Content Curator
Heron's formula is used to calculate the area of a triangle when the length of three sides is given. Heron's formula is also known as hero's formula in geometry. Heron’s formula is used to find different types of triangles, such as scalene, isosceles and equilateral triangles. It consists of many geometric applications: like proving the law of cosines and the law of cotangents. Using this formula we can find the area of the quadrilateral too.
| Table of Content |
Key takeaways: Heron’s Formula, Triangle, Quadrilateral, Pythagoras’ Theorem
Heron's Formula [Click Here for Sample Questions]
[Click Here for Sample Questions]
Heron's Formula was first discovered by a famous Mathematician, Heron of Alexandria. This formula has been used in determining the area of different types of triangles such as equilateral, isosceles, and scalene triangle. The formula does not depend upon the angles of a triangle for finding its area. It mostly depends on the length of the sides of a triangle. Previously it was used in calculating the area of a triangle using the side length of the triangle and further with some modifications it is also used in calculating the area of the quadrilateral.
Triangle
Heron's Formula for a triangle of sides a, b, c can be given as follows.
Area of triangle A = √s(s-a)(s-b)(s-c)
Perimeter, P = a+b+c
Where,
S = Semi Perimeter
S = Perimeter/2 = a+b+c/2
Read more:
Proof of Heron's Formula [Click Here for Sample Questions]
[Click Here for Sample Questions]
Heron's Formula can be proved by two different methods which are given below
- By Pythagoras Theorem
- By Trigonometric Identities
By Pythagoras Theorem
The Heron's Formula can be proved with the help of the Pythagoras theorem, the area of a triangle, and the algebraic expressions. Let's take a triangle ABC having sides a, b and c. Let us assume that a perpendicular line is drawn from a vertex B on the line AC at point M dividing its length into two parts p and q.
Area of the triangle = (1/2) b × h.
Here,
b = base and
h = height of a triangle.
Since b is divided into two parts p and q, then
b = p + q
→ q = b - p….eq(1)
After squaring both sides we get,
→ q² = b² + p² - 2bp….eq(2)
Adding h² on the both sides we get,
q² + h² = b² + p² - 2bp + h²….eq(3)
After applying Pythagoras Theorem in triangle ABM and ACM we get,
h² + q² = a²….eq(4)
p² + h² = c²….eq(5)
Putting the values of equation (4) and (5) in equation (3) we get,
q² + h² = b² + p² - 2bp + h²
→ a² = b² + c² - 2bp
→ p = (b² + c² - a²)/2b….eq(6)
From (5)
p² + h² = c²
→ h² = c² - p² = (c + p) (c - p)....eq(7)
Putting equation (6) in (7) we get,
h² = (c + p) (c - p)
→ h² = (c + (b² + c² - a²)/2b) (c - (b² + c² - a²)/2b)
→ h² = ((2bc + b² + c² - a²)/2b) ((2bc - b² - c² + a²)/2b)
→ h² = ((b + c)2 - a²)/2b) ((a² - (b - c)2)/2b)
→ h² = ((b + c + a)(b + c - a)(a + b - c)(a - b + c))/4b²) eq 8
As perimeter of triangle is P = a + b + c and P = 2s
∴ 2s = a + b + c….eq(9)
Putting eq (9) in (8) we get,
h² = ((b+c+a) (b+c-a) (a+b-c)(a-b+c))/4b²
→ h² = (2s × (2s - 2a) × (2s - 2b) × (2s - 2c))/4b²)
→ h² = (2s × 2(s - a) × 2(s - b) × 2(s - c))/4b²)
After solving the above equation we get h
→ h = √(4s(s - a)(s - b)(s - c)/4b²)
→ h = 2√(s(s - a)(s - b)(s - c))/b. eq.10
Area of triangle ABC, A = (1/2) × base × height
→ A = (1/2) × b × h
A = (1/2) × b × 2√(s(s - a)(s - b)(s - c))/b (From eq.10)
→ A = √(s(s - a)(s - b)(s - c))
Hence, Area of the triangle ABC = √(s(s - a)(s - b)(s - c)) square unit
Read more:
Let there be a triangle ABC having sides a,b and c and have opposite angles α, β, γ to its sides.
