Classwise Concept with Examples
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Class 9th Chapters
1. Number Systems	2. Polynomials	3. Coordinate Geometry
4. Linear Equations In Two Variables	5. Introduction To Euclid’s Geometry	6. Lines And Angles
7. Triangles	8. Quadrilaterals	9. Areas Of Parallelograms And Triangles
10. Circles	11. Constructions	12. Heron’s Formula
13. Surface Areas And Volumes	14. Statistics	15. Probability

Content On This Page
Heron’s Formula Derivation	Area of Equilateral and Isosceles Triangle	Application of Heron’s Formula for Finding Areas of Quadrilaterals

Chapter 12 Heron’s Formula (Concepts)

Welcome to this insightful chapter dedicated to a remarkably elegant and practical tool in geometry: Heron's Formula. While we are well-acquainted with the standard method for calculating the area of a triangle using the formula $Area = \frac{1}{2} \times base \times height$, this approach hinges on knowing or being able to easily determine the triangle's altitude (height) corresponding to a chosen base. However, in many practical and theoretical scenarios, determining the height directly can be cumbersome or impossible, especially if we only know the lengths of the triangle's three sides. This is particularly true for scalene triangles where no simple symmetry aids in finding the height. This chapter introduces a powerful alternative, attributed to the brilliant Greek mathematician and engineer Heron of Alexandria (also known as Hero), which allows us to calculate the area of any triangle using only the lengths of its three sides.

Heron's Formula provides a direct pathway to the area, bypassing the need for angles or altitudes altogether. The process involves two simple steps. Let the lengths of the three sides of the triangle be denoted by $a$, $b$, and $c$.

First, we calculate a crucial intermediate value called the semi-perimeter of the triangle, universally denoted by the letter 's'. The semi-perimeter is simply half the total perimeter of the triangle: $$ s = \frac{a + b + c}{2} $$
Once the semi-perimeter 's' is known, the area (A) of the triangle can be calculated directly using Heron's elegant formula: $$ \mathbf{A = \sqrt{s(s - a)(s - b)(s - c)}} $$

This formula, $A = \sqrt{s(s - a)(s - b)(s - c)}$, is the heart of the chapter. We will focus on mastering its straightforward application: calculating 's' accurately and then carefully substituting the values of 's', 'a', 'b', and 'c' into the formula to compute the area. While the mathematical derivation of Heron's formula itself is quite intricate (often involving sophisticated algebraic manipulation starting from the standard area formula combined with the Law of Cosines or multiple applications of the Pythagorean theorem), understanding or memorizing the derivation is typically not the primary objective at this stage. The emphasis lies firmly on the application of this potent formula.

The true utility of Heron's Formula shines when dealing with triangles where the height is not readily available. It provides a robust method applicable to all types of triangles – scalene, isosceles, or equilateral – provided we know the lengths of all three sides. Furthermore, the application of Heron's formula extends beyond simple triangles. We can strategically employ it to determine the area of more complex polygons, most notably quadrilaterals. The technique involves:

Dividing the quadrilateral into two distinct triangles by drawing one of its diagonals.
Ensuring that the lengths of all four sides of the quadrilateral and the length of the chosen diagonal are known.
Calculating the area of each of the two resulting triangles separately using Heron's Formula.
Adding the areas of the two triangles together to obtain the total area of the quadrilateral.

This demonstrates the formula's versatility. Heron's Formula stands as a testament to the power of algebraic geometry, providing a practical and widely used tool, particularly relevant in fields like surveying, land measurement, architecture, and engineering, where calculating areas from measured side lengths is a common requirement.

Heron’s Formula

We know that the area of a triangle can be calculated using the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$. This formula is very convenient when the height of the triangle, corresponding to the chosen base, is known or can be easily determined.

However, for many triangles, especially those that are not right-angled or where the altitude is difficult to measure directly, finding the height can be challenging. In such cases, if the lengths of all three sides of the triangle are known, we can use a powerful formula called Heron's Formula to calculate the area without needing the height.

Heron's Formula Statement:

Heron's formula provides a direct method to find the area of a triangle given the lengths of its three sides.

