Mix Examples - Heron’s Formula Questions in English

(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: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} \times 8 \times 6 = 24 \, cm^2$.

(B) Let the two equal sides of the isosceles right triangle be $a \, cm$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Since it is an isosceles right triangle,base $= a$ and height $= a$.
Therefore,$\frac{1}{2} \times a \times a = 8$.
$a^{2} = 16$,which gives $a = 4 \, cm$.
By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{a^{2} + a^{2}} = \sqrt{4^{2} + 4^{2}} = \sqrt{16 + 16} = \sqrt{32} \, cm$.

(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.

(B) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Given,$\text{base} = 4 \, \text{cm}$ and $\text{height} = 6 \, \text{cm}$.
Substituting these values into the formula: $\text{Area} = \frac{1}{2} \times 4 \, \text{cm} \times 6 \, \text{cm} = 12 \, \text{cm}^2$.
Since the calculated area is $12 \, \text{cm}^2$ and not $24 \, \text{cm}^2$,the statement is False.

(TRUE) Given that $\Delta ABC$ is a right-angled triangle with $\angle A = 90^{\circ}$,$AB = 4 \, cm$,and $AC = 4 \, cm$.
The area of a right-angled triangle is given by the formula:
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Here,we can take $AC$ as the base and $AB$ as the height.
Area $= \frac{1}{2} \times 4 \, cm \times 4 \, cm$
Area $= \frac{1}{2} \times 16 \, cm^2 = 8 \, cm^2$
Since the calculated area is $8 \, cm^2$,the given statement is True.

(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.

(A) Let $ABCD$ be the rhombus whose one diagonal $AC$ is $16 \, cm$. Each side of the rhombus is $10 \, cm$.
We know that the diagonals of a rhombus bisect each other at right angles,so
$OA = OC = 8 \, cm$ and $OB = OD$.
In $\Delta AOB$,we have $\angle AOB = 90^{\circ}$.
Therefore,$AB^2 = OA^2 + OB^2$
$\Rightarrow OB^2 = AB^2 - OA^2$
$= (10)^2 - 8^2 = 100 - 64 = 36$
$OB = \sqrt{36} = 6 \, cm$.
$DB = 2(OB) = 2 \times 6 = 12 \, cm$.
Hence,the area of the rhombus $= \frac{1}{2} \times \text{Product of diagonals}$
$= \frac{1}{2} \times 16 \times 12 = 96 \, cm^2$.
Hence,the given statement is true.

(B) The base of the parallelogram is $10 \text{ cm}$ and the corresponding altitude is $3.5 \text{ cm}$.
The area of a parallelogram is calculated using the formula: $\text{Area} = \text{base} \times \text{corresponding altitude}$.
Substituting the given values: $\text{Area} = 10 \text{ cm} \times 3.5 \text{ cm} = 35 \text{ cm}^2$.
Since the calculated area is $35 \text{ cm}^2$ and not $30 \text{ cm}^2$,the given statement is False.

(FALSE) The statement is False.
A regular hexagon can be divided into six equilateral triangles by joining its center to each of its vertices.
Since each of these six triangles is an equilateral triangle with side $a$, the total area of the regular hexagon is equal to the sum of the areas of these six equilateral triangles.
Therefore, the area of a regular hexagon of side $a$ is $6 \times (\text{Area of an equilateral triangle of side } a)$, not the sum of the areas of five such triangles.

(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 $= \frac{\text{Total Area}}{\text{Area per bed}} = \frac{180}{0.09} = 2000$.

(N/A) The figure consists of two triangles. The shaded area is the difference between the area of the larger triangle and the smaller triangle.
$1$. For the larger triangle with sides $a = 122 \ m$,$b = 120 \ m$,and $c = 22 \ m$:
Semi-perimeter $s = \frac{122 + 120 + 22}{2} = \frac{264}{2} = 132 \ m$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{132(132-122)(132-120)(132-22)}$
$= \sqrt{132 \times 10 \times 12 \times 110} = \sqrt{1742400} = 1320 \ m^2$.
$2$. For the smaller triangle with sides $a = 24 \ m$,$b = 26 \ m$,and $c = 22 \ m$:
Since it is a right-angled triangle (as indicated in the figure),we can use the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times 24 \times 10 = 120 \ m^2$.
(Note: Using Heron's formula also yields the same result: $s = \frac{24+26+22}{2} = 36$. Area $= \sqrt{36(36-24)(36-26)(36-22)} = \sqrt{36 \times 12 \times 10 \times 14} = \sqrt{60480} \approx 245.9 \ m^2$. However,given the right angle symbol,the base-height method is more accurate).
$3$. Area of the shaded region $= \text{Area of larger triangle} - \text{Area of smaller triangle}$
$= 1320 \ m^2 - 120 \ m^2 = 1200 \ m^2$.

