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Heron's Formula Class 9th Notes - Free NCERT Class 9 Maths Chapter 12 Notes - Download PDF

Heron's Formula Class 9th Notes - Free NCERT Class 9 Maths Chapter 12 Notes - Download PDF

Edited By Ramraj Saini | Updated on Apr 21, 2022 01:59 PM IST

Heron’s Formula class 9 notes:

The Heron’s Formula is the NCERT chapter in which we deal with the area of triangles Egypt's Heron devised a formula for calculating the area of any triangle. Heron's Formula or Hero's Formula is the name given to this formula. The NCERT Class 9 Maths Chapter 12 Notes covers a brief outline of the chapter Heron’s Formula. The main topics covered are What is Triangle, Area of an equilateral triangle, Area of the isosceles triangle, Area of a triangle by heron’s formula in the Heron’s Formula class 9 notes with some FAQs.

The basic equations in the chapter are also covered in the class 9 mathematics chapter 12 notes. All of these concepts are covered in class 9 Heron’s Formula notes. CBSE class 9 maths chapter 12 notes contain important formulas. The class 9 maths chapter 12 notes contains systematic explanations of topics using examples and exercises.

Also, students can refer,

NCERT Class 9 Maths Chapter 12 Notes

Triangle
A triangle is a closed plane figure that has three sides and three angles.

Triangles are classified into three types

Based on their sides

  • Equilateral
  • Isosceles
  • Scalene

Based on angles

  • Acute angled triangle
  • Right-angled triangle
  • Obtuse angled triangle

Area of a triangle:

Area A = ( 1/2 ) ( base ) ( height )1648122606321

If the lengths of the triangles' sides are known, we may apply the Pythagoras theorem to find the height of a triangle is equilateral and isosceles triangles.

Area of an equilateral triangle:

A = ( √3 / 4) a2

Area of an isosceles triangle:

A = ( 1 / 4 ) b √ ( 4 a2 - b2 )

Area of a triangle By Heron’s formula

Heron's formula (also known as Hero's Formula) calculates the area of an ABC given the sides a, b, and c:

C:\Users\GOD IS GREAT\Pictures\heron-s-formula-1624326535.png

For semi perimeter (s) = (a + b + c)/2

Area A = √ [ s ( s - a ) ( s - b ) ( s - c ) ]

This formula can be used to calculate the area of a scalene triangle when the lengths of all of its sides are known.

By Heron's formula, the area of any polygon is calculated.
When one of the diagonal values and the sides of a quadrilateral are known, the area can be computed by breaking the quadrilateral into two triangles and applying Heron's formula.

Significance of NCERT class 9 maths chapter 12 notes

The notes from Heron's Formula class 9th will help you review the chapter and get a sense of the important points presented.

This NCERT class 9 maths chapter 12 will help students to understand the formulas, statements, and rules in detail.

Class 9 mathematics chapter 12 notes pdf download can be used to study when offline.

NCERT solutions of class 9 subject wise

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Notes Chapter wise


Frequently Asked Question (FAQs)

1. Find the area of the Triangle where all sides are equal to a 4 cm as per class 9th maths chapter 12 notes.

As all sides are equal then it must be an equilateral triangle.

So area of an equilateral triangle:

A = ( √3 / 4) a2

; ( √3 / 4) (4)2 = 6.928 cm2

2. Find the semiperimeter of triangle having all side as 1 cm

For semi perimeter (s) = (a + b + c)/2 ;

So, =(1+1+1)/3

=3/3

=1

3. Write down the Herons Formula?

For semi perimeter (s) = (a + b + c)/2 ;

So, =(1+1+1)/3

=3/3

=1

4. Is Heron’s formula applicable in all triangles ?

According to notes for class 9 maths chapter 12,   heron’s formula is applicable to all triangles regardless of side lengths.

5. How does the area of a quadrilateral is calculated using Heron’s formula?

Heron’s formula, the area of a quadrilateral is calculated by dividing it into two triangles.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

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K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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