If the lengths of the sides of a triangle are denoted by $a$, $b$, and $c$, then the area of the triangle is given by the formula:

Area $= \sqrt{s(s-a)(s-b)(s-c)}$

Here, $s$ represents the semi-perimeter of the triangle. The semi-perimeter is half of the perimeter (the sum of the lengths of the three sides):

$s = \frac{a+b+c}{2}$

The formula is named after the ancient Greek mathematician Heron of Alexandria.

Derivation of Heron's Formula

The derivation of Heron's formula is quite elegant but involves a bit of algebraic manipulation, including the Pythagorean theorem. Here's a geometric derivation outline:

Let ABC be a triangle with side lengths BC = $a$, AC = $b$, and AB = $c$. Draw the altitude AD from vertex A to the line containing side BC. Let D be the foot of the perpendicular. Let the length of the altitude AD be $h$. Let the segment BD = $x$. Then the segment DC = $a - x$ (assuming D lies between B and C).

In the right-angled triangle $\triangle \text{ADB}$, by the Pythagorean theorem:

$\text{AB}^2 = \text{AD}^2 + \text{BD}^2$

$c^2 = h^2 + x^2$

... (1)

From this, we can express $h^2$ as $h^2 = c^2 - x^2$.

In the right-angled triangle $\triangle \text{ADC}$, by the Pythagorean theorem:

$\text{AC}^2 = \text{AD}^2 + \text{DC}^2$

$b^2 = h^2 + (a-x)^2$

... (2)

From this, we can express $h^2$ as $h^2 = b^2 - (a-x)^2$.

Equate the two expressions for $h^2$:

$c^2 - x^2 = b^2 - (a-x)^2$

Expand $(a-x)^2$ and simplify:

$c^2 - x^2 = b^2 - (a^2 - 2ax + x^2)$

$c^2 - x^2 = b^2 - a^2 + 2ax - x^2$

Add $x^2$ to both sides:

$c^2 = b^2 - a^2 + 2ax$

Rearrange the terms to solve for $x$:

$2ax = a^2 + c^2 - b^2$

$x = \frac{a^2 + c^2 - b^2}{2a}$

... (3)

Now substitute this value of $x$ back into the equation for $h^2$: $h^2 = c^2 - x^2 = c^2 - \left(\frac{a^2 + c^2 - b^2}{2a}\right)^2$.

$h^2 = c^2 - \frac{(a^2 + c^2 - b^2)^2}{4a^2}$

$h^2 = \frac{4a^2 c^2 - (a^2 + c^2 - b^2)^2}{4a^2}$

The numerator is in the form $A^2 - B^2 = (A-B)(A+B)$, where $A = 2ac$ and $B = a^2 + c^2 - b^2$.

$h^2 = \frac{(2ac - (a^2 + c^2 - b^2))(2ac + (a^2 + c^2 - b^2))}{4a^2}$

$h^2 = \frac{(2ac - a^2 - c^2 + b^2)(2ac + a^2 + c^2 - b^2)}{4a^2}$

Rearrange terms and recognize perfect squares $(a+c)^2 = a^2 + 2ac + c^2$ and $(a-c)^2 = a^2 - 2ac + c^2$:

$h^2 = \frac{(b^2 - (a^2 - 2ac + c^2))((a^2 + 2ac + c^2) - b^2)}{4a^2}$

$h^2 = \frac{(b^2 - (a-c)^2)((a+c)^2 - b^2)}{4a^2}$

Apply the difference of squares formula again: $b^2 - (a-c)^2 = (b - (a-c))(b + (a-c)) = (b-a+c)(b+a-c)$ And $(a+c)^2 - b^2 = ((a+c)-b)((a+c)+b) = (a+c-b)(a+c+b)$.

