1 Meter 
10 Decimeter 
100 centimeter 
1 Decimeter 
10 centimeter 
100 millimeter 
1 Km 
10 Hectometer 
100 Decameter 
1 Decameter 
10 meter 
1000 centimeter 
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 
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 
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
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 }
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 