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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.

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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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

www.edusaksham.com 1

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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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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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.

www.edusaksham.com 3

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

AD = √

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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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

AD = √

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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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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⇒ 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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