1. AREA OF RECTANGLE To find area of rectangle we will count the number of squares, total number of squares will be the area of rectangle which can also be obtained by multiply length and breadth. Area of rectangle=l x b

2. AREA OF PARALLELOGRAM To find the area of parallelogram we will cut the ||gm as shown in figure in two parts and when we place 1 st and 2 nd part as shown in figure it becoms a rectangle and area of rectangle is (l x b ) So Area of ||gm =base x height

3. AREA OF TRIANGLE To find out area of triangle we take two triangles of same size and when we place these two triangles as shown in figure it becomes a ||gm Area of || gm = base x height. Here we have taken two triangles of same size. Area of one triangle = ½ area of ||gm. Area of triangle = ½ base x height

4. AREA OF TRAPEZIUM To find out the area of trapezium we take two trapezium of same size and when we place these two trapezium as shown in figure we get a ||gm. If we name parallel sides as a and b and height h then Area of ||gm= (a+b)h AREA OF TRAPEZIUM=1/2(a+b)h

5. RHOMBUS Diagonals of rhombus bisect each other at rt angles. Area of rhombus is sum of area of triangle ABC and area of triangle ACD. If we add area of these two triangles we get area of rhombus Area of triangle ABC=1/2 ACxOB Area of triangle ACD=1/2 ACxOD Area of rhombus =1/2 ACxOB+1/2 ACxOD =1/2 AC(OB+OD) =1/2 ACxBD =1/2 d 1 x d 2 Where d 1, d 2 are diagonals Area of rhombus =1/2 d 1 x d 2 A B C D d1 d2 o

6. QUADRILATERAL To find area of quadrilateral we meet one diagonal and draw perpendiculars on this diagonal from other two points. Now we get two triangles and area of quadrilateral is sum of area of these two triangles. Area of quadrilateral =1/2 AC(d 1 +d 2 )

7. SUM OF ANGLES OF TRIANGLES We take a triangle and color the angles. If we cut these angles and put on a paper we get a straight line. So sum of angles of triangles=180 o

8. SUM OF ANGLES OF QUADRILATERAL We take a quadrilateral and color the angles.If we cut these angles as shown in figure and put on a paper we get a circle. SUM OF ANGLES OF QUADRILATERAL=360 O ab

9. CUBE Cube can be made by cutting and folding a paper as shown in figure. Lateral surface area of cube is the area of 4 squares on lateral sides except top and bottom,if side of cube is a then Lateral surface area of cube=4a 2 Total surface area is the sum of lateral surface area and two faces on top and bottom which are square. So Total surface area of cube=6a 2 Volume is cube is the no of cube of unit length in the cube. We see that in 2 inch cube 8 cubes of one inch comes and in 3 inch cube 27 cube come. So Volume of cube=a 3

10. Area of circle

11. Perimeter of Square Perimeter of Square is sum of four equal sides so Perimeter of Square=4a Similarly we can find sum of equilateral pentagon as sum of five equal sides and so on. Perimeter of equilateral Pentagon =5a a a a a a a a a a

12. a 2 - b 2 We take a square of side a and make b 2 on one side and make 3 Pieces as shown in figure. we separate these 3 pieces and put one piece b 2 one side and join remaining two pieces as shown in figure. The area of these two pieces is equal to (a + b)(a – b), so if we subtract b 2 from a 2.we get area (a + b ) (a - b ) Hence a 2 - b 2 =(a+ b) (a-b).

13. Area Of 5 pieces Area of 5 pieces is (a + b) 2 in middle we get area (a - b) 2 if we subtract (a-b) 2 From (a + b ) 2 we get the area 4ab Hence (a + b ) 2 – (a- b ) 2 =4ab a a a a b b b b (a- b ) 2 ab

14. (a+b) 2 =a 2 +b 2 +2ab To find (a+b +c) 2 the value of (a+b) 2 =a 2 +b 2 +2ab will remain same and 2ac and 2bc and c 2 will be added as shown in figure:- (a+b +c) 2 =a 2 +b 2 +2ab +c 2 +2ac+2bc To find (a+b +c+d) 2 the value of (a+ b +c) 2 i.e. (a+ b +c) 2 = a 2 +b 2 +2ab +c 2 +2ac+2bc will remain same and d 2,2ad,2bd,2cd will be added as shown in fig (3). (a+b +c+d) 2 = a 2 +b 2 +2ab +c 2 +2ac+2bc +d 2 +2ad+2bd+2cd