(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) The sides of $\Delta BCD$ are $a = 12 \text{ cm}, b = 17 \text{ cm},$ and $c = 25 \text{ cm}.$
$\therefore$ The semi-perimeter $(s)$ of $\Delta BCD$ is:
$s = \frac{a + b + c}{2} = \frac{12 + 17 + 25}{2} = \frac{54}{2} = 27 \text{ cm}.$
$\therefore$ Area of $\Delta BCD = \sqrt{s(s - a)(s - b)(s - c)}$ [By Heron's formula]
$= \sqrt{27(27 - 12)(27 - 17)(27 - 25)}$
$= \sqrt{27 \times 15 \times 10 \times 2}$
$= \sqrt{(9 \times 3) \times (3 \times 5) \times (5 \times 2) \times 2}$
$= \sqrt{9 \times 9 \times 25 \times 4} = 3 \times 3 \times 5 \times 2 = 90 \text{ cm}^2.$
Since a diagonal divides a parallelogram into two triangles of equal area:
Area of parallelogram $ABCD = 2 \times \text{Area of } \Delta BCD = 2 \times 90 = 180 \text{ cm}^2.$
Let the altitude from vertex $A$ to side $DC$ be $h.$
Area of parallelogram $ABCD = \text{Base} \times \text{Altitude}$
$180 = DC \times h$
$180 = 12 \times h$
$h = \frac{180}{12} = 15 \text{ cm}.$
Hence,the area of the parallelogram is $180 \text{ cm}^2$ and the length of the altitude is $15 \text{ cm}.$

(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) To find the area of quadrilateral $ABCD$,we divide it into two triangles: $\Delta ABC$ and $\Delta ACD$.
Since $\angle B = 90^\circ$,$\Delta ABC$ is a right-angled triangle.
By Pythagoras theorem in $\Delta ABC$:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24 \, cm^2$.
Now,for $\Delta ACD$,the sides are $a = 10 \, cm$,$b = 12 \, cm$,and $c = 14 \, cm$.
The semi-perimeter $s = \frac{a + b + c}{2} = \frac{10 + 12 + 14}{2} = \frac{36}{2} = 18 \, cm$.
Using Heron's formula,Area of $\Delta ACD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-10)(18-12)(18-14)} = \sqrt{18 \times 8 \times 6 \times 4} = \sqrt{3456} \approx 58.787 \, cm^2$.
Total area of quadrilateral $ABCD = \text{Area}(\Delta ABC) + \text{Area}(\Delta ACD) = 24 + 58.787 = 82.787 \, cm^2$.

(C) The perimeter of the rhombus is $40 \, cm$.
Since all sides of a rhombus are equal,the side length is $\frac{40}{4} = 10 \, cm$.
The diagonal divides the rhombus into two congruent triangles with sides $10 \, cm, 10 \, cm,$ and $12 \, cm$.
Using Heron's formula for one triangle:
Semi-perimeter $s = \frac{10 + 10 + 12}{2} = \frac{32}{2} = 16 \, cm$.
Area of one triangle $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16(16-10)(16-10)(16-12)} = \sqrt{16 \times 6 \times 6 \times 4} = \sqrt{2304} = 48 \, cm^2$.
Area of the rhombus $= 2 \times 48 = 96 \, cm^2$.
Since both sides are painted,the total area to be painted is $2 \times 96 = 192 \, cm^2$.
The cost of painting at the rate of $Rs \, 5$ per $cm^2$ is $192 \times 5 = Rs \, 960$.