$h^2 = \frac{(b-a+c)(b+a-c)(a+c-b)(a+c+b)}{4a^2}$

Now, let's use the semi-perimeter $s = \frac{a+b+c}{2}$. This gives $2s = a+b+c$. We can express the terms in the numerator using $s$:

$a+b+c = 2s$

$b+c-a = (a+b+c) - 2a = 2s - 2a = 2(s-a)$

$a+c-b = (a+b+c) - 2b = 2s - 2b = 2(s-b)$

$a+b-c = (a+b+c) - 2c = 2s - 2c = 2(s-c)$

Substitute these expressions into the equation for $h^2$:

$h^2 = \frac{[2(s-a)][2(s-c)][2(s-b)][2s]}{4a^2}$

$h^2 = \frac{16 s(s-a)(s-b)(s-c)}{4a^2}$

$h^2 = \frac{4 s(s-a)(s-b)(s-c)}{a^2}$

Take the positive square root to find the height $h$ (since $h$ must be a non-negative length):

$h = \sqrt{\frac{4 s(s-a)(s-b)(s-c)}{a^2}}$

$h = \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)}$

... (4)

Finally, use the basic area formula for the triangle, Area $= \frac{1}{2} \times \text{base} \times \text{height}$. Using base BC $= a$ and height $h$ from equation (4):

Area $= \frac{1}{2} \times a \times h$

Area $= \frac{1}{2} \times a \times \left(\frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)}\right)$

The terms $\frac{1}{2} a$ and $\frac{2}{a}$ cancel out, leaving:

Area $= \sqrt{s(s-a)(s-b)(s-c)}$

This is Heron's formula. This derivation assumes that the foot of the altitude D lies between B and C. The formula, however, is general and holds true even if D lies outside the segment BC (which happens for obtuse or right-angled triangles), as the algebraic steps remain valid due to the properties of squaring and the relationships between side lengths.

Steps for Calculating Area using Heron's Formula

To find the area of a triangle with side lengths $a, b, c$ using Heron's Formula, follow these steps:

Calculate the semi-perimeter, $s$, by adding the lengths of all three sides and dividing by 2: $s = \frac{a+b+c}{2}$.
Calculate the differences between the semi-perimeter and each side length: $(s-a)$, $(s-b)$, and $(s-c)$.
Substitute the values of $s$, $(s-a)$, $(s-b)$, and $(s-c)$ into Heron's formula: Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Calculate the product inside the square root and find the square root to get the area.

The unit of area will be the square of the unit used for the side lengths (e.g., if sides are in cm, area is in $\text{cm}^2$; if sides are in metres, area is in $\text{m}^2$).

Example 1. Find the area of a triangle with sides 5 cm, 12 cm, and 13 cm.

Answer:

Given:

The side lengths of a triangle are $a = 5$ cm, $b = 12$ cm, and $c = 13$ cm.

To Find:

The area of the triangle.

Solution:

First, calculate the semi-perimeter $s$ of the triangle.

$s = \frac{a+b+c}{2}$

Substitute the given side lengths:

$s = \frac{5 + 12 + 13}{2} \text{ cm}$

$s = \frac{30}{2} \text{ cm}$

$s = 15 \text{ cm}$

... (1)

Next, calculate the values of $(s-a)$, $(s-b)$, and $(s-c)$.

$s - a = 15 - 5 = 10 \text{ cm}$

... (2)

$s - b = 15 - 12 = 3 \text{ cm}$

... (3)

$s - c = 15 - 13 = 2 \text{ cm}$

... (4)

Now, apply Heron's formula to find the area of the triangle:

Area $= \sqrt{s(s-a)(s-b)(s-c)}$

Substitute the values from equations (1), (2), (3), and (4):

Area $= \sqrt{15 \times 10 \times 3 \times 2} \text{ cm}^2$

Calculate the product inside the square root:

Area $= \sqrt{30 \times 6} \text{ cm}^2 = \sqrt{180} \text{ cm}^2$

Alternatively, prime factorise the numbers to simplify the square root:

Area $= \sqrt{(3 \times 5) \times (2 \times 5) \times 3 \times 2} \text{ cm}^2$

Area $= \sqrt{2 \times 2 \times 3 \times 3 \times 5 \times 5} \text{ cm}^2$

Area $= \sqrt{2^2 \times 3^2 \times 5^2} \text{ cm}^2$

Take the square root:

Area $= (2 \times 3 \times 5) \text{ cm}^2$

Area $= 30 \text{ cm}^2$

The area of the triangle is 30 cm$^2$.

