Heron’s formula is used to find the area of a triangle when we know the length of all its sides. It is also termed as Hero’s Formula. We don’t have to need to know the angle measurement of a triangle to calculate its area.
\(Area~of~triangle~using~three~sides =\sqrt{s(s-a)(s-b)(s-c)}\)
Semiperimeter, s= Perimeter of triangle/2 = (a+b+c)/2 |
Hero of Alexandria was a great mathematician who derived the formula for the calculation of the area of a triangle using the length of all three sides. It is also termed as Hero’s Formula. He also extended this idea to find the area of quadrilateral and also higher-order polygons. This formula has its huge applications in trigonometry such as proving the law of cosines or law of cotangents, etc.
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Heron’s Formula For Area of Triangle
According to Heron, we can find the area of any given triangle, whether it is a scalene, isosceles or equilateral, by using the formula, provided the sides of the triangle.
Suppose, a triangle ABC, whose sides are a, b and c, respectively. Thus, the area of a triangle can be given by;
\(Area =\sqrt{s(s-a)(s-b)(s-c)}\)
Where “s” is semi-perimeter = (a+b+c) / 2
And a, b, c are the three sides of the triangle.
How to Find the Area Using Heron’s Formula?
To find the area of a triangle using Heron’s formula, we have to follow two steps:
- The first step is to find the value of semi-perimeter of the given triangle.
S = (a+b+c)/2
- The second step is to use Heron’s formula to find the area of a triangle.
Let us understand that with the help of an example.
Example: A triangle PQR has sides 4 cm, 13 cm and 15 cm. Find the area of the triangle.
Semiperimeter of triangle PQR, s = (4+13+15)/2 = 32/2 = 16
By heron’s formula, we know;
A = √[s(s-a)(s-b)(s-c)]
Hence, A = √[16(16-4)(16-13)(16-15)] = √(16 x 12 x 3 x 1) = √576 = 24 sq.cm
This formula is applicable to all types of triangles. Now let us derive the area formula given by Heron.
Heron’s Formula For Quadrilateral
Let us learn how to find the area of quadrilateral using Heron’s formula here.
If ABCD is a quadrilateral, where AB||CD and AC & BD are the diagonals.
AC divides the quad.ABCD into two triangles ADC and ABC.
Now we have two triangles here.
Area of quad.ABCD = Area of ∆ADC + Area of ∆ABC
So, if we know the lengths of all sides of quadrilateral and length of diagonal AC, then we can use Heron’s formula to find the total area.
Hence, we will first find the area of ∆ADC and area of ∆ABC using Heron’s formula and at last, will add them to get the final value.
Heron’s Formula for Equilateral Triangle
As we know the equilateral triangle have all its sides equal. To find the area of equilateral triangle let us first find the semi perimeter of the equilateral triangle will be:
s = (a+a+a)/2
s=3a/2
where a is the length of the side.
Now, as per the heron’s formula, we know;
\(Area =\sqrt{s(s-a)(s-b)(s-c)}\)
Since, a = b = c
Therefore,
A = √[s(s-a)^{3}]
which is the required formula.
Proof
There are two methods by which we can derive the Hero’s formula. First, by using trigonometric identities and cosine rule. Secondly, solving algebraic expression using Pythagoras theorem. Let us see one by one both the proofs or derivation.
Proof using Trigonometric Cosine Rule
Let us prove the result using the law of cosines:
Let a, b, c be the sides of the triangle and α, β, γ are opposite angles to the sides.
We know that, the law of cosines is
\( Cos \gamma =\frac{a^2+b^2-c^2}{2ab}\)
Again, using trig identity, we have
\(Sin \gamma = \sqrt{1-cos^2γ} \)
= \(\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}{2ab}\)
Here, Base of triangle = a
Altitude = b sinγ
Now,
Proof Using Pythagoras Theorem
Area of a Triangle with 3 Sides
Area of ∆ABC is given by
A = 1/2 bh _ _ _ _ (i)
Draw a perpendicular BD on AC
Consider a ∆ADB
x^{2} + h^{2} = c^{2}
x^{2} = c^{2} − h^{2}—(ii)
⇒x = √(c^{2}−h^{2})−−−−−−—(iii)
Consider a ∆CDB,
(b−x)^{2} + h^{2} = a^{2}
(b−x)^{2} = a^{2} − h^{2}
b^{2} − 2bx + x^{2} = a^{2}–h^{2}
Substituting the value of x and x^{2} from equation (ii) and (iii), we get
