Practice Problems on Heron’s Formula with Solution

Problem 1: In a triangle with side lengths of 5 cm, 12 cm, and 13 cm, find the area using Heron’s Formula.

Solution:

Given,

• a = 5cm
• b= 12 cm
• c = 13 cm

According to Heron’s Formula: [Tex]A = \sqrt{s(s-a)(s-b)(s-c)}[/Tex]

s = (a+b+c)/2

s = (5+12+13)/2

s = 30/2 = 15

[Tex]A = \sqrt{s(s-a)(s-b)(s-c)}[/Tex]

A = [Tex]\sqrt{15(15-5)(15-12)(15-13)}[/Tex]

A = [Tex]\sqrt{15 \times 10 \times 3 \times 2}[/Tex]

A = √900

A = 30 cm²

∴ Area of Triangle is 30 cm².

Problem 2: Calculate the area of an equilateral triangle with a side length of 10 meters using Heron’s Formula.

Solution:

Given,

a = b = c = 10 cm For Equilateral Triangle

s = (a+b+c)/2

s = (10+10+10)/2

s = 30/2 = 15 meters

According to Heron’s Formula: [Tex]A = \sqrt{s(s-a)(s-b)(s-c)}[/Tex]

Putting s = 15 in the formula we get,

A = [Tex]\sqrt{15(15-10)(15-10)(15-10)} [/Tex]

A = [Tex]\sqrt{15 \times 5 \times 5 \times 5}[/Tex]

A = √1875 ≈ 43.3 meters²

∴ Area of Equilateral Triangle with a side length of 10 meters is approximately 43.3 m²

Problem 3: A scalene triangle has side lengths of 7 cm, 10 cm, and 12 cm. Find its area using Heron’s Formula.

Solution:

Given,

• a = 7 cm
• b = 10 cm
• c = 12 cm

We know s = (a+b+c)/2

s = (7+10+12)/2

s = 29/2 = 14.5 cm

Putting value of s in the formula we get,

[Tex]A = \sqrt{s(s-a)(s-b)(s-c)}[/Tex]

A = [Tex]\sqrt{14.5(14.5-7)(14.5-10)(14.5-12)}[/Tex]

A = \sqrt{14.5 \times 7.5 \times 4.5 \times 2.5}

A = √1912.8125 cm²

A ≈ 43.74 cm²

∴ Area of Scalene Triangle with side lengths 7 cm, 10 cm, and 12 cm is approximately 43.74 cm².

Problem 4: Find the area of triangle if its two sides are 8 cm and 5 cm resecpectivelu and its perimeter is 20 cm.

Solution:

Let a, b and c be the three sides of the triangle.

a = 8 cm

b = 5 cm

c = ?

Perimeter = 20 cm

Perimeter equals to the sum of the length of three sides of a triangle.

Perimeter = (a + b + c)

⇒ 20 = 8 + 5 + c

⇒ c = 20 – 13

⇒ c = 7 cm

s = perimeter / 2

⇒ s = 20 / 2

⇒ s = 10 cm

As, a = 8 cm, b = 5 cm, c = 7 cm, s = 10 cm

Thus, A = √(s(s – a)(s – a)(s – a)

⇒ A = √(10(10 – 8)(10 – 5)(10 – 7)

⇒ A = 17.32 cm²

Problem 5: Find area of an equilateral triangle with a side of 9 cm.

Solution:

Given,

Side = 9 cm

Area of Equilateral Triangle = √3 / 4 × a₂

⇒ Area of Equilateral Triangle = √3 / 4 × (9)²

⇒ Area of Equilateral Triangle = 20.25 √3 cm²

Heron’s formula is also known as Hero’s formula. It calculates the area of triangles or quadrilaterals based on the lengths of their sides. It does not consider the angles of the shapes, only their side lengths. The formula includes the semi-perimeter, which is symbolized by “s”, which is half the perimeter of the shape. This concept can be extended to find the area of quadrilaterals too.

Heron’s Formula for Area of Triangle

As we know Heron’s formula is used to calculate the area of a triangle,

Therefore, Heron’s Formula is

Area of Triangle ABC = √s(s-a)(s-b)(s-c)

where,

• s = Perimeter/2 = (a + b + c)/2

Area of Triangle Formulas

Formula to calculate different types of triangles is different. In the table given below, we have summarized the formulas for all the Types of Triangles.