# Area of Triangle by Heron's Formula Contributed by: When all 3 sides of a triangle are known, to calculate its area, we need to find its height. But finding its height may be tedious. In this case, we use Heron's formula to find the area of the triangle in geometry.
This formula makes the calculation of finding the area of a triangle simple by eliminating the use of angles and the need for the height of the triangle.
1. Area of Triangle by Heron's Formula
Perimeter: Perimeter of a shape can be defined as the path or the
boundary that surrounds the shape. It can also be defined as the
length of the outline of a shape.
The word perimeter has been derived from the Greek word ‘peri’
meaning around, and ‘metron’ which means measure.
How to find Perimeter
During Christmas, we find people decorating their homes. For
example, people put up decorating lights around their homes like
around the fences of their homes. To find what length of lighting is
required, we have to find the perimeter of the fencing. We often
find the perimeter when putting up Christmas lights around the
house or fencing the field.
Christmas lights fencing the field
If we have to find the length of the track around the oval-shaped
ground, then we have to find the perimeter of the ground.
Soccer Field
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2. For measuring the perimeter of small irregular shape, we can use a
string or thread and place it exactly along the boundary of the
shape, once. Then keeping the thread along the ruler, we find the
length of threads. The total length of the string used along the
boundary is the perimeter of the shape.
Moon Toy
Perimeter of moon toy = 12 cm + 10 cm= 22 cm.
We use a ruler to measure the length of the sides of a polygon. The
perimeter is determined by adding the lengths of the sides/edges
of the shape.
𝐋𝟐 𝐋𝟑
𝐋𝟏 𝐋𝟒
𝐋𝟓
Perimeter = L + L + L + L + L
1 2 3 4 5
Area: Area of an object can be defined as the space occupied by the
two-dimensional object. The area of a figure is the number of unit
squares that cover the surface of a closed figure.
The area is measured in square units such as square centimeters,
square feet, square inches, etc.
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3. 9 cm
B
Area of rectangle ABCD = Length × Breath. A
= 9 cm × 4 cm.
4 cm
= 36 𝐜𝐦𝟐 . D C
Perimeter of rectangle ABCD = AB + BC + CD + DA.
= 9 cm + 4 cm + 9 cm + 4 cm.
= 26 cm
For finding the area of a right-angled triangle, we can directly use
𝟏
the formula (Area of triangle = (base x height)
𝟐
We use the two sides containing the right angle as base and height.
Example 1: Find the area of right triangle ABC whose base and
height are 5 cm and 8 cm respectively. A
Solution: Since, ∆ABC is a right-angled triangle,
in which height AD = 8 cm and base CB = 5 cm.
8 cm
Using the above formula, we can write ∟
C D B
𝟏 5 cm
Area of right-angled triangle ABC = (base × height).
𝟐
Base is the side on which the = 𝟏 (5 cm×8 cm).
𝟐
height (perpendicular) is
drawn. Height is the = (5 cm×4 cm)
perpendicular drawn from
opposite vertex to its base. = 20 𝐜𝐦𝟐 .
When all 3 sides of a triangle are known, to calculate its area, we
need to find its height. But finding its height may be tedious. In this
case, we use Heron's formula to find the area of the triangle in
This formula makes the calculation of finding the area of a triangle
simple by eliminating the use of angles and the need for the height
of the triangle.
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4. The formula given by Heron about the area of a triangle is known as
Heron's formula. It is stated as:
Area of a triangle = √𝐬(𝐬 − 𝐚)(𝐬 − 𝐛)(𝐬 − 𝐜)
Conventionally: Length of A
opposite side of a vertex b
c
denoted by its small letter. For
Example: Length of opposite
side of vertex A is denoted by B a C
a.
Where a, b and c are the sides of the triangle, and S = semi –
𝐚+𝐛 +𝐜
perimeter, i.e., half the perimeter of the triangle = .
𝟐
Example2: Find the area of an equilateral triangle ABC with the side
Solution: Since ABC is an equilateral triangle, So, all sides are equal.
That is, AB = BC = CA = a. A
In equilateral triangle,
By Pythagoras theorem, a a attitude from the vertex
also bisects the base.
