}$
$= \sqrt{13^2 \times (2 \times 3) \times (3^3) \times 5^2}$
$= \sqrt{13^2 \times 2 \times 3^4 \times 5^2}$
$= 13 \times 3^2 \times 5 \times \sqrt{2}$
Wait,let's re-calculate the prime factorization:
$78 = 2 \times 3 \times 13$
$27 = 3^3$
$26 = 2 \times 13$
$25 = 5^2$
$\text{Area} = \sqrt{(2 \times 3 \times 13) \times (3^3) \times (2 \times 13) \times 5^2}$
$= \sqrt{2^2 \times 3^4 \times 13^2 \times 5^2}$
$= 2 \times 3^2 \times 13 \times 5$
$= 2 \times 9 \times 13 \times 5 = 1170 \,m^2$.

(A) Let the sides of the triangle be $4x, 7x,$ and $9x$ meters.
The perimeter of the triangle is given as $500 \, m$.
Therefore,$4x + 7x + 9x = 500$.
$20x = 500 \implies x = 25$.
The sides are $a = 4 \times 25 = 100 \, m$,$b = 7 \times 25 = 175 \, m$,and $c = 9 \times 25 = 225 \, m$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{500}{2} = 250 \, m$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
$s-a = 250 - 100 = 150 \, m$.
$s-b = 250 - 175 = 75 \, m$.
$s-c = 250 - 225 = 25 \, m$.
Area $= \sqrt{250 \times 150 \times 75 \times 25} = \sqrt{(25 \times 10) \times (15 \times 10) \times (25 \times 3) \times 25}$.
Area $= \sqrt{25^2 \times 10^2 \times 3 \times 5 \times 3 \times 5} = \sqrt{25^2 \times 10^2 \times 3^2 \times 5^2} = 25 \times 10 \times 3 \times 5 \sqrt{5} = 3750 \sqrt{5} \, m^2$.

(A) The area of an equilateral triangle with side length $a$ is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Given the side length $a = 50 \, cm$.
Substituting the value of $a$ in the formula:
$\text{Area} = \frac{\sqrt{3}}{4} \times (50)^2$
$\text{Area} = \frac{\sqrt{3}}{4} \times 2500$
$\text{Area} = 625 \sqrt{3} \, cm^2$.
Thus,the correct option is $A$.