RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1 (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1: You can easily prepare for your upcoming Class 9 Maths exams with the RD Sharma Solutions Class 9 Maths. The download link for the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1 will be given in this blog. You can download the PDF for free and access it offline as well. To know more, read the whole blog.

Download RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1 PDF

RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1

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Question 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.

Solution:

We know, Heron’s Formula

Here, a = 150 cm

b = 120 cm

c = 200 cm

Step 1: Find s

s = (a+b+c)/2

s = (150+200+120)/2

s = 235 cm

Question 2: Find the area of a triangle whose sides are respectively 9 cm, 12 cm and 15 cm.

Solution:

We know, Heron’s Formula

Here, a = 9 cm

b = 12 cm

c = 15 cm

Step 1: Find s

s = (a+b+c)/2

s = (9 + 12 + 15)/2

s = 18 cm

Question 3: Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Solution:

Given:

a = 18 cm, b = 10 cm, and perimeter = 42 cm

Let c be the third side of the triangle.

Step 1: Find third side of the triangle, that is c

We know, perimeter = 2s,

2s = 42

s = 21

Again, s = (a+b+c)/2

Put the value of s, we get

21 = (18+10+c)/2

42 = 28 + c

c = 14 cm

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