We can write, 𝐀𝐁 𝟐 = 𝐁𝐃𝟐 + 𝐀𝐃𝟐 . B 𝒂 ∟ C
D 𝒂
𝐀𝐃𝟐 = 𝐀𝐁 𝟐 - 𝐁𝐃𝟐 . 𝟐 𝟐
𝟐 𝒂 𝒂𝟐 𝟒𝐚𝟐 −𝐚𝟐 𝟑𝐚𝟐
=𝐚 - ( )𝟐 = 𝟐
𝐚 - = = .
𝟐 𝟒 𝟒 𝟒
𝟑𝐚𝟐
𝟒
𝐚
= √𝟑 .
𝟐
Therefore, in equilateral triangle ABC, base BC = a and height AD =
𝒂
√𝟑 .
𝟐
𝟏 𝐚 𝐚𝟐
So, the area of an equilateral triangle = (a × √𝟑 )=√𝟑 .
𝟐 𝟐 𝟒
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5. This same question is also solved by Heron's Formula. A
𝐚+𝐚+𝐚 𝟑𝐚
And, s = = . a a
𝟐 𝟐
By Heron's Formula, we can write B C
a
𝟑𝐚 𝟑𝐚 𝟑𝐚 𝟑𝐚
Area of an equilateral triangle = √ ( − 𝐚)( − 𝐚)( − 𝐚).
𝟐 𝟐 𝟐 𝟐
𝟑𝐚 𝐚 𝐚 𝐚
=√ × × × .
𝟐 𝟐 𝟐 𝟐
𝐚𝟐
= √𝟑 . (Where a is the length of a side)
𝟒
𝐚𝟐
Area of an equilateral triangle = √𝟑
𝟒
Example3: The sides of a triangular plot are in the ratio of 3:5:7 and
its perimeter is 450 m. Find its area.
A
5𝒙
3𝒙
B C
7𝒙
Solution: Since sides of a triangular plot are in the ratio 3:5:7. For
exact values, they contain a common factor x, which we have to
Let AB = 3x, AC = 5x and BC = 7x.
By definition of perimeter,
Perimeter of triangular plot = AB + BC + AC.
450 = 3x + 5x + 7x.
⇒ 450 = 15x.
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6. 𝟒𝟓𝟎
⇒ x= .
𝟏𝟓
⇒ x = 30.
Therefore, AB = 3x = 3 × 30 m = 90 m.
AC = 5x = 5 × 30 m = 150 m.
BC = 7x = 7 × 30 m = 210 m.
𝟒𝟓𝟎 𝐦
Now, semi-perimeter of triangular park = .
𝟐
S = 225 m.
So, the area of triangular park = √𝐬(𝐬 − 𝐚)(𝐬 − 𝐛)(𝐬 − 𝐜)
[∵ Area of a triangle = √𝐬(𝐬 − 𝐚)(𝐬 − 𝐛)(𝐬 − 𝐜)]
= √𝟐𝟐𝟓 𝐦 (𝟐𝟐𝟓 𝐦 − 𝟐𝟏𝟎 𝐦)(𝟐𝟐𝟓 𝐦 − 𝟏𝟓𝟎 𝐦)(𝟐𝟐𝟓 𝐦 − 𝟗𝟎 𝐦)
= √𝟐𝟐𝟓 𝐦 × 𝟏𝟓 𝐦 × 𝟕𝟓 𝐦 × 𝟏𝟑𝟓 𝐦.
= √𝟓 × 𝟓 × 𝟑 × 𝟑 × 𝟓 × 𝟑 × 𝟓 × 𝟓 × 𝟓 × 𝟓 × 𝟑 × 𝟑 × 𝟑 𝐦𝟐
------------------------ --------- ----------------- ------------------------
= …….……………... …….……
√𝟓 × 𝟓 × 𝟑 × 𝟑 × 𝟓 × 𝟓 × …….…………… 𝟓×𝟓×𝟓 𝟑 × 𝟑 × 𝟑 × 𝟑 𝐦𝟐
× …….……………...
--------- --------- ………...
--------- ...
--------- --------- ---------
𝟐
= ×3×5×
5 …….…… 5×3×
…….…… 3 √𝟓𝐦…….……
…….…… …….…… …….……
………... ………... ………... ………... ………... ………...
𝟐
= 5×3×5×5×3×3 √𝟓𝐦
= 3,375 √𝟓𝐦𝟐
= 3375 × 2.2360 𝐦𝟐 [∵ √𝟓= 2. 2360]
= 7546.5 𝐦𝟐
Hence, area of triangular park = 7546.5 𝐦𝟐 .
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