Heron’s Formula

Heron’s Formula: Heron of Alexandria was a Greek mathematician. Heron has derived the formula for the area of a triangle when the measure of three sides is given. Unlike the other formula for triangles, we need not calculate angles or other parameters of the triangle while using Heron’s formula.

In this article, we will learn about the area of the triangle whose lengths of all the sides are given. We will also see the area of an equilateral triangle and the area of an isosceles triangle by using Heron’s formula, application of Heron’s formula in finding the areas of quadrilaterals and solved examples.

Also check,

? Area of Triangle: Formulas, Examples

Introduction to Heron’s Formula

There are triangles all around us ranging from nachos, to samosas, to a road sign, to sandwiches, to set squares and so on. We see various triangles almost every day of our lives. They come in different types and sizes too. Some are equilateral like the road sign, some are isosceles like the sandwich, while some others are scalene like the set square that comes in our geometry box.

Have you ever wondered how to calculate the area of a triangle? The most commonly used formula to calculate the area of a triangle is half of the product of the base and its height. But to use this formula we should know the height of the triangle.

In the case of a right triangle, the height is either the length of the base or the perpendicular side. In the case of an equilateral triangle, we can take any side as the base and the height is obtained by drawing a perpendicular line from the vertex opposite to the base. The measure of this height can be found by using the Pythagoras theorem.

For an isosceles triangle, we can take the unequal side as the base and the perpendicular distance from the opposite vertex to this base is taken as the height. We can deduce the height by using the Pythagoras theorem.

Suppose the length of the sides of this scalene triangle is given. Its height is not given and we don’t have a clue to find it. Can we find the area of this triangle by using the same formula?

In such cases, we use the formula known as Heron’s Formula named after the Greek mathematician Heron of Alexandria. It is also known as Hero’s Formula. Heron of Alexandria has derived this famous formula for finding the area of a triangle using only the lengths of its three sides.

Before understanding Heron’s formula in detail, let’s recollect a few facts about the triangle. We know that triangle has three sides, how will you find the perimeter of a triangle??

The perimeter of a triangle is the sum of the lengths of its sides. Let us consider a triangle \(ABC\) whose lengths of the sides are \(a,b,\) and \(c\).

So, perimeter \( = a + b + c\)
Now, let us learn a new measure called the semi-perimeter of a triangle.

What do you mean by the word semi? Yes, semi means half.

So, semi-perimeter \( = \frac{1}{2} \times \) Perimeter
Therefore, the semi-perimeter of the triangle with its sides \(a,b\) and \(c\) is
Semi-perimeter, \(s = \frac{{a + b + c}}{2}\)
After understanding these, now we are in a good position to learn Heron’s Formula.

What is Heron’s Formula?

Its states that, if a triangle has sides of lengths as \(a,b\) and \(c\) then the formula for finding the area of a triangle is,

\({\rm{Area}} = \sqrt {s(s – a)(s – b)(s – c)} \)

where, \(a,b\) and \(c\) are the lengths of sides of the triangle and \(s\) is the semi-perimeter of the triangle, given by \(s = \frac{{a + b + c}}{2}\)

Derivation of Heron’s Formula

Method 1 – By Using Trigonometric Identities or Cosine Rule

Let \(a,b,c\) be the sides of the triangle and \(\alpha ,\beta ,\gamma \) are opposite angles to the sides.

The Law of Cosines is also known as the cosine rule or cosine formula is used to find the third side of a triangle when the other two sides and the angle between them are known. We can also find all the angles of a triangle when all three sides are known.