(A) Draw $RT \perp SP$. From the figure,it is clear that $RT = PQ = 7 \ m$ and $PT = QR = 7 \ m$.
Since $SP = 12 \ m$,we have $ST = SP - PT = 12 \ m - 7 \ m = 5 \ m$.
Now,in the right-angled triangle $STR$,by Pythagoras theorem:
$SR^2 = ST^2 + RT^2$
$13^2 = 5^2 + RT^2$
$169 = 25 + RT^2$
$RT^2 = 169 - 25 = 144$
$RT = 12 \ m$.
Thus,the height of the trapezium is $12 \ m$.
The area of trapezium $PQRS = \frac{1}{2} \times (SP + QR) \times PQ$
$= \frac{1}{2} \times (12 \ m + 7 \ m) \times 12 \ m$
$= \frac{1}{2} \times 19 \ m \times 12 \ m = 19 \times 6 \ m^2 = 114 \ m^2$.

(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 $= \sqrt{25(25 - 13)(25 - 17)(25 - 20)}$
$= \sqrt{25 \times 12 \times 8 \times 5}$
$= \sqrt{25 \times (4 \times 3) \times (4 \times 2) \times 5}$
$= 5 \times 4 \times \sqrt{3 \times 2 \times 5}$
$= 20 \sqrt{30} \, cm^2$.

(A) 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 length of one parallel side be $x \, cm$.
Then,the length of the other parallel side is $(x + 4) \, cm$.
Given,$\text{Area} = 475 \, cm^2$ and $\text{height} = 19 \, cm$.
Substituting these values into the formula:
$475 = \frac{1}{2} \times (x + x + 4) \times 19$
$475 = \frac{1}{2} \times (2x + 4) \times 19$
$475 = (x + 2) \times 19$
$x + 2 = \frac{475}{19} = 25$
$x = 25 - 2 = 23 \, cm$.
Therefore,the two parallel sides are $23 \, cm$ and $(23 + 4) \, cm$,which are $23 \, cm$ and $27 \, cm$.

(B) The length of the rectangular plot is $40 \, m$ and the width (breadth) is $15 \, m$.
According to the rules, a space of $3 \, m$ must be left at the front and back, and a space of $2 \, m$ must be left on each of the other two sides.
The new length of the construction area $= 40 - (3 + 3) = 40 - 6 = 34 \, m$.
The new width of the construction area $= 15 - (2 + 2) = 15 - 4 = 11 \, m$.
The largest area where the house can be constructed $= \text{Length} \times \text{Width} = 34 \, m \times 11 \, m = 374 \, m^2$.

(A) Let the trapezium be $ABCD$ with parallel sides $AB = 90 \, m$ and $CD = 30 \, m$. The side $AC$ is perpendicular to both $AB$ and $CD$.
Draw $DM \perp AB$. Since $ACDM$ forms a rectangle,$AM = CD = 30 \, m$ and $DM = AC$.
Now,$MB = AB - AM = 90 \, m - 30 \, m = 60 \, m$.
In the right-angled triangle $\triangle DMB$,by Pythagoras theorem:
$DM^2 = DB^2 - MB^2 = (100)^2 - (60)^2$
$DM^2 = 10000 - 3600 = 6400$
$DM = \sqrt{6400} = 80 \, m$.
Thus,the height of the trapezium is $80 \, m$.
The area of the trapezium $ABCD = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$= \frac{1}{2} \times (90 + 30) \times 80 = \frac{1}{2} \times 120 \times 80 = 4800 \, m^2$.
Total cost of ploughing = $\text{Area} \times \text{Rate} = 4800 \times 4 = \operatorname{Rs} 19,200$.

(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) $ABCD$ is a rectangle in which $AB = 51 \, cm$ and $BC = 25 \, cm$.
Since the parallel sides $QC$ and $PD$ are in the ratio $9:8$,let $QC = 9x$ and $PD = 8x$.
The height of the trapezium $PQCD$ is the same as the width of the rectangle,which is $25 \, cm$.
The area of the trapezium $PQCD = \frac{1}{2} \times (QC + PD) \times \text{height} = \frac{1}{2} \times (9x + 8x) \times 25 = \frac{1}{2} \times 17x \times 25$.
The area of the rectangle $ABCD = \text{length} \times \text{width} = 51 \times 25$.
It is given that the area of the trapezium $PQCD = \frac{5}{6} \times \text{Area of rectangle } ABCD$.
Therefore,$\frac{1}{2} \times 17x \times 25 = \frac{5}{6} \times 51 \times 25$.
Dividing both sides by $25$ and multiplying by $2$,we get $17x = \frac{5}{6} \times 51 \times 2 = \frac{5}{3} \times 51 = 5 \times 17 = 85$.
Thus,$x = \frac{85}{17} = 5$.
Hence,the length $QC = 9x = 9 \times 5 = 45 \, cm$.
And the length $PD = 8x = 8 \times 5 = 40 \, cm$.

