Area of Triangles using Heron's Formula
Heron’s Formula: Statement and Derivation
Heron's formula, named after Hero of Alexandria (a Greek engineer and mathematician who lived around the 1st century CE), is a powerful and elegant method to calculate the area of a triangle when the lengths of all three sides are known. This formula is particularly useful in situations where the height of the triangle is difficult to measure or calculate directly using the base-height formula ($\frac{1}{2}bh$).
Statement of Heron's Formula
Let a triangle have sides of length $a$, $b$, and $c$.
First, we need to calculate a value called the semi-perimeter of the triangle. The semi-perimeter ('semi' meaning half) is simply half of the perimeter of the triangle.
The perimeter of the triangle is $a+b+c$.
The semi-perimeter ($s$) is given by:
$\mathbf{s = \frac{a+b+c}{2}}$
... (1)
Once the semi-perimeter ($s$) is calculated, the area ($A$) of the triangle is given by Heron's Formula:
$\mathbf{A = \sqrt{s(s-a)(s-b)(s-c)}}$
... (2)
The unit of the area will be the square of the unit used for the side lengths (e.g., if $a, b, c$ are in cm, $A$ will be in $\text{cm}^2$).
Derivation of Heron's Formula (Algebraic Approach)
We can derive Heron's formula using basic geometry (Pythagoras theorem) and algebra, starting from the standard area formula $A = \frac{1}{2} b h$.
Consider a triangle ABC with side lengths $a$, $b$, and $c$ opposite to vertices A, B, and C respectively. Let's choose side AB as the base, so its length is $c$. Draw an altitude (height) $h$ from vertex C to the line containing the base AB, meeting the line at point D.
Let the length of the segment AD be $x$. Assuming D lies between A and B, the length of the segment DB will be $c - x$. (The derivation works even if D falls outside AB; $DB = |c-x|$ or $x-c$, but $DB^2 = (c-x)^2$ in either case, simplifying the algebra).
The area of triangle ABC using the base-height formula is:
$\text{Area} (A) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AB} \times \text{CD}$
$\mathbf{A = \frac{1}{2} c h}$
... (3)
Our goal is to express $h$ in terms of $a$, $b$, and $c$, and then substitute it into equation (3).
Consider the two right-angled triangles formed by the altitude CD:
In right-angled triangle $\triangle \text{ADC}$:
Using the Pythagoras Theorem ($(\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2$):
$\text{AC}^2 = \text{AD}^2 + \text{CD}^2$
$\mathbf{b^2 = x^2 + h^2}$
[Pythagoras Theorem in $\triangle$ADC] ... (4)
From equation (4), we can express $h^2$ as:
$\mathbf{h^2 = b^2 - x^2}$
... (5)
In right-angled triangle $\triangle \text{BDC}$:
Using the Pythagoras Theorem:
$\text{BC}^2 = \text{DB}^2 + \text{CD}^2$
$\mathbf{a^2 = (c-x)^2 + h^2}$
[Pythagoras Theorem in $\triangle$BDC] ... (6)
From equation (6), we can also express $h^2$ as:
$\mathbf{h^2 = a^2 - (c-x)^2}$
... (7)
Now we have two expressions for $h^2$ (equations 5 and 7). Equating them allows us to solve for $x$ in terms of $a$, $b$, and $c$:
$b^2 - x^2 = a^2 - (c-x)^2$
[Equating (5) and (7)]
Expand the term $(c-x)^2$ using the identity $(m-n)^2 = m^2 - 2mn + n^2$:
$b^2 - x^2 = a^2 - (c^2 - 2cx + x^2)$
Remove the parenthesis by distributing the negative sign:
$b^2 - x^2 = a^2 - c^2 + 2cx - x^2$
Add $x^2$ to both sides of the equation:
$b^2 = a^2 - c^2 + 2cx$
