RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions

RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions will help students to learn about the applications of Heron’s Formula. Students will learn about the applications of the Heron’s Formula. For Example- If a landowner wants to find out the land area, which is in a quadrilateral shape. It divides the quadrilateral into triangular parts and uses Heron’s formula for the area of a triangular section. From this chapter, learn about how to apply Heron’s formula.

Although, with the RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions PDF, students can do practice with various questions. The PDF is prepared by ou experts through RD Sharma, Previous Year’s Question Paper, and CBSE Text Book of Class 9. Go through the article and know more information about Heron’s Formula with stepwise definitions and examples.

Learn about RD Sharma Chapter 12 (Heron’s Formula)

Download RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions PDF

Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2

Important Definitions RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions

Heron’s Theorem

Heron’s Theorem was named after Heron of Alexandria, who has discovered the area of triangles can be measured in terms of the length of sides of a triangle. By obtaining Heron’s formula, we can determine the area of a triangle without measuring the angles or another distance. This formula is known for its easy calculation based on the length of three sides of a triangle.

Let’s say, if the length of the three sides are P, Q, and R, then its semi-perimeter is-

S= (P + Q + R)/ 2

Thus, the area will be-

Area = √[S (S – P) (S – Q) (S – R)]

Examples

Ques: A triangle XYZ has sides 4cm, 13cm, and 15cm. Find the area of the triangle.

Solution- Semiperimeter of triangle XYZ, s = (4+ 13+ 15)/ 2 = 32/ 2 = 16

By Heron’s Formula, we know-

Area (A) = √[s (s – a) (s – b) (s – c)]

Hence, A = √[16 (16 – 4) (16 – 13) (16 – 15)] = √(16 x 12 x 3 x 1) = √576 = 24 square cm

In the following points, we will learn how to find the area of quadrilateral using Heron’s formula’-

If PQRS is a quadrilateral, where PQ||RS and PR & QS are the diagonals. (ABCD)

PR divides the quadrilateral PQRS into two triangles PSR and PQR.

Now we have two triangles here.

Areas of quadrilateral PQRS = Area of ∆PSR + Area of ∆PQR

If we know the lengths of all sides of the quadrilateral and the length of diagonal PR, we can apply Heron’s formula to get the total area.

Hence, we will first find the area of ∆PSR and area of ∆PQR applying Heron’s formula and will add them to obtain the final value.

Heron’s Formula for Equilateral Triangle

We know that the equilateral triangle has all equal sides. To find the area of the equilateral triangle, first, we have to find the semi perimeter of the equilateral triangle-

s = (a+ a+ a)/ 2

s= 3a/ 2

where ‘a’ is the length of the side.

Now, apply Heron’s formula, we know-

Area (A) = √s(s − a) (s − b) (s − c)

As, a = b = c

Therefore,

Area (A) = √[s(s-a)3] (required formula).

Frequently Asked Questions (FAQs) of RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions

Ques 1- When we apply Heron’s formula?

Ans- Heron’s formula is applied to find the area of a triangle when all the three side-lengths of a triangle are known to us.

Ques 2- What is Heron’s formula for the equilateral triangle?

Ans- An equilateral triangle has all three sides similar, so Heron’s formula to obtain the area of the equilateral triangle is-

Area (A) = √[s(s-a)3]

Ques 3- Is Heron’s Formula accurate?

Ans- Heron’s formula calculates the area of a triangle if the length of each side is given. If it is a very thin triangle, one where two sides are approx equal and the third side is much smaller, a direct implementation of Heron’s formula may not be accurate.