b^{2} – 2b√(c^{2}−h^{2})+ c^{2}−h^{2} = a^{2} − h^{2}
b^{2} + c^{2} − a^{2} = 2b√(c^{2} − h^{2})
Squaring on both sides, we get;
(b^{2}+c^{2}–a^{2})^{2} = 4b^{2}(c^{2}−h^{2})
\(\frac{(b^2 + c^2 -a^2 )^2}{4b^2} = c^2 – h^2\)
\(h^2~ = ~c^2 ~-~\frac{(b^2 +c^2 -a^2 )^2}{4b^2}\)
\(h^2~ =~\frac{4b^2 c^2 – (b^2 +c^2 -a^2 )^2 }{4b^2}\)
\(h^2~=~\frac{(2bc)^2 -(b^2+c^2- a^2 )^2}{4b^2}\)
\(h^2~=~\frac{[2bc+(b^2 +c^2 -a^2 )][2bc-(b^2+c^2-a^2)]}
{4b^2}\)
\(h^2~ =~\frac{[(b^2+2bc+c^2)-a^2][a^2-(b^2- 2bc+c^2)]}{4b^2}\)
\(h^2~=~\frac{[(b+c)^2 – a^2 ].[a^2 -(b-c)^2 ]}{4b^2}\)
\(h^2~=~\frac{[(b+c)+a][(b+c)-a].[a+(b-c)][a-(b-c)]}{4b^2}\)
\(h^2~=~\frac{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}{4b^2}\)
The perimeter of a ∆ABC is
P= a+b+c
\(⇒~h^2~ =~\frac{P(P – 2a)(P – 2b)(P -2c)}{4b^2}\)
\(⇒~h~ = \sqrt{P(P – 2a)(P – 2b)(P -2c)}{2b}\)
Substituting the value of h in equation (i), we get;
\(A~=~\frac{1}{2} b \frac{\sqrt{P(P – 2a)(P – 2b)(P -2c)}}{2b}\)
\(A~=~\frac{1}{4}\sqrt{(P(P – 2a)(P – 2b)(P -2c)}\)
\(A~=~\sqrt{\frac{1}{16} P(P – 2a)(P – 2b)(P -2c)}\)
\(A~=~\sqrt{\frac{P}{2}\left(\frac{P – 2a}{2}\right)\left(\frac{P – 2b}{2}\right)\left(\frac{P -2c}{2}\right)}\)
Semi perimeter(s) = \(\frac{perimeter}{2}~=~\frac{P}{2}\)
\(⇒~A~=~\sqrt{s(s – a)(s – b)(s – c)}\)
Note: Heron’s formula is applicable to all type of triangles and the formula can also be derived using the law of cosines and the law of Cotangents.
Problems and Solutions
Let us now look into some examples to have a brief insight into the topic:
Example 1: Find the area of a trapezium, length of whose parallel sides is given as 22 cm and 12 cm and the length of other sides are 14 cm each.
Solution: Let PQRS be the given trapezium in which PQ = 22 cm, SR = 12 cm,
PS=QR=14cm.
Constructions: Draw OR||PS
Now, PORS is a parallelogram in which PS||OR and PO||SR
Therefore, PO=SR=12cm
⇒OQ = PQ-PO = 22 -12 = 10cm
In ∆OQR , we have
\(s~ = ~\frac{14+14+10}{2}~ =~\frac{38}{2}~=~19\)
Area of ∆OQR = \(\sqrt{s(s – a)(s – b)(s –C)}\)
= \( \sqrt{ (19(19 – 14)(19 – 14)(19 –10))}\)
= \( \sqrt{4275}\)
= \(15 \sqrt{19} cm^2\) ………(i)
We know that Area = \(\frac{1}{2} \times b \times h\)
\(\Rightarrow 15\sqrt{19} = \frac{1}{2} \times 10 \times h\)
\(\Rightarrow h = 5\sqrt{19}\) …………..(ii)
Area of trapezium = \(\frac{1}{2}~ (PQ+SR) × h \)
= \(\frac{1}{2}(22+12) × 3 \sqrt{19} \)
= \(51 \sqrt{19} cm^2\)
Example 2: Find the area of the triangle whose sides measure 10 cm, 17 cm and 21 cm. Also, determine the length of the altitude on the side which measures 17 cm.
Solution: s = \(\frac{a+b+c}{2}\) = \(\frac{10+17+21}{2}\) = 24
Area of Triangle= \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(\sqrt{24 \times 14 \times 7 \times 3}\)
\(\sqrt{7056}\) = 84 square cm
Taking 17 cm as the base length we need to find the height
Area, A = 1/2 x base x height
1/2 x 17 x h = 84 or h = 168/17 = 9.88 cm (Rounded to the nearest hundredth).
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Frequently Asked Questions – FAQs
What does s represents in Heron’s Formula?
S = (a+b+c)/2
Where a, b and c are three sides of a triangle.
When we use Heron’s formula?
Who gave Heron’s formula?
What is the Heron’s formula for equilateral triangle?
A = √[s(s-a)^{3}]
What happens when S is less than the length of one of the sides (which can happen if you have a triangle with two long sides and 1 very short side). Then you get a negative value for the square root which means there is an imaginary result for the area. This cannot be correct. How is this resolved?
Sides of a triangle is less than the semi-perimeter of triangle.
As we know, the sum of two sides of a triangle is always greater than the third side. Let a,b and c are three sides of a triangle. Hence,
a < b + c Add a on both the sides. a+a < a+b+c 2a < a+b+c a<(a+b+c)/2 (a+b+c)/2 = semi perimeter You can also repeat the same for sides b and c. Thus, you will get; b<(a+b+c)/2 c<(a+b+c)/2 Therefore, semiperimeter is always greater than sides of the triangle and we cannot get any negative value under the root in Heron's formula.