Now, isolate the term containing $x$ ($2cx$) by moving $a^2$ and $-c^2$ to the left side:
$b^2 - a^2 + c^2 = 2cx$
Solve for $x$ by dividing both sides by $2c$:
$\mathbf{x = \frac{b^2 + c^2 - a^2}{2c}}$
... (8)
Now that we have an expression for $x$ in terms of $a$, $b$, and $c$, substitute this value of $x$ back into equation (5) to find $h^2$ in terms of $a$, $b$, and $c$:
$h^2 = b^2 - x^2$
[From (5)]
$h^2 = b^2 - \left( \frac{b^2 + c^2 - a^2}{2c} \right)^2$
[Substituting x from (8)]
Use the difference of squares formula, $p^2 - q^2 = (p-q)(p+q)$, where $p=b$ and $q = \frac{b^2 + c^2 - a^2}{2c}$:
$h^2 = \left( b - \frac{b^2 + c^2 - a^2}{2c} \right) \left( b + \frac{b^2 + c^2 - a^2}{2c} \right)$
Combine terms inside the parentheses by finding a common denominator ($2c$):
$h^2 = \left( \frac{2bc}{2c} - \frac{b^2 + c^2 - a^2}{2c} \right) \left( \frac{2bc}{2c} + \frac{b^2 + c^2 - a^2}{2c} \right)$
$h^2 = \left( \frac{2bc - (b^2 + c^2 - a^2)}{2c} \right) \left( \frac{2bc + (b^2 + c^2 - a^2)}{2c} \right)$
$h^2 = \left( \frac{2bc - b^2 - c^2 + a^2}{2c} \right) \left( \frac{2bc + b^2 + c^2 - a^2}{2c} \right)$
Rearrange the terms in the numerators to recognize perfect squares:
$h^2 = \left( \frac{a^2 - (b^2 - 2bc + c^2)}{2c} \right) \left( \frac{(b^2 + 2bc + c^2) - a^2}{2c} \right)$
Recognize the perfect squares: $(b-c)^2 = b^2 - 2bc + c^2$ and $(b+c)^2 = b^2 + 2bc + c^2$:
$h^2 = \left( \frac{a^2 - (b-c)^2}{2c} \right) \left( \frac{(b+c)^2 - a^2}{2c} \right)$
Apply the difference of squares formula ($p^2 - q^2 = (p-q)(p+q)$) again to both numerators:
$h^2 = \frac{(a - (b-c))(a + (b-c))}{2c} \times \frac{((b+c) - a)((b+c) + a)}{2c}$
$h^2 = \frac{(a - b + c)(a + b - c)(b + c - a)(a + b + c)}{4c^2}$
... (9)
Now, we introduce the semi-perimeter, $s = \frac{a+b+c}{2}$. This definition implies the following relationships:
$a+b+c = 2s$
[From definition of s]
$a+b-c = (a+b+c) - 2c = 2s - 2c = 2(s-c)$
$a+c-b = (a+b+c) - 2b = 2s - 2b = 2(s-b)$
$b+c-a = (a+b+c) - 2a = 2s - 2a = 2(s-a)$
Substitute these expressions into the numerator of equation (9):
$h^2 = \frac{[2(s-b)] \times [2(s-c)] \times [2(s-a)] \times [2s]}{4c^2}$
[Substituting terms using s]
Simplify the numerator ($2 \times 2 \times 2 \times 2 = 16$):
$h^2 = \frac{16 s(s-a)(s-b)(s-c)}{4c^2}$
Simplify the fraction $\frac{16}{4} = 4$:
$\mathbf{h^2 = \frac{4 s(s-a)(s-b)(s-c)}{c^2}}$
... (10)
Now, take the square root of both sides to find an expression for $h$. Since height must be a positive value, we take the positive square root:
$h = \sqrt{\frac{4 s(s-a)(s-b)(s-c)}{c^2}}$
$\mathbf{h = \frac{\sqrt{4}\sqrt{s(s-a)(s-b)(s-c)}}{\sqrt{c^2}} = \frac{2}{c} \sqrt{s(s-a)(s-b)(s-c)}}$
... (11)
Finally, substitute this expression for $h$ (equation 11) back into our original area formula (equation 3):
$A = \frac{1}{2} c h$
[From (3)]
$A = \frac{1}{2} c \left( \frac{2}{c} \sqrt{s(s-a)(s-b)(s-c)} \right)$
[Substituting h from (11)]
Cancel out the terms $\frac{1}{2} \times c \times \frac{2}{c}$:
$A = \cancel{\frac{1}{2}} \cancel{c} \times \frac{\cancel{2}}{\cancel{c}} \sqrt{s(s-a)(s-b)(s-c)}$
$\mathbf{A = \sqrt{s(s-a)(s-b)(s-c)}}$
... (12)
This is Heron's Formula. This derivation shows how the area of a triangle, known from the base and height, can be expressed purely in terms of its side lengths using algebraic manipulation and Pythagoras' theorem.
