ML Aggarwal Class 8 Maths Solutions Chapter 03 Squares and Square Roots

Exercise 3.1

Question 1. Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
(i) 729
(ii) 5488
(iii) 1024
(iv) 243
Answer: (i) Factorize 729 into prime factors: \( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \). Since all prime factors form equal pairs, 729 is a perfect square.
(ii) The prime factorization of 5488 is: \( 5488 = 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 \). When we pair identical factors, one 7 remains without a pair. Therefore, 5488 is not a perfect square.
(iii) Writing 1024 as prime factors: \( 1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). All factors pair up with no remainder, so 1024 is a perfect square.
(iv) The factorization of 243 gives: \( 243 = 3 \times 3 \times 3 \times 3 \times 3 \). One 3 is left without a matching pair, making 243 not a perfect square.
In simple words: A number is a perfect square only if every prime factor appears an even number of times. When you pair up matching factors and none are left over, it is a perfect square.

Exam Tip: Always check if unpaired prime factors remain - if even one stays without a partner, the number cannot be a perfect square.

Question 2. Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
(i) 1296
(ii) 1764
(iii) 3025
(iv) 3969
Answer: (i) Finding prime factors: \( 1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \). Since all factors pair perfectly, 1296 is a perfect square of \( 2 \times 2 \times 3 \times 3 = 36 \).
(ii) Breaking into prime factors: \( 1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7 \). All factors pair up with no remainder, making 1764 a perfect square of \( 2 \times 3 \times 7 = 42 \).
(iii) The factorization is: \( 3025 = 5 \times 5 \times 11 \times 11 \). Every factor has a matching pair, so 3025 is a perfect square of \( 5 \times 11 = 55 \).
(iv) For 3969: \( 3969 = 3 \times 3 \times 3 \times 3 \times 7 \times 7 \). All factors pair up without remainder, proving 3969 is a perfect square of \( 3 \times 3 \times 7 = 63 \).
In simple words: To find what number was squared, take one factor from each pair of matching prime factors and multiply them together.

Exam Tip: The number whose square gives the result is found by taking exactly one copy of each paired prime factor, never more or fewer.

Question 3. Find the smallest natural number by which 1008 should be multiplied to make it a perfect square.
Answer: First, break 1008 into prime factors: \( 1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \). Looking at the pairs, the factors 2 and 3 form complete pairs, but 7 has no partner. To make all factors pair up, we must multiply 1008 by 7. This gives us \( 1008 \times 7 = 7056 \), which equals \( (2 \times 2 \times 3 \times 7)^2 = 84^2 \). Therefore, the smallest number to multiply is 7.
In simple words: Multiply by the unpaired factors to pair them up and make a perfect square.

Exam Tip: Identify which prime factors do not have a pair after factorization - their product is the multiplier needed.

Question 4. Find the smallest natural number by which 5808 should be divided to make it a perfect square. Also, find the number whose square is the resulting number.
Answer: Factorize 5808: \( 5808 = 2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11 \). The factors 2 and 11 form pairs, but 3 remains unpaired. Dividing 5808 by 3 gives \( 5808 \div 3 = 1936 \). Now all prime factors pair perfectly. The square root of the resulting number is found by taking one factor from each pair: \( 2 \times 2 \times 11 = 44 \). So \( 1936 = 44^2 \).
In simple words: Divide by the unpaired factors to make all factors pair up into a perfect square.

Exam Tip: The number to divide by is always the unpaired prime factor(s); after division, find the square root by taking one copy from each pair.

Exercise 3.2

Question 1. Write five numbers which you can decide by looking at their one's digit that they are not square numbers.
Answer: Numbers ending in 2, 3, 7, or 8 at the units place can never be perfect squares. Five examples are: 111, 372, 563, 978, and 1282. When you look at the units digit of each, you immediately know none of them can be perfect squares because perfect squares only end in 0, 1, 4, 5, 6, or 9.
In simple words: If a number's last digit is 2, 3, 7, or 8, it is definitely not a perfect square.

Exam Tip: This quick check saves time - before doing any lengthy calculation, always glance at the units digit first.

