Applications of Heron's Formula

Heron's Formula provides a method to calculate the area of a triangle when the lengths of all three sides are known, without requiring the height or any angles.

Let's say we have a triangle ABC whose sides are of length “a”, “b,” and “c” then area of triangle ABC= \sqrt{s(s-a)(s-b)(s-c)}

where s(semi-perimeter) = \frac{a + b + c}{2}

Finding Area of Triangles

Heron's Formula can be used to calculate the area of different types of triangles when the lengths of all three sides are known.

Equilateral Triangle

In an equilateral triangle, all three sides are equal.
Let the length of each side be a.

Semi-perimeter(s) = 3a/2

Using Heron's Formula, the area of an equilateral triangle

Area = (\frac{\sqrt{3}}{4})a^2

Isosceles Triangle

An isosceles triangle has two equal sides. Let the equal sides be a and the third side be b.

Semi-perimeter, s = (2a + b)/2

Using Heron's Formula, the area of isosceles triangle

Area =\frac{b}{4}(\sqrt{4a^2-b^2 })

Finding Area of Quadrilaterals

Heron's Formula can also be used to find the area of a quadrilateral by dividing it into two triangles using a diagonal and adding the areas of both triangles.

Suppose we have a quadrilateral ABCD. If we draw the diagonal AC, the quadrilateral is divided into two triangles:

• △ABC
• △ADC

Area of quadrilateral ABCD=Area of △ABC + Area of △ADC

Total Area of the Quadrilateral : \text{Area of ABCD} = \sqrt{s_1(s_1-b)(s_1-c)(s_1-e)} + \sqrt{s_2(s_2-a)(s_2-d)(s_2-e)}

where a,b,c,d,e are sides of quadrilateral

Sample Problems

Question 1: There is concrete space in the shape of a triangle that needs to be tiled. The cost of tiling is Rs 20 per square unit. Find the total cost of tiling the area. 

The length of given sides are 10, 10 and 10.

The semi-perimeter "s" = 15

We know the area of the triangle using Heron's formula.

\\ Area = \sqrt{s(s - a)(s - b)(s - c)}

Here, s = 15, a = b = c = 10. 

\\ Area = \sqrt{s(s - a)(s - b)(s - c)} \\     = \sqrt{15(15 - 10)(15 - 10)(15 - 10)} \\     = \sqrt{15(5)(5)(5)} \\     = \sqrt{3(5)(5)(5)(5)} \\     = 25\sqrt{3}

The area of the triangle is 25√3

So, the cost of tiling the area = 25√3 x 20 = 500√3 = 500 (1.73) = 866 (approx)

Thus, cost of tiling the area is Rs. 866. 

Question 2: Find out the area of the quadrilateral given below:

The quadrilateral ABCD can be divided into two triangles whose area we can compute if we join A and C. 

Now we have triangles ACD and ABC. Out of which ACD is a right-angled triangle. 

AC² = AD² + DC² 

⇒ AC² = 9² + 40²

⇒ AC² = 81 + 1600 

⇒ AC² = 1681 

⇒ AC = 41

Area of triangle ADC = \\ \frac{1}{2} \times base \times height \\ = \frac{1}{2} \times 40 \times 9 \\ = 20 \times 9 \\ = 180 

Area of triangle ABC will be calculated using Heron's Formula. The sides of the triangle are 28,15 and 41. 

\\ s = \frac{28 + 15 + 41}{2} \\ = \frac{84}{2} \\ = 42

Area of triangle ABC = \\ \sqrt{42(42 - 41)(42 - 15)(42 - 28)} \\ = \sqrt{42(1)(27)(14)} \\ = 126 

Area of quadrilateral = Area of triangle ADC + Area of triangle ABC  = 180 + 126  = 306 sq units. 

Practice Problem

Problem 1: Calculate the area of a triangle with sides of lengths 5 cm, 12 cm, and 13 cm using Heron's formula.

Problem 2: Find the area of a triangle with sides 7 cm, 24 cm, and 25 cm using Heron's formula.

Problem 3: Determine the area of a triangle with sides 9 cm, 12 cm, and 15 cm using Heron's formula.

Problem 4: Calculate the area of a triangle with sides 8 cm, 15 cm, and 17 cm using Heron's formula.