SQUARES AND SQUARE ROOTS A REVIEW. CONTENTS SQUARES. PERFECT SQUARES. FACTS ABOUT SQUARES. SOME METHODS TO FINDING SQUARES. SOME IMPORTANT PATTERNS.

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Presentation transcript:

SQUARES AND SQUARE ROOTS A REVIEW

CONTENTS SQUARES. PERFECT SQUARES. FACTS ABOUT SQUARES. SOME METHODS TO FINDING SQUARES. SOME IMPORTANT PATTERNS. PYTHAGOREAN TRIPLET.

SQUARES If a whole number is multiplied by itself, the product is called the square of that number. For Examples: 1 x 1 = 1 = 1 2 The square of 1 is 1. 2 x 2 = 4 = 2 2 The square of 2 is

3 x 3 = 9 = x 4 = 16 =

PERFECT SQUARE A natural number ‘x’ is a perfect square, if y 2 = x where ‘y’ is natural number. Examples : 16 and 25 are perfect squares, since 16 = = 5 2

FACTS ABOUT SQUARES A number ending with 2, 3, 7 or 8 is never a perfect square. The squares of even numbers are even. The squares of odd numbers are odd. A number ending with an odd number of zeros is never a perfect square. The ending digits of a square number is 0, 1, 4, 5, 6 or 9 only. Note : it is not necessary that all numbers ending with digits 0, 1, 4, 5, 6 or 9 are square numbers.

SOME METHODS TO FINDING SQUARES USING THE FORMULA ( a + b ) 2 = a 2 + 2ab + b 2 1.(27) 2 = ( ) 2 (20 + 7) 2 = (20) x 20x 7 + (7) 2 = = 729. FIND (32) 2

(a – b ) 2 = a 2 – 2ab + b 2 1.(39) 2 = (40 -1) 2 (40 – 1) 2 = (40) 2 – 2 x 40 x 1 + (1) 2 = 1600 – = FIND (48) 2.

DIAGONAL METHOD FOR SQUARING Example:- Find (72) 2 using the diagonal method. SOLUTION:-

Therefore, (72) 2 =5184. ‘FIND (23) 2 ’

ALTERNATIVE METHOD ALTERNATIVE METHOD ALTERNATIVE METHOD

SOME INTERESTING PATTERNS 1.SQUARES ARE SUM OF CONSECUTIVE ODD NUMBERS. EXAMPLES: = 4 = = 9 = = 16 = = 25 = = =

2. SQUARES OF NUMBERS ENDING WITH DIGIT 5. (15) 2 =1X (1 + 1)X = 1X2X = = 225 (25) 2 = 2X3X = = 625 (35) 2 = (3X4) 25 = 1225 TENSUNITS FIND (45) 2

PYTHAGOREAN TRIPLETS If three numbers x, y and z are such that x 2 + y 2 = z 2, then they are called Pythagorean Triplets and they represent the sides of a right triangle. x z y

Examples (i)3, 4 and 5 form a Pythagorean Triplet = 5 2.( = 25) (ii) 8, 15 and 17 form a Pythagorean Triplet = ( = 289)

Find Pythagorean Triplet if one element of a Pythagorean Triplet is given. For any natural number n, (n>1), we have (2n) 2 + (n 2 -1) 2 = (n 2 +1) 2. such that 2n, n 2 -1 and n 2 +1 are Pythagorean Triplet.

Examples- Write a Pythagorean Triplet whose one member is 12. Since, Pythagorean Triplet are 2n, n 2 -1 and n So, 2n = 12, n = 6. n 2 -1 = (6) 2 -1 = 36 -1= 35 And n 2 +1 = (6) 2 +1= 36+1= 37 Therefore, 12, 35 and 37 are Triplet.

Write a Pythagorean Triplet whose one member is 6.

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