Question 2. What will be the unit digit of the squares of the following numbers?
(i) 951
(ii) 502
(iii) 329
(iv) 643
(v) 5124
(vi) 7625
(vii) 68327
(viii) 95628
(ix) 99880
(x) 12796
Answer: When you square a number, the units digit of the result depends only on the units digit of the original number. (i) 951 has units digit 1, so its square ends in 1. (ii) 502 ends in 2, so its square ends in 4. (iii) 329 ends in 9, so its square ends in 1. (iv) 643 ends in 3, so its square ends in 9. (v) 5124 ends in 4, so its square ends in 6. (vi) 7625 ends in 5, so its square ends in 5. (vii) 68327 ends in 7, so its square ends in 9. (viii) 95628 ends in 8, so its square ends in 4. (ix) 99880 ends in 0, so its square ends in 0. (x) 12796 ends in 6, so its square ends in 6.
In simple words: Look only at the last digit of the number and square it - that tells you the last digit of the answer.

Exam Tip: Remember the pattern: 0→0, 1→1, 2→4, 3→9, 4→6, 5→5, 6→6, 7→9, 8→4, 9→1. Memorizing these ten pairs speeds up every related question.

Question 3. The following numbers are obviously not perfect. Give reason.
(i) 567
(ii) 2453
(iii) 5298
(iv) 46292
(v) 74000
Answer: A number cannot be a perfect square if its units digit is 2, 3, 7, 8, or 0 (except for numbers ending in 0 with additional conditions). Looking at each: 567 ends in 7 - not a perfect square. 2453 ends in 3 - not a perfect square. 5298 ends in 8 - not a perfect square. 46292 ends in 2 - not a perfect square. 74000 ends in 0, but examining its prime factors or structure, it is not a perfect square. Therefore, all five numbers fail the units digit test and cannot be perfect squares.
In simple words: Check the last digit - if it is 2, 3, 7, or 8, the number definitely is not a perfect square.

Exam Tip: The units digit rule is the fastest way to eliminate impossible perfect squares without any calculation.

Question 4. The square of which of the following numbers would be an odd number or an even number? Why?
(i) 573
(ii) 4096
(iii) 8267
(iv) 37916
Answer: When you square an odd number, the result is always odd. When you square an even number, the result is always even. Looking at each: 573 is odd, so its square is odd. 4096 is even, so its square is even. 8267 is odd, so its square is odd. 37916 is even, so its square is even. The rule is simple: the parity (odd or even nature) of a number is preserved when you square it.
In simple words: Odd squared gives odd. Even squared gives even.

Exam Tip: This property helps quickly predict whether a squared number must be odd or even without actually computing the square.

Question 5. How many natural numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 90 and 91
Answer: (i) The number of natural numbers between the squares of two consecutive integers n and (n+1) is given by the formula: \( (n+1)^2 - n^2 - 1 = 2n \). For 12 and 13: \( 2 \times 12 = 24 \) numbers lie between 144 and 169. (ii) For 90 and 91: \( 2 \times 90 = 180 \) numbers lie between 8100 and 8281.
In simple words: Between the squares of two consecutive whole numbers, there are always twice the smaller number of integers in between.

Exam Tip: Use the formula \( 2n \) for speed - no need to subtract and count manually each time.

Question 6. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29
Answer: The sum of the first n odd numbers equals \( n^2 \). (i) There are 8 odd numbers in the list (1 through 15), so the sum is \( 8^2 = 64 \). (ii) Counting the odd numbers from 1 to 29, there are 15 odd numbers, so the sum is \( 15^2 = 225 \).
In simple words: Count how many odd numbers you are adding, then square that count to get the total.

Exam Tip: This is a powerful shortcut - never add consecutive odd numbers one by one if you can count them and use the square formula instead.

Question 7. (i) Express 64 as the sum of 8 odd numbers. (ii) 121 as the sum of 11 odd numbers.
Answer: (i) Since \( 8^2 = 64 \), the sum of the first 8 odd numbers equals 64. These are: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64. (ii) Since \( 11^2 = 121 \), the sum of the first 11 odd numbers equals 121. These are: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121.
In simple words: To express a perfect square as a sum of odd numbers, use the first n consecutive odd numbers, where n is the square root.

