Notes
Assignments
Ncert Solutions
Revision sheet
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
1 Meter

10 Decimeter

100 centimeter

1 Decimeter

10 centimeter

100 millimeter

1 Km

10 Hectometer

100 Decameter

1 Decameter

10 meter

1000 centimeter

Area
 The area of the plane figure is the surface enclosed by its boundary
 It unit is square of length unit. i.e. m^{2} , km^{2}
1 square Meter

100 square Decimeter

10000 square centimeter

1 square Decimeter

100 square centimeter

10000 square millimeter

1 Hectare

100 squareDecameter

10000 square meter

1 square myraimeter

100 square kilometer

10^{8} square meter

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=(1/2)BH

2

Isosceles right angled triangle
Equal side =a

Height=a

A=(1/2)a^{1/2}

3

Any triangle of sides a,b ,c

P=a+b+c

A=[s(sa)(sb)(sc)]^{1/2}
s=(a+b+c)/2
This is called Heron's formula (sometimes called Hero's formula) is named after Hero of Alexandria

4

Square
Side =a



A=a^{2}

5

Rectangle of Length and breath L and B respectively

P=2L +2B

A=LX B

6

Parallelograms
Two sides are given as a and b

P=2a+2b

A= BaseX height
When the diagonal is also given ,say d
Then
A=[s(sa)(sb)(sd)]^{1/2}
s=(a+b+d)/2

7

Rhombus
Diagonal d_{1} and d_{2} are given

p=2(d_{1}^{2}+d_{2}^{2})^{1/2}
Each side=(1/2)(d_{1}^{2}+d_{2}^{2})^{1/2 }

A=(1/2)d_{1}d_{2}

8

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=[s(sa)(sb)(sc)(sd)]^{1/2}
s=(a+b+c+d)/2
b)A=(1/2)d_{1}d_{2}
where d_{1} and d_{2 } are the diagonal
c)A=(1/2)d(h_{1}+h_{2})
where d is diagonal and h_{1} and h_{2} 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 quantites 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 pythogorus 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
1) A right angle traingle has base 20 cm and height as 10 cm, What is the area of the traingle?
Solution
Given values B=20 cm
H=10 cm
Both are in same units
A=(1/2)BH=100 cm
^{2 }
2) Sides of traingles are in the ratio 12:17:25. The perimeter of the traingle is 540 cm. Find out the area of the traingle?
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 =[s(sa)(sb)(sc)]
^{1/2}
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=9000cm
^{2}
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=[(3)
^{1/2 }a
^{2}]/4
Substituting the values given above
A=(3)
^{1/2 }
We can summarize various method to calculate the Area of the Triangle
If you know the altitude and Base

Area =(1/2)BH 
If you all the three sides 
A=[s(sa)(sb)(sc)]^{1/2}
s=(a+b+c)/2

If it is isoceles traingle with equal side a 
A=(1/2)a^{1/2} 
If it is equilateral triangel with equal side a 
A=[(3)^{1/2 }a^{2}]/4 
If it is right angle triangle with Base B and Height H 
Area =(1/2)BH 


Go Back to Class 9 Maths Home page
Go Back to Class 9 Science Home page