Download here the NCERT Solutions for Class 9 Maths Chapter 10 Heron’s Formula in English and Hindi Medium for CBSE Exams 2023-24. Online platforms Tiwari Academy offers features for students to ask questions and seek clarification on specific math problems or concepts using Discussion forum. All the questions of class 9th Maths chapter 10 are updated according to rationalised syllabus and new NCERT book for CBSE 2023-24.

**Class 9 Maths Chapter 10 for CBSE Board**

Class 9 Maths Exercise 10.1 in English

**Class 9 Maths Chapter 10 for State Boards**

Class 9 Maths Chapter 10 Exercise 10.1

Class 9 Maths Chapter 10 Exercise 10.2

Class 9 Maths Chapter 10 Exercise 10.3

Class 9 Maths Chapter 10 Exercise 10.4

Class 9 Maths Chapter 10 Exercise 10.5

Class 9 Maths Chapter 10 Exercise 10.6

This can be particularly helpful for clearing doubts and gaining a deeper understanding. We provide supplementary resources such as video lessons, quizzes, and study tips, which can enhance the overall learning experience and cater to different learning preferences.

**Class 9 Maths Chapter 10 in Hindi Medium**

Class 9 Maths Exercise 10.1 in Hindi

Class: 9 | Mathematics |

Chapter 10: | Heron’s Formula |

Number of Exercises: | 1 (One) |

Content Type: | Images, Text, PDF and Videos |

Educational Session: | CBSE 2023-24 |

Medium: | Hindi and English Medium |

### Class 9 Maths Study Material for 2023-24

### NCERT Solutions for Class 9 Maths Chapter 10

All the solutions for class 9 Mathematics are updated for new academic session 2023-24 to download in PDF format free of cost. Download Prashnavali 10.1 in Hindi Medium also new academic session 2023-24. UP Board students are now officially using NCERT Textbooks. So, download UP Board Solutions for Class 9 Maths Chapter 10 all exercise here in English and Hindi Medium. All NCERT Solutions for standard 9 (High School) for 2023-24 are available Online as well as Offline mode of contents. Class 9 Maths Chapter 10 Solutions in Videos are also available. Online and Offline Apps of Tiwari Academy for 2023-24 in Hindi Medium and English Medium are also available on Play Store as well as App Store.

### Class 9 Maths Chapter 10 Practice Questions with Solution

### What is Heron’s Formula?

Heron’s Formula: The formula given by Heron about the area of a triangle is known as Heron’s formula.

According to this formula, area of triangle = √[s (s – a)(s – b)(s – c)], where a, b and c are three sides of the triangle and s is the semi-perimeter.

This formula is also used for finding the area of quadrilateral. In quadrilateral, we join one diagonal to divide the quadrilateral into two triangles and then find the area of each triangle separately by Heron’s formula.

### What is the formula for Area of an equilateral triangle?

Area of equilateral triangle: Let the side of an equilateral triangle be k. Then, area of an equilateral triangle = (√3/4) k². Square units and altitude = (√3/2) k units.

### 9th Maths Chapter 10 Solutions in English & Hindi Medium

Get here the updated NCERT Solutions for Class 9 Math Chapter 10 Heron’s Formula Exercise 10.1 in Hindi and English Language format. Question Answers are given given below and these are updated for new academic year 2023-24. UP Board Students can also use these solutions for there board exams. Class 9 Maths NCERT Solutions 2023-24 are applicable for CBSE, UP Board, MP Board, Gujrat Board, etc., whoever is following CBSE Syllabus 2023-24.

### Important Questions on 9th Maths Chapter 10

### 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”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Here, the sides of triangle are a = 15 m, b = 11 m and c = 6 m.

So, the semi-perimeter of triangle is given by s = (a + b + c)/2 = (15 + 11 + 6)/2 = 32/2 =16 m

Therefore, using Heron’s formula, the area of triangle

= √(s(s – a)(s – b)(s – c) )

= √(16(16 – 15)(16 – 11)(16 – 6) )

= √(16(1)(5)(10) )

= √(4 × 4 ×(1)(5)(5 × 2) )

= 4 × 5 √2 = 20 √2 m²

Hence, the area painted in colour is 20√2 m².