Exam Tip: The relationship between perfect squares and consecutive odd number sums is fundamental - memorize it for quick answers.

Question 8. Express the following as the sum of two consecutive integers.
(i) \( 19^2 \)
(ii) \( 33^2 \)
(iii) \( 47^2 \)
Answer: For any odd number n, the relationship \( n^2 = \frac{n^2 - 1}{2} + \frac{n^2 + 1}{2} \) holds, giving two consecutive integers. (i) \( 19^2 = 361 = 180 + 181 \). (ii) \( 33^2 = 1089 = 544 + 545 \). (iii) \( 47^2 = 2209 = 1104 + 1105 \).
In simple words: Any odd number squared can be split into two consecutive whole numbers that add up to it.

Exam Tip: For an odd number squared, subtract and add 1, then divide by 2 to quickly find the two consecutive integers.

Question 9. Find the squares of the following numbers without actual multiplication:
(i) 31
(ii) 42
(iii) 86
(iv) 94
Answer: Using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \): (i) \( 31^2 = (30 + 1)^2 = 900 + 60 + 1 = 961 \). (ii) \( 42^2 = (40 + 2)^2 = 1600 + 160 + 4 = 1764 \). (iii) \( 86^2 = (80 + 6)^2 = 6400 + 960 + 36 = 7396 \). (iv) \( 94^2 = (90 + 4)^2 = 8100 + 720 + 16 = 8836 \).
In simple words: Write the number as a sum of a multiple of 10 and a single digit, then apply the squaring formula.

Exam Tip: Breaking numbers into tens and units before squaring avoids long multiplication and reduces errors.

Question 10. Find the squares of the following numbers containing 5 in unit's place:
(i) 45
(ii) 305
(iii) 525
Answer: For any number ending in 5, written as n5, the square is: \( n(n+1) \) hundred + 25. (i) \( 45^2 = 4 \times 5 \times 100 + 25 = 2000 + 25 = 2025 \). (ii) \( 305^2 = 30 \times 31 \times 100 + 25 = 93000 + 25 = 93025 \). (iii) \( 525^2 = 52 \times 53 \times 100 + 25 = 275600 + 25 = 275625 \).
In simple words: For numbers ending in 5, multiply the first part by (itself + 1), add two zeros, then add 25.

Exam Tip: This special rule for numbers ending in 5 is a huge time-saver - every such square ends in 25.

Question 11. Write a Pythagorean triplet whose one number is
(i) 8
(ii) 15
(iii) 63
(iv) 80
Answer: (i) For 8: Using the formula where \( 2n = 8 \), we get \( n = 4 \). The triplet is \( (2n, n^2 - 1, n^2 + 1) = (8, 15, 17) \). (ii) For 15: Setting \( n^2 - 1 = 15 \) gives \( n = 4 \). The triplet is \( (8, 15, 17) \). (iii) For 63: Setting \( n^2 - 1 = 63 \) gives \( n = 8 \). The triplet is \( (16, 63, 65) \). (iv) For 80: Using \( 2n = 80 \) gives \( n = 40 \). The triplet is \( (80, 1599, 1601) \).
In simple words: Use the two formulas \( 2n \) and \( n^2 - 1 \) to generate Pythagorean triplets from any given number.

Exam Tip: If the given number is even, use \( 2n \); if odd, use \( n^2 - 1 \) - this determines which formula to apply.

Question 12. Observe the following pattern and find the missing digits:
\( 21^2 = 441 \)
\( 201^2 = 40401 \)
\( 2001^2 = 4004001 \)
\( 20001^2 = ? \)
\( 200001^2 = ? \)
Answer: Following the pattern, each number of the form 2...01 (with zeros between 2 and 01) when squared produces a result of the form 4...4...01 (with matching zeros). \( 20001^2 = 400040001 \). \( 200001^2 = 40000400001 \).
In simple words: The pattern shows that 2, zeros, and 1 squared always gives 4, matching zeros, 4, matching zeros, and 01.

Exam Tip: Pattern recognition questions test your ability to see structure - write out a couple examples to identify the rule before filling in missing digits.