Calculating Area of a Triangle using Three Sides
To calculate the area of a triangle when only the lengths of its three sides ($a, b, c$) are known, we use Heron's Formula. This method is especially useful when the height of the triangle is not readily available or is difficult to determine.
Steps to Calculate Area Using Heron's Formula
Let the side lengths of the triangle be $a$, $b$, and $c$. The process involves the following systematic steps:
-
Calculate the Semi-perimeter ($s$):
The semi-perimeter is half the total length of the boundary (the perimeter) of the triangle. Add the lengths of the three sides and divide the sum by 2.
$\mathbf{s = \frac{a+b+c}{2}}$
... (1)
Make sure all side lengths are in the same unit before calculating $s$. The unit of $s$ will be the same as the unit of the side lengths (e.g., cm, m).
-
Calculate the Differences:
Calculate the difference between the semi-perimeter $s$ and each of the side lengths. These are $(s-a)$, $(s-b)$, and $(s-c)$.
First difference: $(s-a)$
Second difference: $(s-b)$
Third difference: $(s-c)$
For a triangle to be valid (i.e., for the given side lengths to actually form a triangle), the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be greater than the length of the third side ($a+b>c$, $a+c>b$, $b+c>a$). If this condition is met, then $s$ will always be greater than each individual side length, ensuring that $(s-a)$, $(s-b)$, and $(s-c)$ are positive values.
-
Apply Heron's Formula:
Substitute the calculated values of $s$, $(s-a)$, $(s-b)$, and $(s-c)$ into Heron's formula to find the area ($A$).
$\mathbf{Area} (A) = \mathbf{\sqrt{s(s-a)(s-b)(s-c)}}$
... (2)
The expression under the square root, $s(s-a)(s-b)(s-c)$, must be non-negative. For any valid triangle, it will be positive.
-
Calculate the Final Area:
Compute the product of the four terms under the square root and then find the square root of the result. This value is the area of the triangle. The unit for the area will be the square of the unit used for the side lengths (e.g., $\text{cm}^2$, $\text{m}^2$).
Examples
Example 1. Find the area of a triangle whose sides are $13$ cm, $14$ cm, and $15$ cm.
Answer:
Given:
Side lengths of the triangle are $a = 13$ cm, $b = 14$ cm, and $c = 15$ cm.
To Find:
Area of the triangle using Heron's Formula.
Solution:
Step 1: Calculate the semi-perimeter ($s$)
$s = \frac{a+b+c}{2}$
[Formula (1)]
$s = \frac{13 \$ \text{cm} + 14 \$ \text{cm} + 15 \$ \text{cm}}{2}$
[Substitute values]
$s = \frac{42 \$ \text{cm}}{2}$
[Sum of sides]
$\mathbf{s = 21 \$\$ cm}$
Step 2: Calculate the differences $(s-a), (s-b), (s-c)$
$(s-a) = 21 \$ \text{cm} - 13 \$ \text{cm} = 8 \$ \text{cm}$
[First difference]
$(s-b) = 21 \$ \text{cm} - 14 \$ \text{cm} = 7 \$ \text{cm}$
[Second difference]
$(s-c) = 21 \$ \text{cm} - 15 \$ \text{cm} = 6 \$ \text{cm}$
[Third difference]
Step 3: Apply Heron's Formula
$A = \sqrt{s(s-a)(s-b)(s-c)}$
[Formula (2)]
Substitute the calculated values into the formula:
$A = \sqrt{(21 \$ \text{cm})(8 \$ \text{cm})(7 \$ \text{cm})(6 \$ \text{cm})}$
$A = \sqrt{21 \times 8 \times 7 \times 6 \$ \text{cm}^4}$
Step 4: Calculate the Area
To simplify finding the square root, factorize the numbers into their prime factors and group them in pairs:
$21 = 3 \times 7$
$8 = 2 \times 2 \times 2$
$7 = 7$
$6 = 2 \times 3$
Product under the square root:
$21 \times 8 \times 7 \times 6 = (3 \times 7) \times (2 \times 2 \times 2) \times 7 \times (2 \times 3)$
Group factors:
$= (2 \times 2 \times 2 \times 2) \times (3 \times 3) \times (7 \times 7)$
$= 2^4 \times 3^2 \times 7^2$
Now, calculate the square root:
$A = \sqrt{2^4 \times 3^2 \times 7^2 \$ \text{cm}^4}$
$A = 2^{4/2} \times 3^{2/2} \times 7^{2/2} \$ \text{cm}^{4/2}$
$A = 2^2 \times 3^1 \times 7^1 \$ \text{cm}^2$
$A = 4 \times 3 \times 7 \$ \text{cm}^2$
$A = 12 \times 7 \$ \text{cm}^2$
$\mathbf{A = 84 \$\$ cm^2}$
Therefore, the area of the triangle is 84 square centimetres ($\text{cm}^2$).