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

Here, the sides of triangle are a = 18 cm, b = 10 cm and perimeter is 42 cm.

We know that the perimeter of triangle = a + b + c

⇒ 42 = 18 + 10 + c ⇒ c = 14 cm

So, the semi-perimeter of triangle is given by

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

Therefore, using Heron’s formula, the area of triangle

= √(s(s – a)(s – b)(s – c) )

= √(21(21 – 18)(21 – 10)(21 – 14) )

= √(21(3)(11)(7) )

= √(7 × 3 × (3)(11)(7) )

= 7 × 3√11 = 21√11 cm²

Hence, the area of triangle is 21√11 cm².

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

Perimeter of triangle = 540 cm

The ratio of sides of triangle = 12:17:25

Let, one of the sides of triangle a = 12x

Therefore, remaining two sides are b = 17x and c = 25x.

We know that the perimeter of triangle = a + b + c

⇒ 540 = 12x + 17x + 25x

⇒ 540 = 54x

⇒ x = 540/54 = 10

So, the sides of triangle are

a = 12 × 10 =120 cm,

b = 17 × 10 = 170 cm and

c = 25 × 10 =250 cm.

So, the semi-perimeter of triangle is given by

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

Therefore, using Heron’s formula, the area of triangle

= √(s(s – a)(s – b)(s – c) )

= √(270(270 – 120)(270 – 170)(270 – 250) )

= √(270(150)(100)(20) )

= √81000000

= 9000 cm²

Hence, the area of triangle is 9000 cm².

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

Perimeter of triangle = 30 cm

Two sides of triangle b = 12 cm and c = 12 cm.

Let, the third side = a cm

We know that the perimeter of triangle

= a + b + c ⇒ 30 = a + 12 + 12

⇒ 30 – 24 = a ⇒ a = 6

So, the semi-perimeter of triangle is given by

s = (a + b + c)/2 = 30/2 = 15 cm

Therefore, using Heron’s formula, the area of triangle

= √(s(s – a)(s – b)(s – c) )

= √(15(15 – 6)(15 – 12)(15 – 12) )

= √(15(9)(3)(3) )

= 9√15 cm²

Hence, the area of triangle is 9√15 cm².

#### IMPORTANT FORMULAE AND TERMS ON HERON’S FORMULA

Area of Triangle: The total space inside the boundary of the triangle is known as area of the triangle.

Area of triangle = ½ × base × height

Area of an isosceles triangle: Let B be the base and S be the equal sides of an isosceles triangle, then area of an isosceles triangle = [B√(4S² – B²)]/2 square units.

Perimeter: Perimeter of a triangle is equal to the sum of its three sides. It is denoted by 2s, where s is the semi-perimeter of a triangle.

### How many problems are there in chapter 10 of class 9th Maths?

In chapter 10 of class 9th Maths, there is only one exercise. This exercise (Ex 10.1) has 6 questions and three examples (examples 1, 2, 3).

### What is Heron’s Formula in chapter 10 of grade 9th Maths?

Heron’s Formula is used to find the area of a triangle. This formula is helpful where it is not possible to find the height of the triangle easily.

### Which questions of chapter 10 of grade 9th Maths are compulsory to do for the exams?

Students should do all questions and examples of chapter 10 of class 9th Maths. But examples 2, 3 and questions 2, 3, 5 of exercise 10.1 are compulsory to do for the exams because these are very important questions from an exam point of view. In exams, these questions can come in three marks.

### Is chapter 10 Heron’s Formula of grade 9th Maths easy or complicated to solve?

Chapter 10 of class 9th Maths is not easy and not complicated. It lies in the mid of easy and complicated because some examples and questions of this chapter are easy, and some are complicated. However, the difficulty level of any topic varies from student to student. So, Chapter 10 of class 9th Maths is easy or complicated depends on students also. Some students find it challenging, some find it easy, and some find it in the middle of easy and difficult.

### How much time is needed to complete chapter 10 of class 9th Maths?

Students need a maximum of 1 week to complete chapter 10 of class 9th Maths if they give a minimum of 1-2 hours per day to this chapter. This time also depends on student’s working speed, efficiency, capability, and many other factors.