As a check, note that $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. This indicates that the triangle with sides 5 cm, 12 cm, and 13 cm is a right-angled triangle, with legs 5 cm and 12 cm and hypotenuse 13 cm. The area of a right-angled triangle is also given by $\frac{1}{2} \times \text{product of legs} = \frac{1}{2} \times 5 \text{ cm} \times 12 \text{ cm} = \frac{1}{2} \times 60 \text{ cm}^2 = 30 \text{ cm}^2$, which matches the result from Heron's formula.

Area of Equilateral and Isosceles Triangle

Heron's formula, Area $= \sqrt{s(s-a)(s-b)(s-c)}$, is a general formula applicable to any triangle, irrespective of its shape. This includes special types of triangles such as equilateral and isosceles triangles. While we can always use Heron's formula for these triangles, their specific properties allow us to derive simpler formulas or use basic area calculation methods more directly.

Area of an Equilateral Triangle:

An equilateral triangle is a triangle where all three sides are of equal length. Let the side length of the equilateral triangle be $a$. Thus, the sides are $a, a, a$.

The semi-perimeter ($s$) of an equilateral triangle with side $a$ is:

$s = \frac{\text{Sum of sides}}{2} = \frac{a+a+a}{2} = \frac{3a}{2}$

Now, let's calculate the terms $(s-a)$, $(s-b)$, and $(s-c)$ for Heron's formula. Since all sides are equal ($b=a, c=a$):

$s - a = \frac{3a}{2} - a = \frac{3a - 2a}{2} = \frac{a}{2}$

$s - b = s - a = \frac{a}{2}$

$s - c = s - a = \frac{a}{2}$

Substitute these values into Heron's Formula:

Area $= \sqrt{s(s-a)(s-b)(s-c)}$

Area $= \sqrt{\frac{3a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}$

Multiply the terms inside the square root:

Area $= \sqrt{\frac{3 \times a \times a \times a \times a}{2 \times 2 \times 2 \times 2}}$

Area $= \sqrt{\frac{3a^4}{16}}$

Take the square root:

Area $= \frac{\sqrt{3a^4}}{\sqrt{16}} = \frac{\sqrt{3} \sqrt{a^4}}{4} = \frac{\sqrt{3} a^2}{4}$

Thus, the area of an equilateral triangle with side length $a$ is given by the formula:

Area $= \frac{\sqrt{3}}{4} a^2$

This formula is more direct for equilateral triangles than using Heron's formula from scratch every time.

Area of an Isosceles Triangle:

An isosceles triangle is a triangle with two sides of equal length. Let the length of the two equal sides be $a$, and the length of the base be $b$. The sides are $a, a, b$.

The semi-perimeter ($s$) of an isosceles triangle with equal sides $a$ and base $b$ is:

$s = \frac{\text{Sum of sides}}{2} = \frac{a+a+b}{2} = \frac{2a+b}{2}$

Now, let's calculate the terms for Heron's formula:

$s - a = \frac{2a+b}{2} - a = \frac{2a+b - 2a}{2} = \frac{b}{2}$

$s - b = \frac{2a+b}{2} - b = \frac{2a+b - 2b}{2} = \frac{2a-b}{2}$

$s - c = s - a = \frac{b}{2}$

Using Heron's Formula:

Area $= \sqrt{s(s-a)(s-b)(s-c)}$

Area $= \sqrt{\frac{2a+b}{2} \times \frac{b}{2} \times \frac{2a-b}{2} \times \frac{b}{2}}$

Multiply the terms inside the square root:

Area $= \sqrt{\frac{(2a+b) \times b \times (2a-b) \times b}{2 \times 2 \times 2 \times 2}}$

Area $= \sqrt{\frac{b^2 (2a+b)(2a-b)}{16}}$

Recognize the product $(2a+b)(2a-b)$ as a difference of squares $(2a)^2 - b^2 = 4a^2 - b^2$:

Area $= \sqrt{\frac{b^2 (4a^2 - b^2)}{16}}$

Take the square root:

Area $= \frac{\sqrt{b^2} \sqrt{4a^2 - b^2}}{\sqrt{16}} = \frac{b \sqrt{4a^2 - b^2}}{4}$

Thus, the area of an isosceles triangle with equal sides $a$ and base $b$ is given by the formula:

Area $= \frac{b}{4} \sqrt{4a^2 - b^2}$

Alternate Derivation for Isosceles Triangle Area:

We can also derive the isosceles triangle area formula using the basic Area $= \frac{1}{2} \times \text{base} \times \text{height}$ formula and the Pythagorean theorem. In an isosceles triangle, the altitude from the vertex angle to the base bisects the base and is perpendicular to it.