This can be proved using the law of cosines of a triangle.
The formula of cosine can be written as
Cos γ = a²+b²-c²/2ab
As we know that
Sin γ = √1-Cos²γ
= √1- (a²+b²-c²/2ab)²
= √4a²b² - (a²+b²-c²)²/2ab
Here,
Base of a triangle, b = a and
Altitude, h = b sinγ
As we know
A = (1/2) × b × h
= ½ ab Sin γ
= ¼ √4a²b² - (a²+b²-c²)²
= ¼ √(c²-(a-b)²)((a+b)²-c²))
After rearranging the above equation we can get
=√(a+b+c/2)(b+c-a/2)(a+c-b/2)(a+b-c/2)
= √s(s-a)(s-b)(s-c)
Applications of Heron's Formula [Click Here for Sample Questions]
[Click Here for Sample Questions]
There are mainly two applications of Heron's Formula which are given below
- For determining the area of a triangle if the length of sides of a triangle is given
- For calculating the area of a quadrilateral if the values of sides are given.
Read more:
Things to Remember
[Click Here for Sample Questions]
- Heron's Formula is mainly used for calculating the areas of different types of triangles and quadrilaterals. It does not depend upon the angles of a triangle.
- There are three different types of triangles which are Equilateral Triangle (all sides are equal), Isosceles Triangle (two sides are equal) and Scalene triangle (all sides are different).
- The Heron's Formula of a quadrilateral can be calculated by dividing it into two equal triangles having equal sides.
- This formula can also be used in proving the laws of trigonometry such as the law of cosines and cotangents.
Read more:
Sample Questions
Ques. Calculate the Heron's Formula of an Equilateral Triangle? [3 marks]
Ans. In an equilateral triangle, all sides are equal.
So the semi perimeter of a triangle is S= a+b+c/2
In an equilateral triangle, all sides are equal so we can write it, S= a+a+a/2 i.e, 3a/2.
And we know that the area of triangle= √s(s-a)(s-b)(s-c)
As we said above, in an equilateral triangle all sides are equal so we can put "a" for b and c side also.
And above we have find the value of "s" i.e, 3a/2
So by putting them in the formula we get,
Area of triangle= ?3a/2(3a/2-a)(3a/2-a)(3a/2-a)
After further simplifying the equation we'll get
=?3a/2(a/2)(a/2)(a/2)
=?3a?/6
=√3/4a²
Ques. Calculate the Heron's Formula of an Isosceles Triangle? [4 marks]
Ans. In an isosceles triangle two sides are equal and opposite angles are equal.
The semi perimeter of the triangle,
S=a+b+c/2
In isosceles triangle two sides are equal so we can write semiperimeter as
S=a+a+b/2
Hence S = 2a+b/2
By heron's formula area of triangle=√S(s-a)(s-b)(s-c)
Here we can put the value of S we have found in above and it's two sides are equal so we can replace " c" with "a"
After putting the Value of s and replacing c by a we'll get
Area of triangle = √(2a+b)[2a+b/2-a][2a+b/2-b][2a+b/2-a]
After further simplifying the equation we get
= √2a+b/2[2a+b-2a/2][2a+b-2b/2][2a+b-2a/2]
= √b*b(2a+b)(2a-b)/2*2*2*2
=b/2*2√2a²-b²
=b/4√4a²-b²
Ques. Calculate the Heron's Formula of a Scalene Triangle?[3 marks]
Ans. We can easily find out the area of all triangles using Heron's formula.
Thus, calculating heron's formula for a scalene triangle is given below.
In a scalene triangle all the sides are unequal.
Therefore it's perimeter = a + b + c
semiperimeter( s ) = (a + b + c)/2
Therefore heron's formula to find out the area of a scalene triangle is
Area = √ (a+b+c)/2 [ {( a+b+c)/2 -a}
For a better understanding, a diagram is given below.
Ques. What is the area of a quadrilateral? [4 marks]
Ans. Area of a quadrilateral can be found by dividing it in two triangles.