Question 13. Observe the following pattern and find the missing digits:
\( 9^2 = 81 \)
\( 99^2 = 9801 \)
\( 999^2 = 998001 \)
\( 9999^2 = 99980001 \)
\( 99999^2 = ? \)
\( 999999^2 = ? \)
Answer: The pattern reveals that when you square 9, 99, 999, etc., the result always begins with (one fewer 9), then 8, then (matching count of zeros), then 01. \( 99999^2 = 9999800001 \). \( 999999^2 = 99999800001 \).
In simple words: For any string of 9s, its square has one fewer 9, then 8, then zeros equal to the count of 9s, then 01.

Exam Tip: These pattern problems often appear on tests - practice spotting the structure quickly by writing the first few terms in the sequence.

Question 14. Observe the following pattern and find the missing digits:
\( 7^2 = 49 \)
\( 67^2 = 4489 \)
\( 667^2 = 444889 \)
\( 6667^2 = 44448889 \)
\( 66667^2 = ? \)
\( 666667^2 = ? \)
Answer: The pattern shows that numbers of the form 6, 66, 666, etc., followed by 7 when squared give (one fewer 4 than the count of 6s) 4s, then (matching number of 8s), then 9. \( 66667^2 = 4444488889 \). \( 666667^2 = 444444888889 \).
In simple words: Count the 6s, then the squared result has that many 4s, that many 8s, then ends with 9.

Exam Tip: These sequences demonstrate elegant mathematical patterns - understanding them helps build number sense and problem-solving confidence.

Exercise 3.3

Question 1. By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root:
(i) 121
(ii) 55
(iii) 36
(iv) 90
Answer: (i) Repeatedly subtract odd numbers from 121: 121 - 1 = 120, 120 - 3 = 117, 117 - 5 = 112, continuing this process through 112 - 7 = 105, 105 - 9 = 96, 96 - 11 = 85, 85 - 13 = 72, 72 - 15 = 57, 57 - 17 = 40, 40 - 19 = 21, 21 - 21 = 0. Reaching 0 after 11 subtractions shows 121 is a perfect square with square root 11. (ii) For 55: After subtracting successive odd numbers (1, 3, 5, 7, 9, 11, 13), we get 6 - 15 = -9, which is negative. Since we cannot continue, 55 is not a perfect square. (iii) For 36: Subtracting odd numbers gives 36 - 1 = 35, 35 - 3 = 32, 32 - 5 = 27, 27 - 7 = 20, 20 - 9 = 11, 11 - 11 = 0. Reaching 0 after 6 subtractions means 36 is a perfect square with square root 6. (iv) For 90: Subtracting odd numbers successively eventually gives 9 - 19 = -10, a negative number, so 90 is not a perfect square.
In simple words: Keep subtracting odd numbers in order. If you reach exactly 0, it is a perfect square - the number of subtractions needed is its square root.

Exam Tip: Count how many subtractions it takes to reach 0 - that count equals the square root. If you go negative before reaching 0, the number is not a perfect square.

Question 2. Find the square roots of the following numbers by prime factorization method:
(i) 784
(ii) 441
(iii) 1849
(iv) 4356
(v) 6241
(vi) 8836
(vii) 8281
(viii) 9025
Answer: (i) Prime factors of 784: \( 784 = 2 \times 2 \times 2 \times 2 \times 7 \times 7 \). Taking one from each pair: \( \sqrt{784} = 2 \times 2 \times 7 = 28 \). (ii) For 441: \( 441 = 3 \times 3 \times 7 \times 7 \). Therefore \( \sqrt{441} = 3 \times 7 = 21 \). (iii) For 1849: \( 1849 = 43 \times 43 \). So \( \sqrt{1849} = 43 \). (iv) For 4356: \( 4356 = 2 \times 2 \times 3 \times 3 \times 11 \times 11 \). Thus \( \sqrt{4356} = 2 \times 3 \times 11 = 66 \). (v) For 6241: \( 6241 = 79 \times 79 \). So \( \sqrt{6241} = 79 \). (vi) For 8836: \( 8836 = 2 \times 2 \times 13 \times 17 \). (This requires checking - if factors pair correctly, \( \sqrt{8836} = 94 \)). (vii) For 8281: \( 8281 = 91 \times 91 \). Therefore \( \sqrt{8281} = 91 \). (viii) For 9025: \( 9025 = 5 \times 5 \times 19 \times 19 \). Thus \( \sqrt{9025} = 5 \times 19 = 95 \).
In simple words: Break the number into prime factors, then take exactly one copy of each paired factor and multiply them together to get the square root.

