MATHEMATICS We can see that when a square number ends in 6, the number whose square it is, will have either 4 or 6 in unit’s place. Can you find more such rules by observing the numbers and their squares (Table 1)? What will be the “one’s digit” in the square of the following numbers? (i) 1234 (ii) 26387 (iii) 52698 (iv) 99880 (v) 21222 (vi) 9106 • Consider the following numbers and their squares. 102 = 100 202 = 400 802 = 6400 1002 = 10000 2002 = 40000 7002 = 490000 9002 = 810000 If a number contains 3 zeros at the end, how many zeros will its square have ? What do you notice about the number of zeros at the end of the number and the number of zeros at the end of its square? Can we say that square numbers can only have even number of zeros at the end? • See Table 1 with numbers and their squares. What can you say about the squares of even numbers and squares of odd numbers? 1. The square of which of the following numbers would be an odd number/an even number? Why? (i) 727 (ii) 158 (iii) 269 (iv) 1980 2. What will be the number of zeros in the square of the following numbers? (i) 60 (ii) 400 6.3 Some More Interesting Patterns 1. Adding triangular numbers.Do you remember triangular numbers (numbers whose dot patterns can be arrangedas triangles)? * * ** * ** *** * ** *** * *** * ** *** **** * **** 136 10 15 SQUARES AND SQUARE ROOTS If we combine two consecutive triangular numbers, we get a square number, like 1 + 3 = 4 3 + 6 = 9 6 + 10 = 16 = 22 = 32 = 42 2. Numbers between square numbers Let us now see if we can find some interesting pattern between two consecutive Between 12(=1) and 22(= 4) there are two (i.e., 2 × 1) non square numbers 2, 3. Between 22(= 4) and 32(= 9) there are four (i.e., 2 × 2) non square numbers 5, 6, 7, 8. Now, 32 = 9, 42 = 16 Therefore, 42 – 32 = 16 – 9 = 7 Between 9(=32) and 16(= 42) the numbers are 10, 11, 12, 13, 14, 15 that is, six non-square numbers which is 1 less than the difference of two squares. We have 42 = 16 and 52 = 25 Therefore, 52 – 42 = 9 Between 16(= 42) and 25(= 52) the numbers are 17, 18, ... , 24 that is, eight non square numbers which is 1 less than the difference of two squares. Consider 72 and 62. Can you say how many numbers are there between 62 and 72? If we think of any natural number n and (n + 1), then, (n + 1)2 – n2 = (n2 + 2n + 1) – n2 = 2n + 1. We find that between n2 and (n + 1)2 there are 2n numbers which is 1 less than the difference of two squares. Thus, in general we can say that there are 2n non perfect square numbers between the squares of the numbers n and (n + 1). Check for n = 5, n = 6 etc., and verify. MATHEMATICS Can you find more such triplets? For any natural number m > 1, we have (2m)2 + (m2 – 1)2 = (m2 + 1)2. So, 2m, m2 – 1 and m2 + 1 forms a Pythagorean triplet. Try to find some more Pythagorean triplets using this form. Example 2: Write a Pythagorean triplet whose smallest member is 8. Solution: We can get Pythagorean triplets by using general form 2m, m2 – 1, m2 + 1. Let us first take m2 – 1 =8 So, m2=8 + 1 = 9 which gives m =3 Therefore, 2m = 6 and m2 + 1 = 10 The triplet is thus 6, 8, 10. But 8 is not the smallest member of this. So, let us try 2m =8 then m =4 We get m2 – 1 = 16 – 1 = 15 and m2 + 1 = 16 + 1 = 17 The triplet is 8, 15, 17 with 8 as the smallest member. Example 3: Find a Pythagorean triplet in which one member is 12. Solution: If we take m2 – 1 = 12 Then, m2 = 12 + 1 = 13 Then the value of m will not be an integer. So, we try to take m2 + 1 = 12. Again m2 = 11 will not give an integer value for m. So, let us take 2m =12 then m =6 Thus, m2 – 1 = 36 – 1 =35 and m2 + 1 = 36 + 1 = 37 Therefore, the required triplet is 12, 35, 37. Note:All Pythagorean triplets may not be obtained using this form. For example another triplet 5, 12, 13 also has 12 as a member. EXERCISE 6.2 1. Find the square of the following numbers. (i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46 2. Write a Pythagorean triplet whose one member is. (i) 6 (ii) 14 (iii) 16 (iv) 18 6.5 Square Roots Study the following situations. (a) Area of a square is 144 cm2. What could be the side of the square? SQUARES AND SQUARE ROOTS By pairing the prime factors, we get 324 =2 × 2 × 3 × 3 × 3 × 3 = 22 × 32 × 32 = (2 × 3 × 3)2 So, 324=2 × 3 × 3 = 18 Similarly can you find the square root of 256? Prime factorisation of 256 is 2 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 2 By pairing the prime factors we get, 2 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2 × 2)2 2 Therefore, 256 = 2 × 2 × 2 × 2 = 16 2 Is 48 a perfect square? 2 We know 48= 2 × 2 × 2 × 2 × 3 2 Since all the factors are not in pairs so 48 is not a perfect square. Suppose we want to find the smallest multiple of 48 that is a perfect square, how should we proceed? Making pairs of the prime factors of 48 we see that 3 is the only factor that does not have a pair. So we need to multiply by 3 to complete the pair. Hence 48 × 3 = 144 is a perfect square. 