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Important Worked Out Problems On Heron’s Formula 

Mathematics can be a tricky subject for many students who are preparing for their board exams. At the same time, it is a high scoring subject, so if you get it right, you have a high chance of scoring 100% in your exams. 

Rather than mugging up formulas, mathematics comes easily and effortlessly to students if they practice lots of sums on a given topic. By practising regularly, students are not just able to remember the formulas better, but they also get exposed to many different types of questions that can come in exams. This gives them the confidence to tackle even the most challenging problem in a specific area smoothly.

Heron’s formula is one such topic that holds an important place in the world of mathematics. Heron’s formula is used primarily to determine the area of a triangle where the lengths of all its sides are known. The formula applies to various types of triangles like isosceles (two of its sides are equal in length and two of its angles are the same), equilateral (all three sides are equal in length), or scalene (none of the sides of the triangle is equal in length).

Importance of Heron’s Formula in Real Life

In real life, one comes across many objects shaped like a triangle, such as lands (lands are not always regular in shape like a square or rectangular). The best part about using Heron’s formula for finding the surface area of triangles is that it does not require any angle or distance. In real life, one can not draw a perpendicular from the vertex to the base of a triangle; hence the best way in such cases is to measure the sides of the triangle and then use Heron’s formula to find the area of the triangle.

What is Heron’s Formula

Heron’s formula (also referred to as Heron’s formula) was given by Heron, a famous engineer in Egypt, in 10 AD. Heron is also popularly called the “Hero of Alexandria” due to many of his works in mathematics, vending machines, etc. Heron gave this formula for measuring the area of a triangle in 60 CE.

As per Heron’s formula, if the lengths of the sides of a triangle XYZ is x, y, and z, then the are of the triangle can be given by the below formula:

Area of triangle XYZ = s (s – x) (s – y ) ( s – z), where s is the semi or half perimeter of the triangle. 

The perimeter of a triangle is the sum of all its sides, so s = (x + y + z)/2. 

Heron’s formula allows us to calculate the area of a triangle without knowing the length of its altitudes. Heron’s formula is also used to calculate the area of quadrilaterals.

Important Worked Out Questions on Heron’s Formula

The steps of calculating the area of a triangle based on Heron’s formula are:

  1. Find the length of the sides (x, y, z) from the given information.
  2. Find the half perimeter which is given by s = (x + y + z)/2
  3. Calculate the area with the formula: 

Area of triangle = s (s – x) (s – y ) ( s – z)

Let us now look into a few different ways to get a problem on Heron’s formula in the exam and how to solve these different kinds of sums.

  1. If the perimeter of a triangle is 600 cm and its sides are in the ratio of 12:15:13, find this triangle’s area.

Solution – Let us say the standard ratio between the sides is “x”, so we can say that the triangle sides are 12x, 20x, and 8x.

Now we know that the perimeter of the triangle is 600 cms, so we can write the equation as:

12x + 15x + 13x = 600

40x = 600

x = 600/40 = 15

With x as 15, the sides of the triangle are:

12 * 15 = 180 cms

15 * 15 = 225 cms

13 * 15 = 195 cms

Semiperimeter = perimeter/2 = 600/2 = 300 cms.

Now let us fit all these values ito Heron’s formula to find the area of the triangle:

Area of the triangle = s (s – x) (s – y ) ( s – z) = 300 (300 – 180) (300 – 225 ) ( 300 -195z)

= 300 * 120 * 75 * 105

=28,35,00,000 = 16837.46 cms2

  1. The lengths of two sides of a triangle PQR are 12 cm and 25 cm, and its perimeter is 54 cms. Find the area of the triangle.

Solution – Since the perimeter of the triangle is given, we can find the length of the third side (let us say r) as following:

12 + 25 + r = 54 cms

37 + r = 54 cms

r = 17 cms

Semiperimeter of the triangle = 54/2 = 27 cms

Putting them in Heron’s formula we get:

Area of the triangle = s (s – p) (s – q ) ( s – r) 

        = 27 (27 – 12) (27 – 25 ) ( 27 – 17) 

        = 27 * 15 * 2 * 10

        = 8100 

        = 90 sq. cm.

  1. The lengths of the equal sides of an isosceles triangle are 5 cm, and its base is 6 cm. Use Heron’s formula to obtain the area of the triangle.

Solution – We know that an isosceles triangle is the one where two of its sides are equal. We are given the length of one of the sides as 5 cm, so the length of the other side is also 5 cm.

Perimeter of the triangle = 5 + 5 + 6 = 16 cms

Semiperimeter = 16/2 = 8 cms

As per Heron’s formula, area of the triangle = s (s – x) (s – y ) ( s – z) 

 = 8 (8 – 5) (8 – 5 ) (8 – 6) = 8 * 3 * 3 * 2 = 144) = 12 sq. cm.

Conclusion

While preparing for board exams, mathematics is one of the essential topics for students to secure a high rank. It is very rightly said that “Practice makes it perfect”.” Practising Heron’s formula from reputed resources will give students an edge over others in their exams and many prestigious competitive exams. So it is wise to start preparing now on this topic so that you have mastered all the essential aspects of Heron’s formula by the time you sit for boards.

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