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Activity / class 8 / chap 9 / Algebraic Expression
Topic-Algebraic Identities
Objective: To verify geometrically
A) a  b2  a 2  2ab  b 2
B) a  b2  a 2  2ab  b 2 .
Materials Required: Scale, pencil, colour pencils, set-squares, a pair of scissors, few
sheets of paper, glue.
Procedure A) :
1. Let a = 7 and b = 3.
2. We start with the RHS of the identity.
On a sheet of paper, cut out
i) a square of side 7 cm and color it pink.
ii) a square of side 3 cm and color it yellow.
iii) two rectangles with length 7 cm and breadth 3 cm. Color them blue.
3. Let us place them as shown.
Observation : Area of big square is equal to sum of areas of the two small squares and
areas of two rectangles.
Area of big square = area of sq. I + area of sq. II + area of rect. III + area of rect. IV
(a + b)2 = a2 + b2 + ab + ab = a2 + 2ab + b2
LHS = (a + b)2 = ( 7+3)2 = 102 = 100
RHS = a2 + 2ab + b2 = 72 + 2 x 7 x 3 + 32 = 49 + 42 + 9 = 100.
2
Conclusion: The identity a  b  a 2  2ab  b 2 is valid and has been verified
geometrically.
Copyright with the author.
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Procedure B) :
1. Let a = 7 and b = 3.
2. We start with the RHS of the identity. On a sheet of paper, cut out
i) a square of side 7 and color it pink.
ii) a square of side 3 and color it yellow.
iii) two rectangles with length 7 cm and breadth 3 cm and color them blue.
3. Paste the squares I and II as shown below.
4. Now paste the strips as shown.
Observation: The left over pink area is a square of side 4 cm.
Area of left over square = Area of sq. I + Area of sq. II – Area of sq. III – Area of sq. IV
Side of left over square is a  b  units.
Hence, (a – b)2 = a2 + b2 – ab – ab = a2 – 2ab + b2
LHS = (a – b)2 = ( 7 – 3)2 = 42 = 16
RHS = a2 – 2ab + b2 = 72 – 2 x 7 x 3 + 32 = 49 – 42 + 9 = 16
Conclusion: The identity a  b2  a 2  2ab  b 2 is valid and has been verified
geometrically.
Extended learning: Repeat the activities for some other measurements [say, a = 7 , b = 4
etc. ]
Copyright with the author.