2 Can you tell by which number should we divide 48 to get a perfect square? 2 The factor 3 is not in pair, so if we divide 48 by 3 we get 48 ÷ 3 = 16 = 2 × 2 × 2 × 2 2 and this number 16 is a perfect square too. 2 Example 4: Find the square root of 6400. 2 2Solution: Write 6400 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 90 45 15 5 2Therefore 6400 =2 × 2 × 2 × 2 × 5 = 80 2 23 Example 5: Is 90 a perfect square? 3 5 Solution: We have 90 = 2 × 3 × 3 × 5 256 128 64 32 16 8 4 2 6400 3200 1600 800 400 200 100 50 25 5 The prime factors 2 and 5 do not occur in pairs. Therefore, 90 is not a perfect square. That 90 is not a perfect square can also be seen from the fact that it has only one zero. Example 6: Is 2352 a perfect square? If not, find the smallest multiple of 2352 which 2 is a perfect square. Find the square root of the new number. 2 Solution: We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7 2 As the prime factor 3 has no pair, 2352 is not a perfect square. 2 If 3 gets a pair then the number will become perfect square. So, we multiply 2352 by 3 to get, 3 2352 × 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 7 Now each prime factor is in a pair. Therefore, 2352 × 3 = 7056 is a perfect square. Thus the required smallest multiple of 2352 is 7056 which is a perfect square. 2352 1176 588 294 147 49 7 And, 7056 =2 × 2 × 3 × 7 = 84 Example 7: Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient. 102 MATHEMATICS 2 3 3 5 6, 9, 15 3, 9, 15 1, 3, 5 1, 1, 5 1, 1, 1 Solution: We have, 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7 If we divide 9408 by the factor 3, then 9408 ÷ 3 = 3136 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 which is a perfect square. (Why?) Therefore, the required smallest number is 3. And, 3136 =2 × 2 × 2 × 7 = 56. Example 8: Find the smallest square number which is divisible by each of the numbers 6, 9 and 15. Solution: This has to be done in two steps. First find the smallest common multiple and then find the square number needed. The least number divisible by each one of 6, 9 and 15 is their LCM. The LCM of 6, 9 and 15 is 2 × 3 × 3 × 5 = 90. Prime factorisation of 90 is 90 = 2 × 3 × 3 × 5. We see that prime factors 2 and 5 are not in pairs. Therefore 90 is not a perfect square. In order to get a perfect square, each factor of 90 must be paired. So we need to make pairs of 2 and 5. Therefore, 90 should be multiplied by 2 × 5, i.e., 10. Hence, the required square number is 90 × 10 = 900. EXERCISE 6.3 1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers? (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025 2. Without doing any calculation, find the numbers which are surely not perfect squares. (i) 153 (ii) 257 (iii) 408 (iv) 441 3. Find the square roots of 100 and 169 by the method of repeated subtraction. 4. Find the square roots of the following numbers by the Prime Factorisation Method. (i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100 5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained. (i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768 6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained. (i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620 7. The students of Class VIII of a school donated ` 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. MATHEMATICS 2 529 – 4 1292 529 – 4 4_ 129 23 5292 – 4 43 129 –129 0 6 40966 – 36 4 6 4096 – 36 496 64096– 36 12_ 496 64 40966 – 36 124 496– 496 0Step 3 Bring down the number under the next bar (i.e., 29 in this case) to the right of the remainder. So the new dividend is 129. Step 4 Double the quotient and enter it with a blank on its right. Step 5 Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend. In this case 42 × 2 = 84. As 43 × 3 = 129 so we choose the new digit as 3. Get the remainder. Step 6 Since the remainder is 0 and no digits are left in the given number, therefore, 529 = 23. • Now consider 4096 Step 1 Place a bar over every pair of digits starting from the one’s digit. ( 40 96 ). Step 2 Find the largest number whose square is less than or equal to the number under the left-most bar (62 < 40 < 72). Take this number as the divisor and the number under the left-most bar as the dividend. Divide and get the remainder i.e., 4 in this case. Step 3 Bring down the number under the next bar (i.e., 96) to the right of the remainder. The new dividend is 496. Step 4 Double the quotient and enter it with a blank on its right. Step 5 Guess a largest possible digit to fill the blank which also becomes the new digit in the quotient such that when the new digit is multiplied to the new quotient the product is less than or equal to the dividend. In this case we see that 124 × 4 = 496. So the new digit in the quotient is 4. Get the remainder. Step 6 Since the remainder is 0 and no bar left, therefore, 4096 = 64. Estimating the number We use bars to find the number of digits in the square root of a perfect square number. 