Question

NCERT solution of math 9th class chapter 12​

Answer 1

Answer:

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1.

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

Ex 12.1 Class 9 Maths Question 1.

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula.If its perimeter is 180 cm, what will be the area of the signal board?

Solution:

Let each side of the equilateral triangle be a.

Semi-perimeter of the triangle,



Ex 12.1 Class 9 Maths Question 2.

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?



Solution:

Let the sides of the triangular will be

a = 122m, b = 12cm, c = 22m

Semi-perimeter, s = a+b+c2

(122+120+224)m = 2642 m = 132m

The area of the triangular side wall



Rent for 1 year (i.e. 12 months) per m2 = Rs. 5000

∴ Rent for 3 months per m2 = Rs. 5000 x 312

= Rent for 3 months for 1320 m2 = Rs. 5000 x 312 x 1320 = Rs. 16,50,000.

Ex 12.1 Class 9 Maths Question 3.

There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6m, find the area painted in colour.



Solution:

Let the sides of the wall be

a = 15m, b = 11m, c = 6m

Semi-perimeter,



Thus, the required area painted in colour

= 20√2 m2

Ex 12.1 Class 9 Maths Question 4.

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

Solution:

Let the sides of the triangle be a =18 cm, b = 10 cm and c = x cm

Since, perimeter of the triangle = 42 cm

∴ 18cm + 10 cm + xcm = 42

x = [42 – (18 + 10)cm = 14cm

Now, semi-permimeter, s = 422cm = 21 cm



Thus, the required area of the triangle = 2111−−√ cm2