Example 2. The sides of a triangular field are $50$ m, $80$ m, and $120$ m. Find the area of the field.
Answer:
Given:
Side lengths of the triangular field are $a = 50$ m, $b = 80$ m, and $c = 120$ m.
To Find:
Area of the field using Heron's Formula.
Solution:
Step 1: Calculate the semi-perimeter ($s$)
$s = \frac{a+b+c}{2}$
[Formula (1)]
$s = \frac{50 \$ \text{m} + 80 \$ \text{m} + 120 \$ \text{m}}{2}$
[Substitute values]
$s = \frac{250 \$ \text{m}}{2}$
[Sum of sides]
$\mathbf{s = 125 \$\$ m}$
Step 2: Calculate the differences $(s-a), (s-b), (s-c)$
$(s-a) = 125 \$ \text{m} - 50 \$ \text{m} = 75 \$ \text{m}$
[First difference]
$(s-b) = 125 \$ \text{m} - 80 \$ \text{m} = 45 \$ \text{m}$
[Second difference]
$(s-c) = 125 \$ \text{m} - 120 \$ \text{m} = 5 \$ \text{m}$
[Third difference]
Step 3: Apply Heron's Formula
$A = \sqrt{s(s-a)(s-b)(s-c)}$
[Formula (2)]
Substitute the calculated values:
$A = \sqrt{(125 \$ \text{m})(75 \$ \text{m})(45 \$ \text{m})(5 \$ \text{m})}$
$A = \sqrt{125 \times 75 \times 45 \times 5 \$ \text{m}^4}$
Step 4: Calculate the Area
To simplify the calculation under the square root, factorize the numbers into their prime factors:
$125 = 5^3$
$75 = 3 \times 5^2$
$45 = 3^2 \times 5$
$5 = 5^1$
The product of the terms is:
$125 \times 75 \times 45 \times 5 = (5^3) \times (3 \times 5^2) \times (3^2 \times 5) \times 5^1$
Group the prime factors and sum their exponents:
$= 3^{1+2} \times 5^{3+2+1+1} = 3^3 \times 5^7$
So, $A = \sqrt{3^3 \times 5^7 \$ \text{m}^4}$. To take the square root, express the factors with even exponents:
$A = \sqrt{3^2 \times 3 \times 5^6 \times 5 \$ \text{m}^4}$
Take the square root of the terms with even exponents outside the radical:
$A = \sqrt{3^2} \times \sqrt{5^6} \times \sqrt{3 \times 5} \$ \text{m}^2$
$A = 3^{2/2} \times 5^{6/2} \times \sqrt{15} \$ \text{m}^2$
$A = 3^1 \times 5^3 \times \sqrt{15} \$ \text{m}^2$
$A = 3 \times 125 \times \sqrt{15} \$ \text{m}^2$
$\mathbf{A = 375\sqrt{15} \$\$ m^2}$
Since $\sqrt{15}$ is an irrational number, the exact area is $375\sqrt{15}$ square metres. If an approximate value is needed, we can use $\sqrt{15} \approx 3.873$ (or a more precise value).
$A \approx 375 \times 3.873 \$ \text{m}^2$
$\mathbf{A \approx 1452.375 \$\$ m^2}$
The area of the triangular field is exactly $375\sqrt{15}$ square metres, or approximately $1452.38$ square metres.
Area of Equilateral Triangle using Heron's Formula
An equilateral triangle is a special type of triangle in which all three sides are equal in length. All three interior angles are also equal, each measuring $60^\circ$.
Let the length of each side of the equilateral triangle be '$a$'. So, the side lengths are $a, a,$ and $a$.
We can derive a specific formula for the area of an equilateral triangle using Heron's formula by substituting $a, a, a$ for the side lengths.
