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Class 9 Maths notes for Heron Formula




Table of Content

Mensuration

  • It is branch of mathematics which is concerned about the measurement of length ,area and Volume of plane and Solid figure

Perimeter

  • The perimeter of plane figure is defined as the length of the boundary
  • It units is same as that of length i.e. m ,cm,km
Class 9 Maths notes for Heron Formula
Note:
You have read from Left to right to get the conversion formula
1 m= 10 dm = 100 cm
1 dm= 10 cm = 100 mm
1 km= 10 hm = 100 decameters
1 decameter= 10 m = 1000 cm



Area

  • The area of the plane figure is the surface enclosed by its boundary
  • It unit is square of length unit. i.e. m2 , km2

Note:
You have read from Left to right to get the conversion formula
$1 m^2 = 100 dm^2 = 10^4 cm^2$
$ 1 dm^2 = 100 cm^2 = 10^4 mm^2$
$1 \ hectare = 100 \ square \ Decameter = 10^2 m^2$
$ 1 \ square \ myraimeter=100 km^2 = 10^8 m^2$

Perimeter and Area of Different Figure

N
Shape
Perimeter/height
Area
1
Right angle triangle
Base =b, Height =h
Hypotenuse=d
$P=b+h+d$
Height =h
$A=\frac {1}{2}BH$
2
Isosceles right angled triangle
Equal side =a
$P= a(2+ \sqrt 2)$
$A=\frac {1}{2} a^2$
3
Isosceles triangle
Equal side =a and base as b
$P= 2a +b $
$A=\frac {1}{2} a \sqrt {a^2 - \frac {b^2}{4}}$
4
Any Triangle
Any triangle of sides a,b ,c
$P=a+b+c$
$A=\sqrt {s(s-a)(s-b)(s-c)}$
$s=\frac {a+b+c}{2}$
This is called Heron's formula (sometimes called Hero's formula) is named after Hero of Alexandria
5
Square
Side =a
P=4a $A=a^2$
6
Rectangle of Length and breath L and B respectively
$P=2L +2B$
$A=L \times B$
7
Parallelograms
Two sides are given as a and b
$P=2a+2b$
$A= Base \times height$
When the diagonal is also given ,say d
Then
$A= \sqrt {s(s-a)(s-b)(s-d)}$
$s=\frac {a+b+d}{2}$
8
Rhombus
Diagonal d1 and d2 are given
$p=2 \sqrt {d_1^2+d_2^2}$
Each side=$ \frac {1}{2} \sqrt {d_1^2+d_2^2} $
$A=\frac {1}{2} d_1 d_2$
9
Quadrilateral
a. All the sides are given a,b,c ,d
b. Both the diagonal are perpendicular to each other
c. When a diagonal and perpendicular to diagonal are given
a. $P=a+b+c+d$
a.
$A=\sqrt {s(s-a)(s-b)(s-c)(s-d)}$
$s=\frac {(a+b+c+d)}{2}$
b. $A=\frac {1}{2} d_1 d_2$
where d1 and d2 are the diagonal
c. $A=\frac {1}{2} d(h_1+h_2)$
where d is diagonal and h1 and h2 are perpendicular to that

How to solve the Area and Perimeter problems

1. We must remember the formula for all the common figures as given above the table
2. Find out what all is given in the problem
3. Convert all the given quantities in the same unit
4. Sometimes Perimeter is given and some side is unknown,So you can calculate the sides using the Perimeter
5. If it is a complex figure ,break down into common know figures like square,rectangle,triangle
6. Sometimes we can find another side using Pythagoras theorem in the complex figure
7. If common figure, apply the formula given above and calculate the area.
8. If complex figure, calculate the area for each common figure in it and sum all the area at the end to calculate the total area of the figure

Solved Examples

Question 1.
A right angle triangle has base 20 cm and height as 10 cm, What is the area of the triangle?
Solution
Given values B=20 cm
H=10 cm
Both are in same units
$A=\frac {1}{2}BH=100 \ cm^2$

Question 2.
Sides of triangles are in the ratio 12:17:25. The perimeter of the triangle is 540 cm. Find out the area of the triangle?
Solution
Let the common ration between the sides be y,then sides are 12y,17y,25 y
Now we know the perimeter of the triangle is given by
P=a+b+c
540=12y+17y+25y
or y=10 cm
Now Area of triangle is given as $A=\sqrt {s(s-a)(s-b)(s-c)}$
Where s=(a+b+c)/2
Here s=270 cm
a=120 cm
b=170cm
c=250cm
Substituting all these values in the area equation,we get
A=9000 cm2

Question 3.
A equilateral triangle is having side 2 cm. What is the area of the triangle?
Solution
We know that Are of equilateral triangle is given by
$A= \frac {a^2 \sqrt 3}{4}$
Substituting the values given above
$A=\sqrt 3 \ cm^2$

We can summarize various method to calculate the Area of the Triangle

If you know the altitude and Base
$A =\frac {1}{2}BH$
If you all the three sides $A=\sqrt {s(s-a)(s-b)(s-c)}$
$s=\frac {a+b+c}{2}$
If it is isosceles right angle triangle with equal side a $A=\frac {1}{2}a^2$
If it is isosceles triangle with equal side a and other side as $A=\frac {1}{2} a \sqrt {a^2 - \frac {b^2}{4}}$
If it is equilateral triangle with equal side a $A=\frac {a^2 \sqrt 3}{4}$
If it is right angle triangle with Base B and Height H $A =\frac {1}{2}BH$


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Also Read



Reference Books for class 9 Math

Given below are the links of some of the reference books for class 9 Math.

  1. Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
  2. Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
  3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
  4. Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.



Class 9 Maths Class 9 Science





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