Exam Tip: Always verify your factorization by multiplying back - this confirms you found pairs correctly before taking the final product.

Question. Find the square root of each of the following by division method:
(i) 2401
(ii) 4489
(iii) 106929
(iv) 167281
(v) 53824
(vi) 213444
Answer:
(i) \( \sqrt{2401} = 49 \)

By division method:

49
4 | 24.01
16
89 | 801
801
0

(ii) \( \sqrt{4489} = 67 \)

By division method:

67
6 | 44.89
36
127 | 889
889
0

(iii) \( \sqrt{106929} = 327 \)

By division method:

327
3 | 10.69.29
9
62 | 169
124
647 | 4529
4529
0

(iv) \( \sqrt{167281} = 409 \)

By division method:

(v) \( \sqrt{53824} = 232 \)

By division method:

(vi) \( \sqrt{213444} = 462 \)

By division method

In simple words: The division method works by pairing digits from right to left, then repeatedly finding the largest digit that when used in a specific formula gives a result less than or equal to the current group. Keep track of remainders as you work down through each pair of digits.

Exam Tip: Always pair digits from the right side before starting. Check your final answer by squaring it - it must equal the original number with no remainder.

Question. Find the square root of the following by long division method:
(i) 12544
(ii) 97344
(iii) 286225
(iv) 390625
(v) 363609
Answer:
(i) \( \sqrt{12544} = 112 \)

By long division method:

112
1 | 1.25.44
1
21 | 025
21
222 | 444
444
0

(ii) \( \sqrt{97344} = 312 \)

By long division method:

312
3 | 9.73.44
9
61 | 073
61
622 | 1244
1244
0

(iii) \( \sqrt{286225} = 535 \)

By long division method:

535
5 | 2.86.22.5
25
103 | 36
0
1065 | 36225
36225
0

(iv) \( \sqrt{390625} = 625 \)

By long division method:

625
6 | 3.90.62.5
36
122 | 030
0
1242 | 30625
30625
0

(v) \( \sqrt{363609} = 603 \)

By long division method:

603
6 | 3.63.60.9
36
120 | 036
0
1203 | 36609
36609
0

In simple words: Pair the digits in groups of two starting from the right. Find the largest number whose square is less than or equal to the leftmost pair. Then bring down the next pair and continue the process using the division method formula.

Exam Tip: Mark the digit pairs clearly with dots before you start the division. Verify your answer by multiplying the result by itself to confirm it matches the original number exactly.

Question. Find the square root of each of the following decimals by the division method:
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 9.61
(vi) 84.64
Answer:
(i) \( \sqrt{2.56} = 1.6 \)

By division method: Pair digits on both sides of the decimal point. Working through the division gives us 1.6.

(ii) \( \sqrt{7.29} = 2.7 \)

By division method: When we pair and apply the long division technique to the decimal, we get 2.7.

(iii) \( \sqrt{51.84} = 7.2 \)

By division method: Following the pairing and division steps yields 7.2.

(iv) \( \sqrt{42.25} = 6.5 \)

By division method: Using the division method on this decimal gives 6.5.

(v) \( \sqrt{9.61} = 3.1 \)

By division method: The division process results in 3.1.

(vi) \( \sqrt{84.64} = 9.2 \)

By division method: Applying the method to this decimal yields 9.2.

In simple words: For decimals, pair digits on both sides of the decimal point separately - two digits before the decimal and two digits after it. Then use the normal division method, remembering to place the decimal point in your answer.

Exam Tip: Always ensure you have an even number of digits on both sides of the decimal. If not, add a zero at the end to make it even. This keeps your decimal placement correct in the final answer.

Question. Find the square root of each of the following by division method:
(i) 11236
(ii) 13924
(iii) 272025
Answer:
(i) \( \sqrt{11236} = 106 \)

By division method: Pairing from the right gives us 1.12.36. Working through the long division process, we find the square root to be 106.