Let the equal sides be AC = AB = $a$, and the base BC = $b$. Let AD be the altitude from A to BC, where D is on BC. Then D is the midpoint of BC, so BD = DC $= \frac{b}{2}$. $\triangle \text{ADB}$ is a right-angled triangle with hypotenuse AB = $a$, one leg BD $= \frac{b}{2}$, and the other leg AD = $h$ (the height).

$\text{AB}^2 = \text{AD}^2 + \text{BD}^2$

(Pythagorean Theorem)

$a^2 = h^2 + \left(\frac{b}{2}\right)^2$

$a^2 = h^2 + \frac{b^2}{4}$

Solve for $h^2$:

$h^2 = a^2 - \frac{b^2}{4} = \frac{4a^2 - b^2}{4}$

Take the square root to find $h$:

$h = \sqrt{\frac{4a^2 - b^2}{4}} = \frac{\sqrt{4a^2 - b^2}}{\sqrt{4}} = \frac{\sqrt{4a^2 - b^2}}{2}$

Now use the basic area formula with base $b$ and height $h$:

Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h$

Area $= \frac{1}{2} \times b \times \frac{\sqrt{4a^2 - b^2}}{2}$

Area $= \frac{b \sqrt{4a^2 - b^2}}{4}$

This matches the formula derived from Heron's method, demonstrating consistency.

Example 1. Find the area of an equilateral triangle with side length 6 cm.

Answer:

Given:

Side length of an equilateral triangle $a = 6$ cm.

To Find:

The area of the equilateral triangle.

Solution:

Using the derived formula for the area of an equilateral triangle with side $a$:

Area $= \frac{\sqrt{3}}{4} a^2$

Substitute the given side length $a = 6$ cm:

Area $= \frac{\sqrt{3}}{4} (6 \text{ cm})^2$

Area $= \frac{\sqrt{3}}{4} \times 36 \text{ cm}^2$

Simplify the expression by dividing 36 by 4:

Area $= \sqrt{3} \times \frac{36}{4} \text{ cm}^2$

Area $= 9\sqrt{3} \text{ cm}^2$

The area of the equilateral triangle is $9\sqrt{3}$ cm$^2$.

Example 2. Find the area of an isosceles triangle with equal sides 10 cm each and base 12 cm.

Answer:

Given:

Equal sides of an isosceles triangle $a = 10$ cm.

Base of the isosceles triangle $b = 12$ cm.

To Find:

The area of the isosceles triangle.

Solution (Using the derived formula):

Using the derived formula for the area of an isosceles triangle with equal sides $a$ and base $b$:

Area $= \frac{b}{4} \sqrt{4a^2 - b^2}$

Substitute the given values $a = 10$ cm and $b = 12$ cm:

Area $= \frac{12}{4} \sqrt{4(10)^2 - (12)^2} \text{ cm}^2$

Simplify the expression:

Area $= 3 \sqrt{4(100) - 144} \text{ cm}^2$

Area $= 3 \sqrt{400 - 144} \text{ cm}^2$

Area $= 3 \sqrt{256} \text{ cm}^2$

The square root of 256 is 16 (since $16 \times 16 = 256$).

Area $= 3 \times 16 \text{ cm}^2$

Area $= 48 \text{ cm}^2$

The area of the isosceles triangle is 48 cm$^2$.

Alternative method using Heron's Formula directly:

The side lengths of the isosceles triangle are 10 cm, 10 cm, and 12 cm. Let $a=10, b=10, c=12$.

