# RD‌ ‌Sharma‌ ‌Chapter‌ ‌12 ‌Class‌ ‌9‌ ‌Maths‌ ‌Exercise‌ ‌12.1 ‌Solutions‌

RD‌ ‌Sharma‌ ‌Chapter‌ ‌12 ‌Class‌ ‌9‌ ‌Maths‌ ‌Exercise‌ ‌12.1 ‌Solutions‌ is concerned with Heron’s formula. The students will study “how to find the area of a triangle using heron’s formula?” with the help of several solved problems. The experts have used the step-by-step problem-solving methods of the examples mentioned in this article.

Students can freely download RD‌ ‌Sharma‌ ‌Chapter‌ ‌12 ‌Class‌ ‌9‌ ‌Maths‌ ‌Exercise‌ ‌12.1 ‌Solutions‌ PDF and grind their skills. The PDF consists of varieties of questions based on Heron’s Formula. The questions were explained stepwise, which is easily understandable to the students. By practicing through PDF, learners will get to know the several types of questions that can be formed. Practicing with these problems help students to score well in the exam.

## Download RD‌ ‌Sharma‌ ‌Chapter‌ ‌12 ‌Class‌ ‌9‌ ‌Maths‌ ‌Exercise‌ ‌12.1 ‌Solutions‌ PDF

Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1

## Important Definitions RD‌ ‌Sharma‌ ‌Chapter‌ ‌12 ‌Class‌ ‌9‌ ‌Maths‌ ‌Exercise‌ ‌12.1 ‌Solutions‌

We can calculate any triangle if we know the length of all three (3) sides of that triangle by using Heron’s Formula, which has been known for approximately 2000 years. It is known as “Heron’s Formula” after Hero of Alexandria.

### Heron’s Formula

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

Semi Perimeter= s=(a+b+c)2, where, a, b, and c are the sides of the triangle.

### Examples of Heron’s Formula RD‌ ‌Sharma‌ ‌Chapter‌ ‌12 ‌Class‌ ‌9‌ ‌Maths‌ ‌Exercise‌ ‌12.1 ‌Solutions

Ques- In a triangle PQR, PQ = 15cm, QR = 13cm, and PR = 14cm. Find the area of a triangle PQR and hence its altitude on PR.

Solution-

Let the sides of the given triangle be PQ = p, QR = q, PR = r, respectively.

Here,

= p = 15 cm

= q = 13 cm

= r = 14 cm

From Heron’s Formula;

Area of a triangle= √s (s-p) (s-q) (s-r)

Semi Perimeter= s=(p+ q+ r)2

Where, p, q, and r are sides of a triangle.

= s= (15 + 13 + 14)/ 2

= s= 21

= Area= √21 (21-15) (21-13) (21-14)

= √21 (6 x 7 x 8)

= √7056

= 84

= Area = 84 cm2

= Let, QT is a perpendicular on PR

Now, area of triangle = ½ x Base x Height

= ½ × QT × PR = 84

= QT = 12cm

= The altitude is 12 cm (Hence Proved).

## Frequently Asked Questions (FAQs) of RD‌ ‌Sharma‌ ‌Chapter‌ ‌12 ‌Class‌ ‌9‌ ‌Maths‌ ‌Exercise‌ ‌12.1 ‌Solutions‌

Ques- What is S in a Triangle of Hero’s Formula?

Ans- The other is Heron’s formula, which gives the area in terms of the three sides of a triangle, specifically, as the square root of the product s(s – a)(s – b)(s – c), where, ‘s’ is the semi perimeter of a triangle. So, s = (a + b + c)/2.

Ques- What is the semi perimeter of a triangle?

Ans- A semi perimeter of a triangle is equivalent to the perimeter of its medial triangle.

Ques- Who gave Heron’s formula?

Ans- Hero of Alexandria, a great mathematician who derived the formula to calculate the area of the triangle using the length of all three (3) sides.