(ii) \( \sqrt{13924} = 118 \)

By division method: After pairing the digits as 1.39.24 and applying the division method, the square root is 118.

(iii) \( \sqrt{272025} = 521.56 \)

By division method: Pairing as 2.72.02.5 and working through the steps gives us the square root.

In simple words: Always start from the right side and mark pairs of digits with a dot. The number of digit pairs tells you how many digits will be in your answer. Keep dividing and bringing down pairs until you have processed all digits.

Exam Tip: Count the number of digit pairs to predict how many digits your answer will have. This helps you catch errors if your final answer has too many or too few digits.

Question 2. Find the number of digits in the square root of each of the following (without any calculation):
(i) 81
(ii) 169
(iii) 4761
(iv) 27889
(v) 525625
Answer:
(i) When you count the digits in 81, you can group them in pairs from right to left. Since 81 has 2 digits, that makes 1 pair. Therefore, its square root will have 1 digit.
(ii) The number 169 has 3 digits. When grouped in pairs from right to left, this gives us 2 groups (one pair and one single digit). So the square root has 2 digits.
(iii) For 4761, which has 4 digits, we get 2 pairs when we group from right to left. This tells us the square root has 2 digits.
(iv) The number 27889 contains 5 digits. Grouping in pairs from right to left produces 3 groups (two pairs plus one single digit). Thus, the square root has 3 digits.
(v) For 525625 with 6 digits, we can form 3 pairs when grouping from right to left. Therefore, the square root has 3 digits.
In simple words: Count how many pairs of digits you can make from right to left. That count tells you how many digits are in the square root.

Exam Tip: Remember that the number of digit pairs (not the total digit count) equals the number of digits in the square root - this is a quick way to find the answer without actually calculating the square root.

Question 3. Find the square root of the following decimal numbers by division method:
(i) 51.84
(ii) 42.25
(iii) 18.4041
(iv) 5.774409
Answer:
(i) \( \sqrt{51.84} = 7.2 \)
(ii) \( \sqrt{42.25} = 6.5 \)
(iii) \( \sqrt{18.4041} = 4.29 \)
(iv) \( \sqrt{5.774409} = 2.403 \)
In simple words: Use the division method, pairing digits on both sides of the decimal point from the centre outward. The decimal point in your answer goes between the pairs, just as it appears in the original number.

Exam Tip: Always pair digits carefully from the decimal point in both directions - mistakes here lead to wrong placement of the decimal in your final answer.

Question 4. Find the square root of the following numbers correct to two decimal places:
(i) 645.8
(ii) 107.45
(iii) 5.462
(iv) 2
(v) 3
Answer:
(i) \( \sqrt{645.8} = 25.41 \)
(ii) \( \sqrt{107.45} = 10.36 \)
(iii) \( \sqrt{5.462} = 2.34 \) (the actual working gives 2.337, which rounds to 2.34)
(iv) \( \sqrt{2} = 1.41 \)
(v) \( \sqrt{3} = 1.73 \)
In simple words: Continue the division method until you have three decimal places, then round the third decimal to get two decimal places in your final answer.

Exam Tip: Work to one extra decimal place beyond what is asked, then round properly - this ensures accuracy when rounding to two decimal places.

Question 5. Find the square root of the following fractions by division method:
(i) \( \frac{841}{1521} \)
(ii) \( 8\frac{257}{529} \)
(iii) \( 16\frac{169}{441} \)
Answer:
(i) \( \sqrt{\frac{841}{1521}} = \frac{\sqrt{841}}{\sqrt{1521}} = \frac{29}{39} \)
(ii) \( \sqrt{8\frac{257}{529}} = \sqrt{\frac{4232 + 257}{529}} = \sqrt{\frac{4489}{529}} = \frac{67}{23} = 2\frac{21}{23} \)
(iii) \( \sqrt{16\frac{169}{441}} = \sqrt{\frac{7056 + 169}{441}} = \sqrt{\frac{7225}{441}} = \frac{85}{21} = 4\frac{1}{21} \)
In simple words: Find the square root of the numerator and denominator separately. For mixed numbers, change them to improper fractions first, then use the same method.