First, calculate the semi-perimeter $s$:

$s = \frac{10+10+12}{2} = \frac{32}{2} = 16 \text{ cm}$

Next, calculate the terms $(s-a)$, $(s-b)$, and $(s-c)$:

$s-a = 16-10 = 6 \text{ cm}$

$s-b = 16-10 = 6 \text{ cm}$

$s-c = 16-12 = 4 \text{ cm}$

Now, apply Heron's Formula:

Area $= \sqrt{s(s-a)(s-b)(s-c)}$

Area $= \sqrt{16 \times 6 \times 6 \times 4} \text{ cm}^2$

Simplify the expression:

Area $= \sqrt{16} \times \sqrt{6 \times 6} \times \sqrt{4} \text{ cm}^2$

Area $= 4 \times 6 \times 2 \text{ cm}^2$

Area $= 48 \text{ cm}^2$

Both methods give the same result, confirming the area is 48 cm$^2$.

Application of Heron’s Formula for Finding Areas of Quadrilaterals

Heron's formula is specifically designed for calculating the area of a triangle. However, we can extend its application to find the area of any polygon, such as a quadrilateral, by dividing the polygon into triangles. The key requirement is that we must know the lengths of all sides of the triangles we form.

Method to Find the Area of a Quadrilateral using Heron's Formula:

To find the area of a quadrilateral using Heron's formula, we can follow these steps:

Divide the Quadrilateral: Divide the given quadrilateral into two triangles by drawing one of its diagonals.

Determine Side Lengths: Find the lengths of all three sides for each of the two triangles formed. The lengths of the four sides of the quadrilateral are usually given. The length of the chosen diagonal must also be known or calculated using other geometric properties (like the Pythagorean theorem if there's a right angle, or the Law of Cosines if angles are known).
Calculate Triangle Areas: Use Heron's formula to calculate the area of each of the two triangles separately. For each triangle, identify its three side lengths, calculate its semi-perimeter, and then apply Heron's formula.
Sum the Areas: The total area of the quadrilateral is the sum of the areas of the two triangles it was divided into.

Area(Quadrilateral) = Area(Triangle 1) + Area(Triangle 2)

Important Note:

This method requires knowing the lengths of all four sides of the quadrilateral AND the length of at least one of its diagonals. If only the four side lengths are provided, the area of the quadrilateral cannot be uniquely determined. This is because a quadrilateral is not a rigid figure like a triangle; it can be reshaped (deformed) while keeping the side lengths constant, resulting in different areas (unless it's a special case like a square or rectangle where angles are fixed).

Example 1. A park is in the shape of a quadrilateral ABCD, where $\angle \text{C} = 90^\circ$, AB = 9 m, BC = 12 m, CD = 5 m, and AD = 8 m. Find the area of the park.

Answer:

Given:

A quadrilateral park ABCD with side lengths AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m, and $\angle \text{C} = 90^\circ$.

To Find:

The area of the park (quadrilateral ABCD).

Solution:

To find the area of the quadrilateral ABCD using Heron's formula, we divide it into two triangles by drawing a diagonal. Since we are given that $\angle \text{C} = 90^\circ$, drawing the diagonal BD is convenient because it forms a right-angled triangle $\triangle \text{BCD}$, whose diagonal length can be easily calculated using the Pythagorean theorem.

Step 1: Find the length of the diagonal BD.

Consider $\triangle \text{BCD}$. We are given that $\angle \text{C} = 90^\circ$. So, $\triangle \text{BCD}$ is a right-angled triangle with legs BC and CD, and hypotenuse BD.

By the Pythagorean Theorem (Theorem 6.9), in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$\text{BD}^2 = \text{BC}^2 + \text{CD}^2$

(By Pythagoras Theorem in $\triangle \text{BCD}$)

Substitute the given lengths BC = 12 m and CD = 5 m:

$\text{BD}^2 = (12)^2 + (5)^2 \text{ m}^2$

$\text{BD}^2 = 144 + 25 \text{ m}^2$

$\text{BD}^2 = 169 \text{ m}^2$

Take the square root of both sides to find the length of BD. Since BD is a length, it must be positive.

$\text{BD} = \sqrt{169} \text{ m}$

$\text{BD} = 13 \text{ m}$

... (1)

The length of the diagonal BD is 13 metres.

Step 2: Calculate the area of $\triangle$BCD.