By the law of cosines, we have

\(\cos \,\alpha  = \frac{{{b^2} + {c^2} – {a^2}}}{{2bc}}\)
\(\cos \,\beta  = \frac{{{a^2} + {c^2} – {b^2}}}{{2ac}}\)
\(\cos \,\gamma  = \frac{{{a^2} + {b^2} – {c^2}}}{{2ab}}\)

Again, by using trigonometric identity, we have

\({\rm{sin}}\gamma = \sqrt {1 – {{\cos }^2}\gamma } = \frac{{\sqrt {4{a^2}{b^2} – {{\left( {{a^2} + {b^2} – {c^2}} \right)}^2}} }}{{2ab}}\)
The height of the triangle with base \(a\) is \(b\sin \gamma, \)
Area \( = \frac{1}{2} \times {\rm{Base}} \times {\rm{Height}}\)
\( = \frac{1}{2} \times a \times b{\rm{sin}}\gamma = \frac{1}{2}ab{\rm{sin}}\gamma \)
\( = \frac{1}{4}\sqrt {4{a^2}{b^2} – {{\left( {{a^2} + {b^2} – {c^2}} \right)}^2}} \)
\( = \frac{1}{4}\sqrt {\left( {2ab – \left( {{a^2} + {b^2} – {c^2}} \right)} \right)\left( {2ab + \left( {{a^2} + {b^2} – {c^2}} \right)} \right)} \)
\( = \frac{1}{4}\sqrt {\left( {{c^2} – {{\left( {a – b} \right)}^2}} \right)\left( {{{\left( {a + b} \right)}^2} – {c^2}} \right)} \)
\( = \sqrt {\frac{{\left( {c – \left( {a – b} \right)} \right)\left( {c + \left( {a – b} \right)} \right)\left( {\left( {a + b} \right) – c} \right)\left( {\left( {a + b} \right) + c} \right)}}{{16}}} \)
\( = \sqrt {\frac{{\left( {b + c – a} \right)}}{2}\frac{{\left( {a + c – b} \right)}}{2}\frac{{\left( {a + b – c} \right)}}{2}\frac{{\left( {a + b + c} \right)}}{2}} \)
\( = \sqrt {\frac{{\left( {a + b + c} \right)}}{2}\frac{{\left( {b + c – a} \right)}}{2}\frac{{\left( {a + c – b} \right)}}{2}\frac{{\left( {a + b – c} \right)}}{2}} \)
\( = \sqrt {s\left( {s – a} \right)\left( {s – b} \right)\left( {s – c} \right)} \)
where \(s\) is the semi-perimeter of the triangle.

Method 2: By Using the Pythagoras Theorem

Let us consider a \(\Delta ABC\) with sides \(BC = a,AC = b\) and \(AB = c\)

Area of \(\Delta ABC\) is given by
\({\rm{Area = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times Base \times Height}} = \frac{1}{2}bh\)………..\(\left( 1 \right)\)
Draw a perpendicular line segment from \(B\) on \(AC\) intersecting at point \(D.\)

In a right triangle, △\(ADB\)
By using the Pythagoras theorem, we get
\({x^2} + {h^2} = {c^2}\)
\({x^2} = {c^2} – {h^2}\) ………… \(\left( 2 \right)\)
\( \Rightarrow x = \sqrt {{c^2} – {h^2}} \)…………… \(\left( 3 \right)\)

Now, in △\(CDB,\)
\({(b – x)^2} + {h^2} = {a^2}\)
\({(b – x)^2} = {a^2} – {h^2}\)
\({b^2} – 2bx + {x^2} = {a^2} – {h^2}\)

Substitute the values of \(x\) and \({x^2}\) from equation \(\left( 2 \right)\) and \(\left( 3 \right)\) we get
\({b^2} – 2b\sqrt {\left( {{c^2} – {h^2}} \right)}  + {c^2} – {h^2} = {a^2} – {h^2}\)
\({b^2} + {c^2} – {a^2} = 2b\sqrt {\left( {{c^2} – {h^2}} \right)} \)

