NCERT Solutions Class 9 Maths Chapter 10 Herons Formula Exercise 10.1

NCERT Solutions Class 9 Maths Chapter 10 Exercise 10.1, Heron's Formula, shows students how to figure out a triangle's area when they know how long the sides are. Heron’s Formula is one of the most useful methods to calculate the area of a triangle when you know the lengths of all three sides. Instead of needing the height or altitude, this formula allows you to find the area quickly and accurately using just the side measures. NCERT Solutions Class 9 Maths Chapter 10 Heron's Formula Exercise 10.1 builds on earlier geometry ideas and helps students get better at solving problems they might face in real life.

In Exercise 10.1, you’ll practice solving problems where you calculate the area of triangles using Heron’s Formula step by step. These questions help you build confidence, understand the concept thoroughly, and prepare effectively for exams

1.0Download Class 9 Maths Chapter 10 Ex 10.1 NCERT Solutions PDF

You can find and get the full well-organized NCERT Solutions Class 9 Maths Chapter 10 Herons Formula Exercise 10.1 as a PDF. This helps you learn and review more . Experts have created these solutions to match what CBSE wants students to learn.

Key Concepts of Exercise 10.1

• Introduction to Heron's Formula and Key Points of Exercise 10.1
• Learning the semi-perimeter of a triangle
• Calculating the area in terms of side lengths rather than height
• Heron's Formula for different kinds of triangles
• Further application problems based on the area of a triangle

2.0NCERT Class 9 Maths Chapter 10 Exercise 10.1 : Detailed Solutions

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?

Sol. The equilateral triangle has each side = a

Its semi-perimeter = (a+a+a)/2 = 3a/2

By Heron's formula, the area of the triangle = √[ (3a/2) × (3a/2 − a) × (3a/2 − a) × (3a/2 − a) ]

= √[ (3a/2) × (a/2) × (a/2) × (a/2) ]

= √[ (3a⁴)/16 ]

= (√3 / 4)a²

When the perimeter of the triangle is 180 cm, we have 3a = 180 cm, i.e., a = 60 cm.

Then the area of the triangle = (√3 / 4) × (60)² cm²

= (√3 / 4) × 3600 cm²

= 900√3 cm²

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.). The advertisements yield an earning of Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Sol. Sides of the two equal triangular walls below the bridge are 122 m, 22 m and 120 m.

s = (122 m + 22 m + 120 m)/2 = 264 m / 2 = 132 m

Area of one triangular wall = √[132 × (132−122) × (132−22) × (132−120)] m²

= √[132 × 10 × 110 × 12] m²

= √[1742400] m²

= 1320 m²

Company hired only one wall for 3 months.

Thus, earning from advertisements for 3 months at the rate of Rs. 5000 per m² per year = Rs. 5000 × (3/12) × 1320

= Rs. 5000 × (1/4) × 1320

= Rs. 1250 × 1320

= Rs. 16,50,000

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

Sol. The sides of the triangular wall be 15 m, 11 m and 6 m.

s = (15 m + 11 m + 6 m)/2 = 32 m / 2 = 16 m

Area of the wall = √[16 × (16−15) × (16−11) × (16−6)] m²

= √[16 × 1 × 5 × 10] m²

= √[800] m²

= 20√2 m²

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

Sol. a=18 cm, b=10 cm

Perimeter = 42 cm

We have a+b+c = 42

18 + 10 + c = 42

28 + c = 42

c = 14 cm

s = (a+b+c)/2 = 42/2 = 21 cm

Area of Δ = √[s(s−a)(s−b)(s−c)]

= √[21(21−18)(21−10)(21−14)]

= √[21 × 3 × 11 × 7]

= √[3 × 7 × 3 × 11 × 7]

= √[3² × 7² × 11]

= 3 × 7 × √11

= 21√11 cm²

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Sol. Let the sides of triangle be 12k, 17k, 25k.

Perimeter = 12k + 17k + 25k = 54k.

Given perimeter = 540 cm.

54k = 540

k = 10

So, a = 12 × 10 = 120 cm

b = 17 × 10 = 170 cm

c = 25 × 10 = 250 cm

s = (a+b+c)/2 = 540/2 = 270 cm

Area = √[s(s−a)(s−b)(s−c)]

= √[270 × (270−120) × (270−170) × (270−250)]

= √[270 × 150 × 100 × 20]

= √[27 × 10 × 15 × 10 × 10 × 10 × 2 × 10]

= √[3³ × 10 × 3 × 5 × 10 × 10 × 10 × 2 × 10]

= √[3⁴ × 2² × 5² × 10⁴]

= 3² × 2 × 5 × 10²

= 9 × 10 × 100

= 9000 cm²

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Sol.

Let the equal sides be a = 12 cm and b = 12 cm.

Perimeter = 30 cm

a + b + c = 30

12 + 12 + c = 30

24 + c = 30

c = 6 cm

s = 30/2 = 15 cm

Area = √[s(s−a)(s−b)(s−c)]

= √[15 × (15−12) × (15−12) × (15−6)]

= √[15 × 3 × 3 × 9]

= √[3 × 5 × 3 × 3 × 3²]

= √[3⁴ × 5]

= 3²√15

= 9√15 cm²

3.0Key Features and Benefits: Class 9 Maths Chapter 10 Exercise 10.1

• Comprehensive Step-by-Step Answers: Learn every step with understanding not by rote learning. 
• Improves Application Skills: Learn how to apply formulas in real life without measuring heights.
• Develops Geometric Reasoning: Develops knowledge of triangle properties through computation based learning.
• Exams-Focused Approach: All solutions are in harmony with CBSEpatterns so students can safely make choices in questions with higher accuracy and exam confidence. 
• Enables Self Learning: Well structured information helps students revise through constant reminders and charts their progress.