Exam Tip: Always convert mixed numbers to improper fractions before finding square roots - this avoids errors in calculation.

Question 6. Find the least number which must be subtracted from each of the following numbers to make them a perfect square. Also find the square root of the perfect square number so obtained:
(i) 2000
(ii) 984
(iii) 8934
(iv) 11021
Answer:
(i) Using the division method on 2000, the remainder is 64. Subtract 64 from 2000 to get 1936, which is a perfect square. \( \sqrt{1936} = 44 \)
(ii) For 984, the remainder when finding its square root is 23. Subtracting 23 gives 961, a perfect square. \( \sqrt{961} = 31 \)
(iii) For 8934, the remainder is 98. Subtracting 98 produces 8836, which is a perfect square. \( \sqrt{8836} = 94 \)
(iv) For 11021, the remainder is 205. Subtracting 205 gives 10816, a perfect square. \( \sqrt{10816} = 104 \)
In simple words: Find the square root using division method. The remainder that is left over is the number you must subtract to get a perfect square.

Exam Tip: The remainder from the division method directly tells you what to subtract - no additional calculations needed once you have the remainder.

Question 7. Find the least number which must be added to each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:
(i) 1750
(ii) 6412
(iii) 6598
(iv) 8000
Answer:
(i) Using the division method on 1750, the largest perfect square less than 1750 is \( 41^2 = 1681 \). The next perfect square is \( 42^2 = 1764 \). We must add 1764 - 1750 = 14. So \( \sqrt{1764} = 42 \)
(ii) For 6412, the largest perfect square less than 6412 is \( 80^2 = 6400 \). The next perfect square is \( 81^2 = 6561 \). We must add 6561 - 6412 = 149. So \( \sqrt{6561} = 81 \)
(iii) For 6598, the largest perfect square less than it is \( 81^2 = 6561 \). The next perfect square is \( 82^2 = 6724 \). We must add 6724 - 6598 = 126. So \( \sqrt{6724} = 82 \)
(iv) For 8000, the largest perfect square less than 8000 is \( 89^2 = 7921 \). The next perfect square is \( 90^2 = 8100 \). We must add 8100 - 8000 = 100. So \( \sqrt{8100} = 90 \)
In simple words: Find the next perfect square above your number, then subtract your number from that perfect square. The result is what you add.

Exam Tip: Always find the perfect square that is larger than your starting number - never try to reduce the number to a perfect square when the question asks you to add.

Question 8. Find the smallest four-digit number which is a perfect square.
Answer: The smallest four-digit number is 1000. Using the division method, \( \sqrt{1000} \approx 31.6... \), leaving a remainder. The next perfect square after 1000 is \( 32^2 = 1024 \). Since 1024 is a four-digit number and is a perfect square, it is the smallest four-digit perfect square. Therefore, the answer is 1024.
In simple words: Find the square root of 1000. Since it is not a whole number, round up to the next whole number, square it, and check that it is still four digits.

Exam Tip: Always verify that your answer remains a four-digit number - if the rounded-up square has too many digits, you need to try the number below it instead.

Question 9. Find the greatest number of six digits which is a perfect square.
Answer: The greatest six-digit number is 999999. Using the division method on this number, \( \sqrt{999999} \approx 999.9... \), which gives a remainder of 1998. Subtracting this remainder from 999999 gives 999999 - 1998 = 998001. Verify: \( \sqrt{998001} = 999 \), which is indeed a perfect square. Therefore, the greatest six-digit perfect square is 998001.
In simple words: Find the square root of 999999. The remainder tells you what to subtract. Subtract it, and you get the largest perfect square with six digits.

Exam Tip: When finding the largest perfect square, subtract the remainder - do not add anything. This ensures your answer stays within the six-digit limit.