On squaring both sides, we get

\({\left( {{b^2} + {c^2} – {a^2}} \right)^2} = 4{b^2}\left( {{c^2} – {h^2}} \right)\)
\(\frac{{{{\left( {{b^2} + {c^2} – {a^2}} \right)}^2}}}{{4{b^2}}} = \left( {{c^2} – {h^2}} \right)\)
\({h^2} = {c^2} – \frac{{{{\left( {{b^2} + {c^2} – {a^2}} \right)}^2}}}{{4{b^2}}}\)
\({h^2} = \frac{{4{b^2}{c^2} – {{\left( {{b^2} + {c^2} – {a^2}} \right)}^2}}}{{4{b^2}}}\)
\({h^2} = \frac{{{{(2bc)}^2} – {{\left( {{b^2} + {c^2} – {a^2}} \right)}^2}}}{{4{b^2}}}\)
\({h^2} = \frac{{\left( {2bc – \left( {{b^2} + {c^2} – {a^2}} \right)} \right)\left( {2bc + \left( {{b^2} + {c^2} – {a^2}} \right)} \right)}}{{4{b^2}}}\)
\({h^2} = \frac{{\left( {{{(b + c)}^2} – {a^2}} \right)\left( {{a^2} – {{(b – c)}^2}} \right)}}{{4{b^2}}}\)
\({h^2} = \frac{{((b + c) + a)((b + c) – a)(a + (b – c))(a – (b – c))}}{{4{b^2}}}\)
\({h^2} = \frac{{(a + b + c)(b + c – a)(a + b – c)(a – b + c)}}{{4{b^2}}}\)

The perimeter of △\(ABC\)  is
\(P = a + b + c\)
\( \Rightarrow {h^2} = \frac{{P(P – 2a)(P – 2b)(P – 2c)}}{{4{b^2}}}\)
\(h = \frac{{\sqrt {P(P – 2a)(P – 2b)(P – 2c)} }}{{2b}}\)
Substitute the value of \(h\) in equation \(\left( 1 \right)\) we get
\({\rm{Area}} = \frac{1}{2}b\frac{{\sqrt {P(P – 2a)(P – 2b)(P – 2c)} }}{{2b}}\)
\({\rm{Area}} = \frac{1}{4}\sqrt {P(P – 2a)(P – 2b)(P – 2c)} \)
\({\rm{Area}} = \sqrt {\frac{1}{{16}}P(P – 2a)(P – 2b)(P – 2c)} \)
\({\rm{Area}} = \sqrt {\frac{P}{2}\left( {\frac{P}{2} – a} \right)\left( {\frac{P}{2} – b} \right)\left( {\frac{P}{2} – c} \right)} \)
Semi-perimeter of △\(ABC\) is
\(s = \frac{P}{2}\)
Therefore, \({\rm{Area}} = \sqrt {s(s – a)(s – b)(s – c)} \)

Area of Equilateral Triangle Using Heron’s Formula

Heron’s formula can be used to derive a special formula applicable to calculate the area of an equilateral triangle.

In an equilateral triangle, all three sides are equal in length.

So, in this case \(a = b = c\)
So, \(s = \frac{{a + b + c}}{2} = \frac{{a + a + a}}{2} = \frac{{3a}}{2}\)
So, the \( = \sqrt {s(s – a)(s – b)(s – c)}  = \sqrt {\frac{{3a}}{2} \times \left( {\frac{{3a}}{2} – a} \right) \times \left( {\frac{{3a}}{2} – a} \right) \times \left( {\frac{{3a}}{2} – a} \right)} \)
\(=\sqrt {\frac{{3a}}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}  = \sqrt {\frac{{3{a^4}}}{{16}}}  = \frac{{\sqrt 3 }}{4}{a^2}\)
where \(a\) is the length of the side of the equilateral triangle.
Hence, the area of an equilateral triangle \( = \frac{{\sqrt 3 }}{4} \times {a^2} = \frac{{\sqrt 3 }}{4} \times {({\rm{ side }})^2}\)

Area of an Isosceles Triangle Using Heron’s Formula

Let the two equal sides of an isosceles triangle \(ABC\) be \(AB = AC = a\) and the length of the base be \(BC = b\)
Draw \(AD \bot BC\) . So, \(D\) bisects \(AB\)
Hence, \(BD = \frac{b}{2}.\)

Applying Pythagoras theorem on \(ABD\) , it can be written that \(A{D^2} + B{D^2} = A{B^2}\)
\( \Rightarrow A{D^2} + {\left( {\frac{b}{2}} \right)^2} = {(a)^2}\)
\( \Rightarrow A{D^2} = {a^2} – \frac{{{b^2}}}{4}\)
\( \Rightarrow AD = \sqrt {{a^2} – \frac{{{b^2}}}{4}} \)