Derivation using Heron's Formula
Let the sides of the triangle be $a_1=a$, $a_2=a$, and $a_3=a$. We apply the steps of Heron's Formula:
Step 1: Calculate the semi-perimeter ($s$)
The semi-perimeter is half the sum of the side lengths:
$\mathbf{s = \frac{a_1+a_2+a_3}{2}}$
[Formula for semi-perimeter]
Substitute the side lengths $a, a, a$:
$\mathbf{s = \frac{a+a+a}{2} = \frac{3a}{2}}$
... (1)
Step 2: Calculate the differences $(s-a_1), (s-a_2), (s-a_3)$
Calculate the difference between the semi-perimeter and each side length:
$(s-a_1) = \frac{3a}{2} - a = \frac{3a - 2a}{2} = \mathbf{\frac{a}{2}}$
$(s-a_2) = \frac{3a}{2} - a = \frac{3a - 2a}{2} = \mathbf{\frac{a}{2}}$
$(s-a_3) = \frac{3a}{2} - a = \frac{3a - 2a}{2} = \mathbf{\frac{a}{2}}$
So, we have $s=\frac{3a}{2}$, and $(s-a_1)=(s-a_2)=(s-a_3)=\frac{a}{2}$.
Step 3: Apply Heron's Formula
Heron's formula for the area ($A$) of a triangle is:
$\mathbf{A = \sqrt{s(s-a_1)(s-a_2)(s-a_3)}}$
[Heron's Formula]
Substitute the expressions for $s$ and the differences:
$\mathbf{A = \sqrt{\left(\frac{3a}{2}\right) \left(\frac{a}{2}\right) \left(\frac{a}{2}\right) \left(\frac{a}{2}\right)}}$
... (2)
Step 4: Calculate the Area (Simplify the expression)
Multiply the terms under the square root:
"$A = \sqrt{\frac{3a \times a \times a \times a}{2 \times 2 \times 2 \times 2}}$"
$\mathbf{A = \sqrt{\frac{3a^4}{16}}}$
... (3)
To find the square root, we can take the square root of the numerator and the denominator separately:
"$A = \frac{\sqrt{3a^4}}{\sqrt{16}}$"
Simplify the square roots. $\sqrt{16} = 4$, and $\sqrt{a^4} = a^{4/2} = a^2$:
"$A = \frac{\sqrt{3} \times a^2}{4}$"
Rearrange the terms to get the standard form:
$\mathbf{A_{\text{equilateral}} = \frac{\sqrt{3}}{4} a^2}$
... (4)
Thus, the standard formula for the area of an equilateral triangle with side length '$a$' is $\mathbf{\frac{\sqrt{3}}{4} a^2}$. This formula is typically faster to use for equilateral triangles than applying Heron's formula steps every time.
Example
Example 1. Find the area of an equilateral triangle with a side length of $10$ cm using Heron's formula. Verify your answer using the standard formula for an equilateral triangle.
Answer:
Given:
Equilateral triangle with side length, $a = 10$ cm.
To Find:
Area of the triangle using Heron's formula, and verify using the standard formula.
Solution using Heron's Formula Steps:
Let the side lengths be $a_1=10$ cm, $a_2=10$ cm, $a_3=10$ cm.
Step 1: Calculate the semi-perimeter ($s$)
"$s = \frac{a_1+a_2+a_3}{2} = \frac{10+10+10}{2} \$ \text{cm}$"
"$s = \frac{30}{2} \$ \text{cm} = 15 \$ \text{cm}$"
Step 2: Calculate the differences $(s-a_1), (s-a_2), (s-a_3)$
"$(s-a_1) = 15 \$ \text{cm} - 10 \$ \text{cm} = 5 \$ \text{cm}$"
"$(s-a_2) = 15 \$ \text{cm} - 10 \$ \text{cm} = 5 \$ \text{cm}$"
"$(s-a_3) = 15 \$ \text{cm} - 10 \$ \text{cm} = 5 \$ \text{cm}$"
Step 3: Apply Heron's Formula
"$A = \sqrt{s(s-a_1)(s-a_2)(s-a_3)}$"
"$A = \sqrt{(15 \$ \text{cm})(5 \$ \text{cm})(5 \$ \text{cm})(5 \$ \text{cm})}$"
"$A = \sqrt{15 \times 5 \times 5 \times 5 \$ \text{cm}^4}$"
Step 4: Calculate the Area
Simplify the product under the square root. Factorize 15 as $3 \times 5$:
"$A = \sqrt{(3 \times 5) \times 5 \times 5 \times 5 \$ \text{cm}^4}$"
Group the factors in pairs:
"$A = \sqrt{3 \times (5 \times 5) \times (5 \times 5) \$ \text{cm}^4}$"
