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NCERT Class 9 Maths Textbook Chapter 12 With Answer Book PDF Free Download
Chapter 12: Heron’s Formula
12.1 Introduction
You have studied in earlier classes about figures of different shapes such as squares, rectangles, triangles, and quadrilaterals.
You have also calculated perimeters and the areas of some of these figures like rectangles, squares, etc.
For instance, you can find the area and the perimeter of the floor of your classroom. Let us take a walk around the floor along its sides once; the distance we walk is its perimeter. The size of the floor of the room is its area.
12.2 Area of a Triangle — by Heron’s Formula
Heron was born in about 10AD possibly in Alexandria in Egypt. He worked in applied mathematics. His works on mathematical and physical subjects are so numerous and varied that he is considered to be an encyclopedic writer in these fields.
His geometrical works deal largely with problems on mensuration written in three books.
Book I deals with the area of squares, rectangles, triangles, trapezoids (trapezia), various other specialized quadrilaterals, the regular polygons, circles, surfaces of cylinders, cones, spheres, etc.
In this book, Heron has derived the famous formula for the area of a triangle in terms of its three sides
| Author | NCERT |
| Language | English |
| No. of Pages | 11 |
| PDF Size | 908 KB |
| Category | Mathematics |
| Source/ Credits | ncert.nic.in/ |
NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula
1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with a 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:
Given,
Side of the signal board = a
Perimeter of the signal board = 3a = 180 cm
∴ a = 60 cm
Semi perimeter of the signal board (s) = 3a/2
By using Heron’s formula,
Area of the triangular signal board will be =
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 Fig. 12.9). 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:
The sides of the triangle ABC are 122 m, 22 m, and 120 m respectively.
Now, the perimeter will be (122+22+120) = 264 m
Also, the semi perimeter (s) = 264/2 = 132 m
Using Heron’s formula,
Area of the triangle =
=1320 m2
We know that the rent of advertising per year = 5000 per m2
∴ The rent of one wall for 3 months = Rs. (1320×5000×3)/12 = Rs. 1650000
3. There is a slide in a park. One of its side walls has been painted in some color with the message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m, and 6 m, find the area painted in color.
Solution:
It is given that the sides of the wall as 15 m, 11 m, and 6 m.
So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m
Using Heron’s formula,
Area of the message =
= √[16(16-15)(16-11) (16-6)] m2
= √[16×1×5×10] m2 = √800 m2
= 20√2 m2
NCERT Class 9 Maths Textbook Chapter 12 With Answer Book PDF Free Download