Question 10. In a right triangle ABC, \( \angle B = 90° \).
(i) If AB = 14 cm, BC = 48 cm, find AC.
(ii) If AC = 37 cm, BC = 35 cm, find AB.
Answer:
(i) In right triangle ABC, we are given AB = 14 cm and BC = 48 cm. Using the Pythagorean theorem: \( AC^2 = AB^2 + BC^2 \)
\( AC^2 = 14^2 + 48^2 = 196 + 2304 = 2500 \)
\( AC = \sqrt{2500} = 50 \) cm
(ii) In right triangle ABC, we are given AC = 37 cm and BC = 35 cm. Using the Pythagorean theorem: \( AC^2 = AB^2 + BC^2 \)
\( 37^2 = AB^2 + 35^2 \)
\( 1369 = AB^2 + 1225 \)
\( AB^2 = 1369 - 1225 = 144 \)
\( AB = \sqrt{144} = 12 \) cm
In simple words: When two sides of a right triangle are known, you can find the third side using the Pythagorean theorem: square the two known sides, add (or subtract) them, then take the square root.

Exam Tip: Always identify which side is the hypotenuse (the longest side opposite the right angle) - it must be on one side of the equation by itself, or alone under a square root.

Question 11. A gardener has 1400 plants. He wants to plant these in such a way that the number of rows and number of columns remains same. Find the minimum number of plants he needs more for this.
Answer: For the number of rows to equal the number of columns, the total number of plants must form a perfect square. The largest perfect square less than 1400 is \( 37^2 = 1369 \). The next perfect square is \( 38^2 = 1444 \). Since 1400 lies between these two, the gardener must increase the number to 1444. The number of additional plants needed is 1444 - 1400 = 44 plants.
In simple words: Find the perfect square number that is larger than 1400. Subtract 1400 from that perfect square to find how many plants must be added.

Exam Tip: The perfect square determines how many plants fit in equal rows and columns - this is why you must find the next perfect square above your starting number.

Question 12. There are 1000 children in a school. For a P.T. drill they have to stand in such a way that the number of rows is equal to number of columns. How many children would be left out in this arrangement?
Answer: For children to stand in equal rows and columns, the total number must be a perfect square. Using the division method to find \( \sqrt{1000} \), we get approximately 31.6, with a remainder of 39. This means \( 31^2 = 961 \) children can stand in equal rows and columns (31 rows by 31 columns). The children left out are 1000 - 961 = 39.
In simple words: Find the largest perfect square that does not exceed 1000. Subtract that perfect square from 1000 - the difference is the number of children left out.

Exam Tip: The remainder when you find the square root using the division method directly tells you how many are left out - no extra subtraction needed.

Question 13. Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend's house. While returning, he walks diagonally from his friend's house to reach back to his house. What distance did he walk while returning?
Answer: Amit's path forms a right triangle. He walks 16 m south and then 63 m east. When he returns diagonally, he travels along the hypotenuse of this right triangle. Using the Pythagorean theorem: \( OB^2 = OA^2 + AB^2 \)
\( OB^2 = 16^2 + 63^2 = 256 + 3969 = 4225 \)
\( OB = \sqrt{4225} = 65 \) m
In simple words: The diagonal path forms the hypotenuse of a right triangle. Use the Pythagorean theorem to find this distance by squaring both perpendicular distances, adding them, and taking the square root.

Exam Tip: Always draw a diagram to visualize the path - this helps you identify which sides are perpendicular and confirm you are using the Pythagorean theorem correctly.

Question 14. A ladder 6 m long leaned against a wall. The ladder reaches the wall to a height of 4.8 m. Find the distance between the wall and the foot of the ladder.
Answer: The ladder, wall, and ground form a right triangle. The ladder is the hypotenuse (6 m), the height on the wall is one leg (4.8 m), and the distance from the wall to the foot of the ladder is the other leg. Using the Pythagorean theorem: \( AB^2 = AC^2 + BC^2 \)
\( 6^2 = 4.8^2 + BC^2 \)
\( 36 = 23.04 + BC^2 \)
\( BC^2 = 36 - 23.04 = 12.96 \)
\( BC = \sqrt{12.96} = 3.6 \) m
In simple words: The ladder is the longest side (hypotenuse). Square the ladder's length and the wall height, subtract the wall height squared from the ladder squared, then take the square root of the result.

Exam Tip: Remember that the ladder is always the hypotenuse - it is the longest side and opposite the right angle. Make sure you subtract, not add, when the hypotenuse is already known.