"$A = \sqrt{3 \times 5^2 \times 5^2 \$ \text{cm}^4}$"
Take the square root of the terms with even exponents:
"$A = 5^{2/2} \times 5^{2/2} \times \sqrt{3} \$ \text{cm}^{4/2}$"
"$A = 5 \times 5 \times \sqrt{3} \$ \text{cm}^2$"
"$\mathbf{A = 25\sqrt{3} \$\$ cm^2}$"
Verification using Standard Formula:
The standard formula for the area of an equilateral triangle with side '$a$' is $A = \frac{\sqrt{3}}{4} a^2$. Using formula (4) derived above:
"$A = \frac{\sqrt{3}}{4} a^2$"
Substitute the given side length $a=10$ cm:
"$A = \frac{\sqrt{3}}{4} (10 \$ \text{cm})^2$"
[Substituting $a=10$ cm]
"$A = \frac{\sqrt{3}}{4} (100 \$ \text{cm}^2)$"
"$A = \frac{100\sqrt{3}}{4} \$ \text{cm}^2$"
Simplify the fraction $\frac{100}{4} = 25$:
"$\mathbf{A = 25\sqrt{3} \$\$ cm^2}$"
Both methods yield the same result. The area of the equilateral triangle is $25\sqrt{3}$ square centimetres ($\text{cm}^2$). If an approximate value is needed, use $\sqrt{3} \approx 1.732$.
"$A \approx 25 \times 1.732 \$ \text{cm}^2$"
"$\mathbf{A \approx 43.3 \$\$ cm^2}$"
Area of Isosceles Triangle using Heron's Formula
An isosceles triangle is a triangle in which exactly two sides are of equal length. The third side, which has a different length, is often referred to as the base of the isosceles triangle, especially in contexts related to angles or symmetry, although any side can be considered a base for area calculation using height.
Let the lengths of the two equal sides be '$a$', and the length of the third side (the base) be '$b$'. So, the side lengths are $a, a,$ and $b$.
We can derive a specific formula for the area of an isosceles triangle using Heron's formula by substituting $a, a, b$ for the side lengths.
Derivation using Heron's Formula
Let the sides of the triangle be $a_1=a$, $a_2=a$, and $a_3=b$. We apply the steps of Heron's Formula:
Step 1: Calculate the semi-perimeter ($s$)
The semi-perimeter is half the sum of the side lengths:
$\mathbf{s = \frac{a_1+a_2+a_3}{2}}$
[Formula for semi-perimeter]
Substitute the side lengths $a, a, b$:
$\mathbf{s = \frac{a+a+b}{2} = \frac{2a+b}{2}}$
... (1)
Step 2: Calculate the differences $(s-a_1), (s-a_2), (s-a_3)$
Calculate the difference between the semi-perimeter and each side length:
$(s-a_1) = s - a = \frac{2a+b}{2} - a = \frac{2a+b - 2a}{2} = \mathbf{\frac{b}{2}}$
$(s-a_2) = s - a = \frac{2a+b}{2} - a = \frac{2a+b - 2a}{2} = \mathbf{\frac{b}{2}}$
$(s-a_3) = s - b = \frac{2a+b}{2} - b = \frac{2a+b - 2b}{2} = \mathbf{\frac{2a-b}{2}}$
So, we have $s=\frac{2a+b}{2}$, $(s-a_1)=\frac{b}{2}$, $(s-a_2)=\frac{b}{2}$, and $(s-a_3)=\frac{2a-b}{2}$.
Step 3: Apply Heron's Formula
Heron's formula for the area ($A$) of a triangle is:
$\mathbf{A = \sqrt{s(s-a_1)(s-a_2)(s-a_3)}}$
[Heron's Formula]
Substitute the expressions for $s$ and the differences:
$\mathbf{A = \sqrt{\left(\frac{2a+b}{2}\right) \left(\frac{b}{2}\right) \left(\frac{b}{2}\right) \left(\frac{2a-b}{2}\right)}}$
... (2)
Step 4: Calculate the Area (Simplify the expression)
Multiply the terms under the square root:
"$A = \sqrt{\frac{(2a+b) \times b \times b \times (2a-b)}{2 \times 2 \times 2 \times 2}}$"
$\mathbf{A = \sqrt{\frac{b^2 (2a+b)(2a-b)}{16}}}$
... (3)
In the numerator, recognize the product $(2a+b)(2a-b)$. This is in the form $(x+y)(x-y)$, which simplifies to $x^2 - y^2$. Here, $x=2a$ and $y=b$.