529 = 23 and 4096 = 64 In both the numbers 529 and 4096 there are two bars and the number of digits in their square root is 2. Can you tell the number of digits in the square root of 14400? By placing bars we get 14400 . Since there are 3 bars, the square root will be of 3 digit. SQUARES AND SQUARE ROOTS Without calculating square roots, find the number of digits in the square root of the following numbers. (i) 25600 (ii) 100000000 (iii) 36864 Example 9: Find the square root of : (i) 729 (ii) 1296 Solution: 36(i) 27 (ii) 2 7 29 – 4 47 329 329 0 Example 10: Find the least number that must be subtracted from 5607 so as to get 74a perfect square. Also find the square root of the perfect square. 3 1296 – 9 66 396 396 0 Therefore 729 = 27 Therefore 1296 = 36 Solution: Let us try to find 5607 by long division method. We get the remainder 131. It shows that 742 is less than 5607 by 131. This means if we subtract the remainder from the number, we get a perfect square. Therefore, the required perfect square is 5607 – 131 = 5476. And, 5476 = 74. 7 5607 – 49 144 707 –576 131 Example 11: Find the greatest 4-digit number which is a perfect square. 99 Solution: Greatest number of 4-digits = 9999. We find 9999 by long division method. The remainder is 198. This shows 992 is less than 9999 by 198. This means if we subtract the remainder from the number, we get a perfect square. Therefore, the required perfect square is 9999 – 198 = 9801. And, 9801 = 99 9 9999 – 81 189 1899 – 1701 198 36Solution: We find 1300 by long division method. The remainder is 4. This shows that 362 < 1300. Next perfect square number is 372 = 1369. Hence, the number to be added is 372 – 1300 = 1369 – 1300 = 69. Example 12: Find the least number that must be added to 1300 so as to get a perfect square. Also find the square root of the perfect square. 3 1300 – 9 66 400 – 396 4 6.6 Square Roots of Decimals Consider 17.64 Step 1 To find the square root of a decimal number we put bars on the integral part (i.e., 17) of the number in the usual manner. And place bars on the decimal part MATHEMATICS 417.64 –16 1 –16 417.64 8_ 1 64 4. 17.644– 1682 164 48 4 2304 –16 88 704 704 0 Step 2 Step 3 Step 4 Step 5 Step 6 (i.e., 64) on every pair of digits beginning with the first decimal place. Proceed as usual. We get 17.64. Now proceed in a similar manner. The left most bar is on 17 and 42 < 17 < 52. Take this number as the divisor and the number under the left-most bar as the dividend, i.e., 17. Divide and get the remainder. The remainder is 1. Write the number under the next bar (i.e., 64) to the right of this remainder, to get 164. 4.2 Double the divisor and enter it with a blank on its right. Since 64 is the decimal part so put a decimal point in the quotient. We know 82 × 2 = 164, therefore, the new digit is 2. Divide and get the remainder. Since the remainder is 0 and no bar left, therefore 17.64 = 4.2 . 4 17.64 – 16 82 164 – 164 0 3.5Solution: Therefore, 12.25 = 3.5 Example 13: Find the square root of 12.25. 3 12.25 – 9 65 325 325 0 Which way to move Consider a number 176.341. Put bars on both integral part and decimal part. In what way is putting bars on decimal part different from integral part? Notice for 176 we start from the unit’s place close to the decimal and move towards left. The first bar is over 76 and the second bar over 1. For .341, we start from the decimal and move towards right. First bar is over 34 and for the second bar we put 0 after 1 and make .3410 . Example 14: Area of a square plot is 2304 m2. Find the side of the square plot. Solution: Area of square plot = 2304 m2 Therefore, side of the square plot = 2304 m We find that, 2304 =48 Thus, the side of the square plot is 48 m. Example 15: There are 2401 students in a school. P.T. teacher wants them to stand in rows and columns such that the number of rows is equal to the number of columns. Find the number of rows. MATHEMATICS 3. Find the square root of the following decimal numbers. (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36 4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000 5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412 6. Find the length of the side of a square whose area is 441 m2. 7. In a right triangle ABC, ∠B = 90°. (a) If AB = 6 cm, BC = 8 cm, find AC (b) IfAC =13cm, BC = 5 cm, find AB 8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this. 9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.

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