Heron’s Formula: To score good marks in exams, students should practice important questions related to the given topics. In this section, we will discuss some of the most important questions on the concept of Heron’s formula.

Before going on to the questions, let’s understand Heron’s formula and its application in our day-to-day lives. At the end of the section, there are some of Heron’s formula extra questions that can help you test your understanding of the formula and its application.

**Heron’s Formula: Introduction**

Heron’s formula is part of the geometry used to calculate the area of a triangle. This formula is also known as Heron’s formula and is named after the Hero of Alexandria.

Don’t we already have a formula for calculating the area of a triangle? Yes, we do; the formula is: (Base×Height)/ 2.

Why do we need a new formula then? Heron’s formula comes into use when we have the measurement of all three sides of a triangle.

Heron’s formula helps in avoiding the stress of finding the angles or other distances first. We can simply calculate the area of the triangle with the help of the measurements of all three sides by using Heron’s formula.

**Heron’s formula**

Area of triangle= √s (s-a) (s-b) (s-c)

Here, a, b and c are the measurements of the sides of a triangle, and “s” is the semi perimeter of the same triangle.

Semi perimeter of a triangle can be calculated as follows:

( a+b+c)/2.

Let’s take an example for a better understanding of the formula:

- Find the area of a triangle whose sides are a,b and c are of 3 cm, 4 cm, and 5 cm, respectively.

**First, we will have to calculate the semi-perimeter of the triangle.**

s= (a+b+c)/2= ( 3+4+5 )/2= 12/2= 6

Hence, s = 6

**Now, using heron’s formula:**

Area of triangle = √s (s-a) (s-b) (s-c)

= √6 (6-3) (6-4) (6-5)

=√6(3) (2) (1)= √6(6)= √36

=6 cm

Therefore, the area of the given triangle is 6cm.

**Application of Heron’s Formula **

- Heron’s formula can easily help us calculate the area of any kind of triangle if we have the measurements of all the sides of the given triangle.
- Heron’s formula can also come in use in calculating the area of a quadrilateral, providing all the sides of the quadrilateral are given.

**Important questions based on Heron’s Formula **

**1. Find the area of the triangle in which**

- a = b = c = 4 cm

Solution: Area of equilateral triangle=√3/4× a²= √3/4 × 4² = 4√3 cm²

- A right-angled triangle whose hypotenuse is 13 cm and the base is 5 cm.

Solution: Area of right-angled triangle = (Base× Height) /2

Given, Hypotenuse= 13 cm and base= 5 cm.

We need to find the height; let’s find out with the help of the Pythagoras theorem.

Hypotenuse²= Base ²+ Perpendicular²

**( Perpendiculars are the heights of the right-angled triangles) **

13²= 5² + Height ²

Therefore, 169 – 25= Height²

Therefore, Height² = 144

Applying roots on both the sides

Height = 12 cm.

Now, area of triangle=( 12 × 5 ) / 2

= 60/2= 30cm²

**2. Find the area of a triangle whose sides are 11 m, 60 m, and 61 m.**

Solution: First, let’s find the value of s.

s= ( 11+ 60 + 61 ) / 2

s= (132)/ 2 = 66 m

As now we have all three sides and the semi-perimeter, we can find the area of the triangle using Heron’s formula.

Area of triangle= √66(66-11)(66-60)(66-61)

=√66(55)(6)(5) = √108900= 330 m²

Therefore, the area of the given triangle is 330 sq. m.

**3. The sides of a triangular plot are in the ratio of 6: 7: 8, and its perimeter is 420 m. Find its area.**

Solution: The ratio of the sides of the triangle is given as 6:7:8

Now, let the common ratio between the sides of the triangle be “x.”

∴ The sides are 6x, 7x and 8x

It is also given that the perimeter of the triangle = 420 cm.

6x + 7x + 8x = 420 cm

=> 21x= 420cm

So, x = 20

Now, the sides of the triangle are 120 cm, 140 cm, 160 cm.

So, the semi perimeter of the triangle (s) = 420/2 = 210 cm

**Using Heron’s formula,**

Area of the triangle= √210( 210-120)(210-140)(210-160)

= √210(90)(70)(50)

= 8133. 26 cm² appx.

Hence, the area of the given triangle is 8133.26 cm² appx.

The above-solved questions must’ve helped you understand the application of Heron’s formula in practical questions. Now, you are ready to practice the questions given below to test your knowledge.

**Extra questions based on Heron’s formula for practice (Unsolved) **

- Find the area of an isosceles triangle with two equal sides as 5 cm each and the third side as 8 cm.
- The sides of a triangle are in the ratio of 12: 17: 25, and its perimeter is 540cm. Find its area.
- A triangular park has sides given sides as 50 m, 80 m, and 120 m. A gardener needs to put a fence around it to plant grass inside. Determine the area in which he needs to plant.

**Also**, find the total cost of fencing it using a barbed wire that comes at the rate of 20 rupees per meter. Note: gardener has to leave a 3 m wide gap for a gate on 1 side.

4. Determine the area of the field. A field is formed in a trapezium with parallel sides of length 10 m and 25 m and. Also, the non-parallel sides are of length 13 m and 13 m.

**Conclusion:**

Geometry has so many topics that need enough attention from students. However, some topics just need the practice of concept, and Heron’s formula is one of them.

The simple formula √s(s-a)(s-b)(s-c) is easily recognisable. Questions based on Heron’s formula can be mastered through the practice of multiple questions of different types.

Heron’s formula proves itself very useful in calculating the area of a triangle; it ignores the process of considering angles and sides related to it. The area of a triangle can be easily calculated with the help of Heron’s formula when the measurement of all the sides is given.

HussaiN is a full-time professional blogger from India. He is passionate about content writing, Tech enthusiast & computer technologies. Apart from content writing on the internet, he likes reading various tech magazines and several other blogs on the internet.

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