We can divide it by draw an diagonal
As we know area of triangle= 1/2* base * height
Now suppose a quadrilateral having sides ABCD.
Draw a diagonal through BD
Now we have two triangles i.e, ?ABD and ?BCD.
We can get the height of the triangle by drawing perpendicular to its base.
Now we can easily get the area of quadrilateral by adding the area of two triangles i.e,
Area of quad. = Area ?ABD + area ?BCD
= 1/2 * BD*h¹+ 1/2* BD*h²
=1/2*BD(h¹+h²)
=1/2*d*(h1+h²)
Hence the area of quadrilateral=1/2*d*(h1+h²)
Here d= diagonal of the triangle and h¹,h² is the height of the triangle
Ques. Find the area of an Equilateral Triangle having all its sides with a length of 6cm? [4 marks]
Ans. As we know that in an equilateral triangle , all sides are equal.
Here we are asked to find out the area of an equilateral triangle with sides 6 cm.
Now let's find out the perimeter.
Perimeterv = 3× side
=3× 6cm
Perimeter =18 cm
Now , S (semiperimeter) = Perimeter/2
=18/2 = 9
Therefore we get ,
S = 9 cm
a= 6cm b= 6cm c= 6 cm
On the application of heron's formula,
we get,
Area of triangle = √S(s-a)(s-b)(s-c)
= √ 9 (9-6) (9-6) (9-6)
= √ 9 ×3 ×3 × 3
= √ 243
= 15.6 cm²
Ques. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD. [3 marks]
Ans.
In ?ABC, we have
AB = AC [ABC is an isosceles triangle]
∴ ∠ABC = ∠ACB …(1)
[Angles opposite to equal sides of a ? are equal]
Again, in ?BDC, we have
BD = CD [BDC is an isosceles triangle]
∴ ∠CBD = ∠BCD …(2)
[Angles opposite to equal sides of a A are equal]
Adding (1) and (2), we have
∠ABC + ∠CBD = ∠ACB + ∠BCD
⇒ ∠ABD = ∠ACD
Ques. ?ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle. [5 marks]
Ans.
AB = AC [Given] …(1)
AB = AD [Given] …(2)
From (1) and (2), we have
AC = AD
Now, in ?ABC, we have
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒ 2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of A are equal)]
Similarly, in ?ACD,
∠ADC + ∠ACD + ∠CAD = 180°
⇒ 2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° – 180° = 180°
⇒ ∠BCD = 180°/2 = 90°
Thus, ∠BCD = 90°
Ques. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). [3 marks]
Show that
(i) ?ABE ≅ ?ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.
Ans.
(i) In ?ABE and ?ACE, we have
∠AEB = ∠AFC
[Each 90° as BE ⊥ AC and CF ⊥ AB]
∠A = ∠A [Common]
BE = CF [Given]
∴ ?ABE ≅ ?ACF [By AAS congruency]
(ii) Since, ?ABE ≅ ?ACF
∴ AB = AC [By C.P.C.T.]
⇒ ABC is an isosceles triangle.
Ques. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal. [3 marks]
Ans.
?ABC is an isosceles triangle
∴ AB = AC
⇒ ∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]
⇒ ∠BCE = ∠CBF
Now, in ?BEC and ?CFB
∠BCE = ∠CBF [Proved above]
∠BEC = ∠CFB [Each 90°]
BC = CB [Common]
∴ ?BEC ≅ ?CFB [By AAS congruency]
So, BE = CF [By C.P.C.T.]
Ques. In ?ABC, AD is the perpendicular bisector of BC (see figure). Show that ? ABC is an isosceles triangle in which AB = AC. [3 marks]
Ans. Since AD is the bisector of BC.
∴ BD = CD
Now, in ?ABD and ?ACD, we have
AD = DA [Common]
∠ADB = ∠ADC [Each 90°]
BD = CD [Proved above]
∴ ?ABD ≅ ?ACD [By SAS congruency]
⇒ AB = AC [By C.P.C.T.]
Thus, ?ABC is an isosceles triangle.
Read more:
Comments