Hence, the area of the isosceles triangle \(= \frac{1}{2} \times {\rm{Base \times Height}} = \frac{1}{2} \times BC \times AD\)
\( = \frac{1}{2} \times b \times \sqrt {{a^2} – \frac{{{b^2}}}{4}} \)
\( = \frac{b}{4}\sqrt {4{a^2} – {b^2}} \)

Solved Examples – Heron’s Formula

Q.1. Find the area of an isosceles triangle whose base is \({ \rm{2\;cm}}\) and length of its equal sides is \({ \rm{4\;cm}}\)

Ans: Given, \(a = 2\;{\rm{cm}},b = 4\;{\rm{cm}}\) and \(c = 4\;{\rm{cm}}\)
According to Heron’s formula, the area of a triangle is given by,
\({\rm{Area}} = \sqrt {s(s – a)(s – b)(s – c)} \)
Where, \(a,b\) and \(c\)  are the lengths of sides of the triangle and \(s\) is the semi-perimeter of the triangle, given by \(s = \frac{{a + b + c}}{2}\)
\(s = \frac{{2 + 4 + 4}}{2} = \frac{{10}}{2} = 5\)
\({\rm{Area}} = \sqrt {5(5 – 2)(5 – 4)(5 – 4)} \)
\(= \sqrt {5 \times 3 \times 1 \times 1} \)
\( = \sqrt {15} \)
Hence, the area of an isosceles triangle is \(\sqrt {15} \;{\rm{c}}{{\rm{m}}^2}\)

Q.2. Find the area of a triangle whose lengths of the sides are \(3\;{\rm{cm}},4\;{\rm{cm}}\) and \({\rm{5\;cm}}\).

Ans: Given, \(a = 3\;{\rm{cm}},b = 4\;{\rm{cm}}\) and \(c = 5\;{\rm{cm}}\)
According to Heron’s formula, the area of a triangle is given by,
 \({\rm{Area}}= \sqrt {s(s – a)(s – b)(s – c)} \)
Where \(a,b\) and \(c\) are the lengths of sides of the triangle and \(s\) is the semi-perimeter of the triangle, given by \(s = \frac{{a + b + c}}{2}\)
So, \(s = \frac{{a + b + c}}{2} = \frac{{3 + 4 + 5}}{2} = \frac{{12}}{2} = 6\)
Hence, the area of the given triangle \( = \sqrt {s(s – a)(s – b)(s – c)} \)
\( = \sqrt {6(6 – 3)(6 – 4)(6 – 5)} \)
\( = \sqrt {6 \times 3 \times 2 \times 1}  = \sqrt {36}  = 6\;{\rm{c}}{{\rm{m}}^2}\)
Hence, the area of the triangle \(6\;{\rm{c}}{{\rm{m}}^2}\)

Q.3. The sides of a triangular board are \({\rm{6\;m,8\;m}}\), and \({\rm{10\;m}}\) Find the cost of painting it at the rate of \(9\) paise per \({{\rm{m}}^2}\).

Ans: Given, \(a = 6\;{\rm{m}},b = 8\;{\rm{m}}\) and \(c = 10\;{\rm{m}}\)
According to this formula, the area of a triangle is given by,
\({\rm{Area}}= \sqrt {s(s – a)(s – b)(s – c)} \)
where, \(a,b\) and \(c\) are the lengths of sides of the triangle and \(s\) is the semi-perimeter of the triangle, given by \(s = \frac{{a + b + c}}{2}\)
\(s = \frac{{6 + 8 + 10}}{2} = \frac{{24}}{2} = 12\)
\({\rm{Area}} = \sqrt {12(12 – 6)(12 – 8)(12 – 10)} \)
\({\rm{Area}} = \sqrt {12 \times 6 \times 4 \times 2} \)
\({\rm{Area  = 24}}\)
Hence, the area of the triangle is \(24\;{{\rm{m}}^2}\)
Therefore, the cost of painting at the rate of \(9\) paise per \({{\rm{m}}^2} = (24 \times 0.09) = 2.16\)

Q.4. Find the area of the isosceles triangle if the perimeter is \({\rm{11\;cm}}\) and the base is \({\rm{5\;cm}}\).

