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PowerPoint Presentation : TOPIC SQUARES AND SQUARE ROOTS
PowerPoint Presentation : PRESENTED BY H. P. PATEL TGT MATHS, JNV- AMARKANTAK
CONTENTS : CONTENTS SQUARES. PERFECT SQUARES. FACTS ABOUT SQUARES. SOME METHODS TO FINDING SQUARES. SOME IMPORTANT PATTERNS. PYTHAGOREAN TRIPLET.
SQUARES : 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 4 2 2 1 1
3 x 3 = 9 = 32 4 x 4 = 16 = 42 : 3 x 3 = 9 = 3 2 4 x 4 = 16 = 4 2 3 3 4 4
PERFECT SQUARE : 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 = 4 2 25 = 5 2
FACTS ABOUT SQUARES : 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 : SOME METHODS TO FINDING SQUARES USING THE FORMULA ( a + b ) 2 = a 2 + 2ab + b 2 (27) 2 = (20 + 7 ) 2 (20 + 7) 2 = (20) 2 + 2 x 20x 7 + (7) 2 = 400 + 280 + 49 = 729 . FIND (32) 2
(a – b )2 = a2 – 2ab + b2 : (a – b ) 2 = a 2 – 2ab + b 2 (39) 2 = (40 -1) 2 (40 – 1) 2 = (40) 2 – 2 x 40 x 1 + (1) 2 = 1600 – 80 + 1 = 1521 . FIND (48) 2 .
DIAGONAL METHOD FOR SQUARING : DIAGONAL METHOD FOR SQUARING Example:- Find (72) 2 using the diagonal method. SOLUTION:-
PowerPoint Presentation : Therefore, (72) 2 =5184. ‘FIND (23) 2 ’
SOME INTERESTING PATTERNS : SOME INTERESTING PATTERNS SQUARES ARE SUM OF CONSECUTIVE ODD NUMBERS. EXAMPLES: 1 + 3 = 4 = 2 2 1 + 3 + 5 = 9 = 3 2 1+3+5+7 = 16 = 4 2 1+3+5+7+9 = 25 = 5 2 1+3+5+7+9+11 = ------- = -------
2. SQUARES OF NUMBERS ENDING WITH DIGIT 5. (15)2 =1X (1 + 1)X 100 +25 = 1X2X100 + 25 = 200 + 25 = 225 (25)2 = 2X3X100 + 25 = 600 + 25 = 625 (35)2 = (3X4) 25 = 1225 : 2. SQUARES OF NUMBERS ENDING WITH DIGIT 5. (15) 2 =1X (1 + 1)X 100 +25 = 1X2X100 + 25 = 200 + 25 = 225 (25) 2 = 2X3X100 + 25 = 600 + 25 = 625 (35) 2 = ( 3X4 ) 25 = 1225 TENS UNITS FIND (45) 2
PYTHAGOREAN TRIPLETS : 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 : Examples 3, 4 and 5 form a Pythagorean Triplet. 3 2 + 4 2 = 5 2 .( 9 + 16 = 25) (ii) 8, 15 and 17 form a Pythagorean Triplet. 8 2 +15 2 = 17 2 . (64 +225 = 289)
Find Pythagorean Triplet if one element of a Pythagorean Triplet is given. : 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- : Examples- Write a Pythagorean Triplet whose one member is 12. Since, Pythagorean Triplet are 2n, n 2 -1 and n 2 +1. 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. : Write a Pythagorean Triplet whose one member is 6.
PowerPoint Presentation : My sincere thanks to :- NVS R. O. Bhopal and our Principal Ms. Kavita Singh for providing me an opportunity to prepare a PPT, and also to Mrs. Anju Pandey , TGT Eng, For her cooperation. Acknowledgement
PowerPoint Presentation : T hanks