SQUARES AND SQUARE ROOTS A REVIEW
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SQUARES AND SQUARE ROOTS A REVIEW. welcome. TO POWERPOINT PRESENTATION. TOPIC. SQUARES AND SQUARE ROOTS. CONTENTS. SQUARES. PERFECT SQUARES. FACTS ABOUT SQUARES. SOME METHODS TO FINDING SQUARES. SOME IMPORTANT PATTERNS. PYTHAGOREAN TRIPLET. SQUARES. 1. 1.
SQUARES AND SQUARE ROOTS A REVIEW
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welcome TO POWERPOINT PRESENTATION
TOPIC SQUARES AND SQUARE ROOTS
CONTENTS • SQUARES. • PERFECT SQUARES. • FACTS ABOUT SQUARES. • SOME METHODS TO FINDING SQUARES. • SOME IMPORTANT PATTERNS. • PYTHAGOREAN TRIPLET.
SQUARES 1 1 If a whole number is multiplied by itself, the product is called the square of that number. For Examples: 1x 1 = 1 = 12 The square of 1 is 1. 2 x 2 = 4 = 22 The square of 2 is 4 2 2
3 3 3 x 3 = 9 = 324 x 4 = 16 = 42 4 4
PERFECT SQUARE A natural number ‘x’ is a perfect square, if y2 = x where ‘y’ is natural number. Examples : 16 and 25 are perfect squares, since 16 = 42 25 = 52
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 = a2 + 2ab + b2 (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 (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 Example:- Find (72)2 using the diagonal method. SOLUTION:-
‘FIND (23)2’ • Therefore,(72)2 =5184.
ALTERNATIVE METHOD ALTERNATIVE METHOD
SOME INTERESTING PATTERNS SQUARES ARE SUM OF CONSECUTIVE ODD NUMBERS. EXAMPLES: 1 + 3 = 4 = 22 1 + 3 + 5 = 9 = 32 1+3+5+7 = 16 = 42 1+3+5+7+9 = 25 = 52 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 TENS UNITS FIND (45)2
PYTHAGOREAN TRIPLETS If three numbers x, y and z are such that x2 + y2 = z2, then they are called Pythagorean Triplets and they represent the sides of a right triangle. x z y
Examples 3, 4 and 5 form a Pythagorean Triplet. 32 + 42 = 52.( 9 + 16 = 25) (ii) 8, 15 and 17 form a Pythagorean Triplet. 82+152 = 172. (64 +225 = 289)
Find Pythagorean Triplet if one element of a Pythagorean Triplet is given. For any natural number n, (n>1), we have (2n)2 + (n2-1)2 = (n2+1)2. such that 2n, n2-1 and n2+1 are Pythagorean Triplet.
Examples- Write a Pythagorean Triplet whose one member is 12. Since, Pythagorean Triplet are 2n, n2-1 and n2+1. So, 2n = 12, n = 6. n2-1 = (6)2-1 = 36 -1= 35 And n2+1 = (6)2+1= 36+1= 37 Therefore, 12, 35 and 37 are Triplet.
EVALUATION • EXCEL QUIZ