$\triangle \text{BCD}$ is a right-angled triangle with base BC = 12 m and corresponding height CD = 5 m (or vice versa). We can use the basic area formula for a right triangle.

Area($\triangle \text{BCD}$) $= \frac{1}{2} \times \text{base} \times \text{height}$

Area($\triangle \text{BCD}$) $= \frac{1}{2} \times 12 \text{ m} \times 5 \text{ m}$

Area($\triangle \text{BCD}$) $= \frac{1}{2} \times 60 \text{ m}^2$

Area($\triangle \text{BCD}$) $= 30 \text{ m}^2$

... (2)

The area of $\triangle \text{BCD}$ is 30 m$^2$.

Step 3: Calculate the area of $\triangle$ABD using Heron's Formula.

The side lengths of $\triangle \text{ABD}$ are AB = 9 m, AD = 8 m, and BD = 13 m (from Step 1).

First, calculate the semi-perimeter $s$ for $\triangle \text{ABD}$.

$s = \frac{\text{AB} + \text{AD} + \text{BD}}{2}$

Substitute the side lengths:

$s = \frac{9 + 8 + 13}{2} \text{ m}$

$s = \frac{30}{2} \text{ m}$

$s = 15 \text{ m}$

... (3)

Now, calculate the values of $(s-\text{AB})$, $(s-\text{AD})$, and $(s-\text{BD})$.

$s - \text{AB} = 15 - 9 = 6 \text{ m}$

... (4)

$s - \text{AD} = 15 - 8 = 7 \text{ m}$

... (5)

$s - \text{BD} = 15 - 13 = 2 \text{ m}$

... (6)

Use Heron's Formula to find the area of $\triangle \text{ABD}$:

Area($\triangle \text{ABD}$) $= \sqrt{s(s-\text{AB})(s-\text{AD})(s-\text{BD})}$

Substitute the values from equations (3), (4), (5), and (6):

Area($\triangle \text{ABD}$) $= \sqrt{15 \times 6 \times 7 \times 2} \text{ m}^2$

Calculate the product inside the square root. It's helpful to prime factorise the numbers to simplify the square root:

Area($\triangle \text{ABD}$) $= \sqrt{(3 \times 5) \times (2 \times 3) \times 7 \times 2} \text{ m}^2$

Area($\triangle \text{ABD}$) $= \sqrt{2^2 \times 3^2 \times 5 \times 7} \text{ m}^2$

Take the square root. Factors that are squared can be brought outside the square root:

Area($\triangle \text{ABD}$) $= \sqrt{2^2} \times \sqrt{3^2} \times \sqrt{5 \times 7} \text{ m}^2$

Area($\triangle \text{ABD}$) $= (2 \times 3) \sqrt{35} \text{ m}^2$

Area($\triangle \text{ABD}$) $= 6\sqrt{35} \text{ m}^2$

... (7)

The area of $\triangle \text{ABD}$ is $6\sqrt{35}$ m$^2$. (Note: $\sqrt{35}$ is approximately 5.916). So, $6\sqrt{35} \approx 6 \times 5.916 = 35.496$ m$^2$.

Step 4: Calculate the total area of the quadrilateral ABCD.

The area of the quadrilateral ABCD is the sum of the areas of the two triangles $\triangle \text{BCD}$ and $\triangle \text{ABD}$.

Area(ABCD) = Area($\triangle \text{BCD}$) + Area($\triangle \text{ABD}$)

Substitute the areas from equation (2) and equation (7):

Area(ABCD) $= 30 \text{ m}^2 + 6\sqrt{35} \text{ m}^2$

Area(ABCD) $= (30 + 6\sqrt{35}) \text{ m}^2$

If an approximate numerical value is required, we can substitute the approximation for $\sqrt{35}$:

Area(ABCD) $\approx (30 + 6 \times 5.916) \text{ m}^2$

Area(ABCD) $\approx (30 + 35.496) \text{ m}^2$

Area(ABCD) $\approx 65.496 \text{ m}^2$

Rounding to one decimal place, the area is approximately 65.5 m$^2$.

The area of the park is $(30 + 6\sqrt{35})$ m$^2$.