Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Long Question Answer: Heron’s Formula

Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

Q1: Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Ans:

Let a = 11 m, b = 60 m and c = 61 m :
Firstly, to calculate semi perimeter (s)
Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

Q2: Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Ans:

Here, perimeter of the rhombus is 400 m.
∴ Side of the rhombus = 400/4 = 100 m
Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.

Class 9 Maths Chapter 10 Question Answers - Heron’s Formula
Hence, area of land allotted to two sons for their crops is 4800 m2 each.

Q3: The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Ans:

Let the sides of the triangle be 3x, 4x and 5x
∴ The perimeter of the triangular field = 144 m
⇒ 3x + 4x + 5x = 144m
⇒ 12x = 144m
Class 9 Maths Chapter 10 Question Answers - Heron’s FormulaClass 9 Maths Chapter 10 Question Answers - Heron’s Formula

Q4: Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Ans:

Perimeter of given triangle = 180 cm
Two sides are 18 cm and 80 cm
∴ Third side = 180 – 18 – 80 = 82 cm
Class 9 Maths Chapter 10 Question Answers - Heron’s Formula
Hence, area of triangle is 720 cm2 and altitude of the triangle corresponding to the shortest side is 80 cm. 


Q5: Calculate the area of the shaded region.
Class 9 Maths Chapter 10 Question Answers - Heron’s FormulaAns:

Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

= 2 × 2 × 3 × 7 = 84 cm2
Area of shaded region = Area of ∆ABC – Area of ∆AOB
= 84 cm2 – 30 cm2 = 54 cm2

Q6: The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Ans:

The sides of the triangular park are 8 m, 10 m and 6 m.
Class 9 Maths Chapter 10 Question Answers - Heron’s Formula
Radius of the circle = 2/2 = 1 m
Area of the circle = πr2 = 3.14 × 1 × 1 = 3.14 m2
∴ Area to be used for growing roses = Area of the park – area of the circle
=> 24 – 3.14 = 20.86 m2

The document Class 9 Maths Chapter 10 Question Answers - Heron’s Formula is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
44 videos|412 docs|54 tests

Top Courses for Class 9

FAQs on Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

1. What is Heron’s Formula and how is it derived?
Ans.Heron’s Formula is a method for calculating the area of a triangle when the lengths of all three sides are known. It is derived from the semi-perimeter of the triangle, which is calculated as \(s = \frac{a + b + c}{2}\), where \(a\), \(b\), and \(c\) are the lengths of the sides. The area \(A\) can then be found using the formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\).
2. How can I use Heron’s Formula to find the area of a triangle with side lengths 7, 8, and 9?
Ans.To find the area using Heron’s Formula, first calculate the semi-perimeter: \(s = \frac{7 + 8 + 9}{2} = 12\). Then, apply the formula: \[ A = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720} \approx 26.83. \] So, the area of the triangle is approximately 26.83 square units.
3. Can Heron’s Formula be applied to any type of triangle?
Ans.Yes, Heron’s Formula can be applied to any triangle, regardless of whether it is acute, obtuse, or right-angled, as long as the lengths of all three sides are known. It is a versatile tool for calculating the area in various geometric situations.
4. What are the limitations of using Heron’s Formula?
Ans.The main limitation of Heron’s Formula is that it requires knowledge of all three side lengths of the triangle. If only angles or one side length is known, the formula cannot be used directly. Additionally, if the side lengths do not satisfy the triangle inequality theorem, the formula will not yield a valid area.
5. Is there a relationship between Heron's Formula and other area calculation methods for triangles?
Ans.Yes, Heron’s Formula is related to other methods for calculating the area of triangles, such as the base-height method. While the base-height method requires knowing the height corresponding to a chosen base, Heron’s Formula offers a way to find the area purely based on the side lengths, making it particularly useful when height is not easily obtainable.
44 videos|412 docs|54 tests
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

shortcuts and tricks

,

Exam

,

Semester Notes

,

pdf

,

mock tests for examination

,

Viva Questions

,

study material

,

Extra Questions

,

Objective type Questions

,

Sample Paper

,

past year papers

,

Previous Year Questions with Solutions

,

Free

,

practice quizzes

,

Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

,

Summary

,

video lectures

,

ppt

,

Important questions

,

MCQs

,

Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

,

Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

;