Heron’s formula

It used to find the area of Triangle when all three dimension of triangle is given.

2 S = (a + b + c)

Area = √ s (s – a) (s – b) (s – c)

Perimeter = Sum of all Sides

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Problems based on Heron's formula

1. A board in a shape of triangle having equal dimensions ‘a’. Find the area of board using Heron’s formula. If its perimeter is 180 cm. What will be the area of board ?

Sol. We know that

Perimeter = Sum of all Sides

P = a + a + a => P = 3 a

3 a = 180

So a = 60 cm

S = (a + b + c) / 2

S = ( 60 + 60 + 60 ) / 2

S = 180 / 2 => s = 90 cm

Now Area = √ s (s – a) (s – b) (s – c)

Area = √ 90 (90 – 60) (90 – 60) (90 – 60)

A = √ 90 * 30 * 30 * 30

A = √ 3 * 30 * 30 * 30 * 30

A = 30 * 30 √ 3

A = 900 √ 3 cm^2

2. A board in a shape of triangle having equal dimensions ‘a’. Find the area of board using Heron’s formula. If its perimeter is 120 cm. What will be the area of board ?

Sol. We know that

Perimeter = Sum of all Sides

P = a + a + a => P = 3 a

3 a = 120

So a = 40 cm

S = (a + b + c) / 2

S = ( 40 + 40 + 40 ) / 2

S = 120 / 2 => s = 60 cm

Now Area = √ s (s – a) (s – b) (s – c)

Area = √ 60 (60 – 40) (60 – 40) (60 – 40)

A = √ 60 * 20 * 20 * 20

A = √ 3 * 20 * 20 * 20 * 20

A = 20 * 20 √ 3

A = 400 √ 3 cm^2

3. The triangle board having the side walls of 122 cm , 22 cm and 120 cm . The rent of advertisement on board is 6000 rs per cm^2 per year . A company hired this walls for their advertisement for 3 months how much this company have to pay .

Sol. Given a = 122 cm b = 22 cm c = 120 cm

S = (a + b + c) / 2

S = ( 122 + 22 + 120 ) / 2

S = 264 / 2 => s = 132 cm

Now Area = √ s (s – a) (s – b) (s – c)

Area = √ 132 (132 - 122) (132 – 22) (132 – 120)

A = √ 132 * 10 * 110 * 12

A = √ 2 * 2 * 3 * 11 * 10 * 11 * 10 * 2 * 2 * 3

A = 2 * 2 * 3 * 10 * 11

A = 1320 cm^2

Earning by board in 12 months ( one year ) = 6000 * 1320

Earning by board in 1 months = ( 6000 * 1320 ) / 12

Earning by board in 1 months = ( 6000 * 1320 * 3 ) / 12 = 19,80,000 Rs

4. The triangle board having the side walls of 16 cm , 18 cm and 18 cm . The rent of advertisement on board is 6000 rs per cm^2 per year . A company hired this walls for their advertisement for 3 months how much this company have to pay .

Sol. Given a = 16 cm b = 18 cm c = 18 cm

S = (a + b + c) / 2

S = ( 16 + 18 + 18 ) / 2

S = 52 / 2 => s = 26 cm

Now Area = √ s (s – a) (s – b) (s – c)

Area = √ 26 (26 - 16) (26 - 18) (26 - 18)

A = √ 26 * 10 * 8 * 8

A = √ 2 * 13 * 2 * 5 * 2 * 2 *2 * 2 * 2 * 2

A = 2 * 2 * 2 * 2 √13 * 5

A = 16 √65 cm^2 = 16 * 8.06 = 129 cm^2

Earning by board in 12 months ( one year ) = 6000 * 129

Earning by board in 1 months = ( 6000 * 129 ) / 12

Earning by board in 1 months = ( 6000 * 129 * 3 ) / 12 = 1,93,500 Rs

5. The sides of a park is 15 cm , 11 cm and 6 cm. Find the area of this park.

Sol. Given a = 15 cm b = 11 cm c = 6 cm

S = (a + b + c) / 2

S = ( 15 + 11 + 6 ) / 2

S = 32 / 2 => s = 16 cm

Now Area = √ s (s – a) (s – b) (s – c)

Area = √ 16 (16 – 15) (16 – 11) (16 – 6)

A = √ 16 * 1 * 5 * 10

A = √ 2 * 2 * 2 * 2 * 5 * 2 * 5

A = 2 * 2 * 5 √ 2

A = 20 √ 2 cm^2 => 28.28 cm^2

6. The sides of a park is 30 cm , 22 cm and 12 cm. Find the area of this park.

Sol. Given a = 30 cm b = 22 cm c = 12 cm

S = (a + b + c) / 2

S = ( 30 + 22 + 12 ) / 2

S = 64 / 2 => s = 32 cm

Now Area = √ s (s – a) (s – b) (s – c)

Area = √ 32 (32 – 30) (32 – 22) (32 – 12)

A = √ 32 * 2 * 10 * 20

A = √ 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 2 * 2 * 5

A = 2 * 2 * 2 * 2 * 5 √ 2

A = 80 √ 2 cm^2 => 113.14 cm^2

Milan Tomic

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