"$(2a+b)(2a-b) = (2a)^2 - b^2 = 4a^2 - b^2$"
Substitute this back into the expression for $A$:
$\mathbf{A = \sqrt{\frac{b^2 (4a^2 - b^2)}{16}}}$
... (4)
To find the square root, we can take the square root of the numerator and the denominator separately:
"$A = \frac{\sqrt{b^2 (4a^2 - b^2)}}{\sqrt{16}}$"
Simplify the square roots. $\sqrt{16} = 4$, and $\sqrt{b^2} = b$ (assuming $b>0$ as it's a side length):
"$A = \frac{\sqrt{b^2} \sqrt{4a^2 - b^2}}{4} = \frac{b \sqrt{4a^2 - b^2}}{4}$"
Thus, a formula for the area of an isosceles triangle with equal sides of length '$a$' and base of length '$b$' is:
$\mathbf{A_{\text{isosceles}} = \frac{b}{4} \sqrt{4a^2 - b^2}}$
... (5)
This derived formula can be used directly for isosceles triangles, but one can always calculate the area using the general steps of Heron's formula as shown in the example below.
Example
Example 1. Find the area of an isosceles triangle with equal sides of $5$ cm and a base of $8$ cm using Heron's formula. Verify the result using the derived formula for an isosceles triangle.
Answer:
Given:
Isosceles triangle with equal sides $a=5$ cm and base $b=8$ cm.
The side lengths are $a_1=5$ cm, $a_2=5$ cm, and $a_3=8$ cm.
To Find:
Area of the triangle using Heron's formula, and verify using the derived formula for isosceles triangles.
Solution using Heron's Formula Steps:
Step 1: Calculate the semi-perimeter ($s$)
"$s = \frac{a_1+a_2+a_3}{2} = \frac{5+5+8}{2} \$ \text{cm}$"
"$s = \frac{18}{2} \$ \text{cm} = 9 \$ \text{cm}$"
Step 2: Calculate the differences $(s-a_1), (s-a_2), (s-a_3)$
"$(s-a_1) = 9 \$ \text{cm} - 5 \$ \text{cm} = 4 \$ \text{cm}$"
"$(s-a_2) = 9 \$ \text{cm} - 5 \$ \text{cm} = 4 \$ \text{cm}$"
"$(s-a_3) = 9 \$ \text{cm} - 8 \$ \text{cm} = 1 \$ \text{cm}$"
Step 3: Apply Heron's Formula
"$A = \sqrt{s(s-a_1)(s-a_2)(s-a_3)}$"
"$A = \sqrt{(9 \$ \text{cm})(4 \$ \text{cm})(4 \$ \text{cm})(1 \$ \text{cm})}$"
"$A = \sqrt{9 \times 4 \times 4 \times 1 \$ \text{cm}^4}$"
Step 4: Calculate the Area
Simplify the product under the square root:
"$A = \sqrt{9 \times 16 \$ \text{cm}^4}$"
"$A = \sqrt{144 \$ \text{cm}^4}$"
Take the square root:
"$\mathbf{A = 12 \$\$ cm^2}$"
Verification using the derived formula:
The derived formula for the area of an isosceles triangle with equal sides $a$ and base $b$ is $A = \frac{b}{4} \sqrt{4a^2 - b^2}$. Using formula (5) with $a=5$ cm and $b=8$ cm:
"$A = \frac{8}{4} \sqrt{4(5^2) - 8^2} \$ \text{cm}$"
[Substituting $a=5$ cm and $b=8$ cm]
"$A = 2 \sqrt{4(25) - 64} \$ \text{cm}^2$"
[Simplifying $\frac{8}{4}$ and $5^2$]
"$A = 2 \sqrt{100 - 64} \$ \text{cm}^2$"
[Calculating $4 \times 25$]
"$A = 2 \sqrt{36} \$ \text{cm}^2$"
[Calculating $100-64$]
Take the square root of 36:
"$A = 2 \times 6 \$ \text{cm}^2$"
"$\mathbf{A = 12 \$\$ cm^2}$"
Both methods yield the same result. The area of the isosceles triangle is 12 square centimetres ($\text{cm}^2$).