(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 $1868\, cm^2$.

(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 $A$.

(A) To find the area of a triangle with sides $a = 9 \, cm$,$b = 10 \, cm$,and $c = 17 \, cm$,we use Heron's formula.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{9 + 10 + 17}{2} = \frac{36}{2} = 18 \, cm$.
Now,use the area formula $\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$:
$\text{Area} = \sqrt{18(18 - 9)(18 - 10)(18 - 17)}$
$\text{Area} = \sqrt{18 \times 9 \times 8 \times 1}$
$\text{Area} = \sqrt{1296} = 36 \, cm^2$.

(C) Given sides of the triangle are $a = 360 \, m$,$b = 450 \, m$,and $c = 450 \, m$.
First,calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{360 + 450 + 450}{2} = \frac{1260}{2} = 630 \, m$.
Using Heron's formula,Area = $\sqrt{s(s - a)(s - b)(s - c)}$.
Area = $\sqrt{630(630 - 360)(630 - 450)(630 - 450)}$.
Area = $\sqrt{630 \times 270 \times 180 \times 180}$.
Area = $180 \times \sqrt{630 \times 270}$.
Area = $180 \times \sqrt{(90 \times 7) \times (90 \times 3)}$.
Area = $180 \times 90 \times \sqrt{7 \times 3}$.
Area = $16200 \sqrt{21} \, m^2$.

(D) The sides of the triangular plot are $a = 200\,m$,$b = 210\,m$,and $c = 290\,m$.
First,calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{200 + 210 + 290}{2} = \frac{700}{2} = 350\,m$.
Using Heron's formula,the area $(A)$ is:
$A = \sqrt{s(s-a)(s-b)(s-c)}$
$A = \sqrt{350(350-200)(350-210)(350-290)}$
$A = \sqrt{350 \times 150 \times 140 \times 60}$
$A = \sqrt{35 \times 10 \times 15 \times 10 \times 14 \times 10 \times 6 \times 10}$
$A = \sqrt{35 \times 15 \times 14 \times 6 \times 10000}$
$A = \sqrt{(5 \times 7) \times (3 \times 5) \times (2 \times 7) \times (2 \times 3) \times 10000}$
$A = \sqrt{2^2 \times 3^2 \times 5^2 \times 7^2 \times 100^2} = 2 \times 3 \times 5 \times 7 \times 100 = 21000\,m^2$.
The cost of raising the lawn is $Area \times Rate = 21000 \times 5 = \text{Rs. } 105000$.

(A) Given sides are $a = 15\,cm$ and $b = 28\,cm$. The perimeter $P = 84\,cm$.
The third side $c = P - (a + b) = 84 - (15 + 28) = 84 - 43 = 41\,cm$.
Semi-perimeter $s = P / 2 = 84 / 2 = 42\,cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{42(42-15)(42-28)(42-41)}$.
Area $= \sqrt{42 \times 27 \times 14 \times 1}$.
Area $= \sqrt{(14 \times 3) \times (9 \times 3) \times 14 \times 1} = \sqrt{14^2 \times 3^2 \times 3^2} = 14 \times 3 \times 3 = 126\,cm^2$.

(B) Let the equal sides be $a = 17\,cm$ and $b = 17\,cm$.
The perimeter of the triangle is $P = 50\,cm$.
Let the third side be $c$. Then,$a + b + c = 50$.
$17 + 17 + c = 50 \implies 34 + c = 50 \implies c = 16\,cm$.
The semi-perimeter $s = P / 2 = 50 / 2 = 25\,cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{25(25-17)(25-17)(25-16)}$.
Area $= \sqrt{25 \times 8 \times 8 \times 9}$.
Area $= 5 \times 8 \times 3 = 120\,cm^2$.