15. (a-b) 2 To find (a-b) 2 we take a square board or paper of side a we take one length equal to b on two sides of board or paper and make 3 pieces as shown in figure These 3 Pieces area is a 2 We separate these three pieces One piece is (a-b) 2 one is ab and one is b(a-b) If we subtract two pieces ab,b(a-b) from three pieces we get (a-b) 2 and so (a-b) 2 =a 2 –ab-b(a-b) =a 2 –ab-ab+b 2 (a-b) 2 =a 2 –2ab+b 2

16. Cuboid Lateral surface area of cuboid is the area of 4 sides except top and bottom Lateral surface area of cuboid=2lh+2bh Total surface area of cuboid is lateral surface area and area of top and bottom which is equal to 2lb Total surface area of cuboid=2lh+2bh+2lb Volume of cuboid is the no. of cubes of unit length that comes in cuboid. If length of cuboid 5 breadth 3 height 2 we see that 30 cuboid of unit length comes in the cuboid so volume is 30 and product of 5,3,2=30 so Volume of cuboid=lbh

17. Cylinder

19. Perimeter of quadrilateral Perimeter of quadrilateral=sum of all four sides Perimeter of quadrilateral=a+b+c+d

20. Perimeter Perimeter:- Perimeter of any shape is the sum of all outer sides. If we add all outer sides we get the perimeter. If sides are not given we can measure sum of outer sides with the help of thread by placing the thread on one end and revolving the thread upto that end.Perimeter of triangle is sum of all sides as shown in figer. Perimeter of equilateral triangle is sum of three equal sides so Perimeter of Equilateral Triangle=3a Perimeter of Triangle=a+b+c

21. Area of square To find area students will count number of squares of unit length as we see in figure first square is of length 1, second of length 2,third of length 3 and so on Area of square will be number of squares made in each square. Area of square = a 2 or side square

22. a 3 +b 3 To find the formula of a 3 +b 3 Let a=3, b=2,take a 3 (27 cubes) b 3 (8 cubes ) place these cubes as shown in figure(1). Now take upper 22 cubes and put them as shown in figure (2). In rectangle form whose one side is a+b (3+2=5) other side is a 2 +b 2 –ab(3 2 +2 2 -3x2=7 ) Area of rectangle is (a+b)(a 2 +b 2 –ab ) 3 3 +2 3 =(3+2)(3 2 +2 2 - 3x2=7) a 3 +b 3 =(a+b)( a 2 +b 2 –ab)

23. a 3 -b 3 To find the formula of a 3 -b 3 Let a=3, b=2,take a 3 (27 cubes) from these 27 cubes take b 3 (8 cubes) and place one side the remaining cubes will be a 3 -b 3 (27-8=19 ) These 19 cubes make a rectangle whose one side is a-b (3-2=1) other side is a 2 +b 2 +ab(3 2 +2 2 +3x2=19) Area of rectangle is (a-b)(a 2 +b 2 +ab ) 3 3 -2 3 =(3-2)(3 2 +2 2 +3x2=19) a 3 -b 3 =(a-b)( a 2 +b 2 +ab)

24. (a-b) 3 Let a=5,b=2 place a 3 (5 3 =125) cubes as shown in figure. Take upper b 3 i.e.(2 3 =8)cubes and put them one side. Take (a-b) 3 i.e.[(5-2) 3 =27 ]cubes and put them one side. The remaining cubes are (125-8- 27=90) These 90 cubes make a rectangle whose one side is 3ab(3x5x2=30) And other side is a-b(5-2=3) Whose one side is 3ab(a-b) Now, 125=8+27+30x3 a 3 = b 3 +(a- b) 3 +3ab(a-b) So, (a-b) 3 =a 3 - b 3 -3ab(a-b )

25. (a+b) 3 Let a=3,b=2 place (a+b) 3 i.e (5 3 =125) cubes as shown in figure. Take upper a 3 i.e.(27cubes) and put them one side. Take b 3 (8cubes) and put them one side. The remaining cubes will be (125- 8-27=90) These 90 cubes make a rectangle whose one side is 3ab(3x3x2=18) And other side is (a+b)i.e(3+2=5) Whose area is 3ab(a+b) Now,5 3 = 3 3 + 2 3 +3x3x2(3+2) So,(a+b) 3 =a 3 + b 3 +3ab(a+b )