Ans: Let the equal sides of the isosceles triangle be \(‘a’\) and the base of the triangle be \(‘b’\).
Perimeter of a triangle \({\rm{11\;cm}}\)
\(\Rightarrow 5 + a + a = 11\)
\(\Rightarrow 2a = 11 – 5\)
\(\Rightarrow 2a = 6\)
\(\Rightarrow a = 3\)
So, the length of the equal sides is \(3\;{\rm{cm}}.\)
Area of an isosceles triangle \( = \frac{b}{4}\sqrt {4{a^2} – {b^2}} \)
\( = \frac{5}{4}\sqrt {4{{(3)}^2} – {5^2}} \)
\( = \frac{5}{4}\sqrt {11} \;{\rm{c}}{{\rm{m}}^2}\)

Q.5. What is the cost of laying grass in an isosceles triangular field of edges \(65\;{\rm{cm}},65\;{\rm{cm}}\) and \({\rm{50\;cm}}\) at the rate of \(₹10\;{\rm{per\;c}}{{\rm{m}}^2}\)?

Ans: Given, \(a = 65\;{\rm{cm}},\;{\rm{b}} = 65\;{\rm{cm}}\) and \(c = 50\;{\rm{cm}}\)
According to this formula, the area of a triangle is given by,
\({\rm{Area}}= \sqrt {s(s – a)(s – b)(s – c)} \)
where, \(a,b\) and \(c\) are the lengths of sides of the triangle and \(s\) is the semi-perimeter of the triangle, given by \(s = \frac{{a + b + c}}{2}\)
\(s = \frac{{65 + 65 + 50}}{2} = \frac{{180}}{2} = 90\)
\({\rm{Area}}= \sqrt {90(90 – 65)(90 – 65)(90 – 50)} \)
\({\rm{Area}}= \sqrt {90 \times 25 \times 25 \times 40} \)
\({\rm{Area}} = 1500\;{\rm{c}}{{\rm{m}}^2}\)
Hence, the area of the triangle is \(1500\;{\rm{c}}{{\rm{m}}^2}\)
The cost of laying grass at the rate of \(₹10\;{\rm{per\;c}}{{\rm{m}}^2} = (1500 \times 10) = ₹15000\)

Summary

In this article, we have discussed how to find the area of a triangle if the lengths of all three sides are given by using Heron’s formula. We have also derived Heron’s formula by using two methods, first, by using the law of cosines and second by using the Pythagoras theorem. We have also deduced the area of the equilateral and isosceles triangle by using Heron’s formula.

FAQs

Q.1. What is the formula of semi-perimeter?
Ans: The semi-perimeter of the triangle with its sides \(a,b\) and \(c\) is
Semi-perimeter
\(s = \frac{{a + b + c}}{2}.\)

Q.2. How do you calculate the area of the triangle if the lengths of all three sides are known?
Ans: If the lengths of all the three sides of a triangle are known, then, the area of the triangle is calculated by using Heron’s formula, which says
\({\rm{Area}} = \sqrt {s(s – a)(s – b)(s – c)} \)
where \(a,b\) and \(c\) are the length of sides of the triangle and (s) is the semi-perimeter of the triangle, given by \(s = \frac{{a + b + c}}{2}\)

Q.3. What is (s) in the area of a triangle?
Ans: (s) is the semi-perimeter of the triangle in Heron’s formula for calculating the area of a triangle.

Q.4. What is Heron’s formula?
Ans: Heron’s formula is given by
\({\rm{Area\;of\;triangle}} = \sqrt {s(s – a)(s – b)(s – c)} \)
where \(a,b\) and \(c\) are the lengths of sides of the triangle and (s) is the semi-perimeter of the triangle, given by \(s = \frac{{a + b + c}}{2}\)

Q.5. Who gave Heron’s formula?
Ans: Heron of Alexandria, a Greek mathematician, had derived the formula for the area of the triangle when three sides are given.

Q.6. Is Heron’s formula accurate?
Ans: Yes. Heron’s formula is used to calculate the area of a triangle when the length of each side is given.

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