CBSE Class 8 Mathematics Square And Square Roots Worksheet Set 05

SQUARES AND SQUARE ROOTS

Question 1. Find the square root of 6400
Answer: \(\sqrt{6400} = \sqrt{64 \times 100} = \sqrt{8^2 \times 10^2} = 8 \times 10 = 80\)
In simple words: Break the number into factors of 100 to easily find the square root.

Exam Tip: Remove the trailing zeros in pairs, find the square root of the remaining number, and put half the number of zeros back.

Question 2. Is 90 a perfect square
Answer: No, 90 is not a perfect square.
Prime factorization of \(90 = 2 \times 3^2 \times 5\). Since the prime factors 2 and 5 do not occur in pairs, 90 is not a perfect square.
In simple words: A number is a perfect square only if all its prime factors can be grouped into pairs.

Exam Tip: Check the prime factorization; any factor without a pair means it is not a perfect square.

Question 3. Is 2352 a perfect square? If not find the smallest multiple of 2352 which is a perfect square. Find the square root of the new number.
Answer: Prime factorization of \(2352 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7 = 2^4 \times 3 \times 7^2\).
Since the factor 3 is left unpaired, 2352 is not a perfect square.
To make it a perfect square, we multiply 2352 by 3.
New number = \(2352 \times 3 = 7056\)
Square root of the new number = \(\sqrt{7056} = 2^2 \times 7 \times 3 = 4 \times 21 = 84\)
In simple words: Multiply by the leftover unpaired prime factor to complete the pair.

Exam Tip: Group prime factors in pairs; the lonely factor is always the multiplier.

Question 4. 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.
Answer: Prime factorization of \(9408 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7 = 2^6 \times 3 \times 7^2\).
The prime factor 3 has no pair.
Thus, the smallest number by which 9408 must be divided is 3.
Quotient = \(\frac{9408}{3} = 3136\)
Square root of the quotient = \(\sqrt{3136} = 2^3 \times 7 = 8 \times 7 = 56\)
In simple words: Divide by the unpaired prime factor to remove it.

Exam Tip: Dividing to make a perfect square uses the exact same prime factorization grouping as multiplication.

Question 5. Without doing any calculation, find the numbers which are surely not perfect square.
i.) 153 ii.) 257 iii) 408 iv.) 441
Answer:
(i) 153 ends in 3, so it is not a perfect square.
(ii) 257 ends in 7, so it is not a perfect square.
(iii) 408 ends in 8, so it is not a perfect square.
(iv) 441 ends in 1, which is a possible ending digit, so it may be a perfect square (in fact, \(21^2 = 441\)).
Therefore, 153, 257, and 408 are surely not perfect squares.
In simple words: Numbers ending in 2, 3, 7, or 8 are never perfect squares.

Exam Tip: Memorize the rule: perfect squares never end in 2, 3, 7, or 8.

Question 6. Find the square root of the following numbers by the prime factorisation method.
i.) 400 ii.) 9604 iii.) 8100 iv.) 1764 v.) 5929 vi.) 9216
Answer:
(i) \(400 = 2^4 \times 5^2 \implies \sqrt{400} = 2^2 \times 5 = 20\)
(ii) \(9604 = 2^2 \times 7^4 \implies \sqrt{9604} = 2 \times 7^2 = 98\)
(iii) \(8100 = 2^2 \times 3^4 \times 5^2 \implies \sqrt{8100} = 2 \times 3^2 \times 5 = 90\)
(iv) \(1764 = 2^2 \times 3^2 \times 7^2 \implies \sqrt{1764} = 2 \times 3 \times 7 = 42\)
(v) \(5929 = 7^2 \times 11^2 \implies \sqrt{5929} = 7 \times 11 = 77\)
(vi) \(9216 = 2^{10} \times 3^2 \implies \sqrt{9216} = 2^5 \times 3 = 32 \times 3 = 96\)
In simple words: Factorize completely, pair up identical factors, and take one from each pair.

Exam Tip: Take half the exponent of each prime factor to find its square root.

Question 7. For each of the following numbers, find the smallest whole number by which it should be multiplied 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.) 2028 v.) 1458 vi.) 768
Answer:
(i) \(252 = 2^2 \times 3^2 \times 7\). Multiplier = 7. New number = 1764, Square root = 42.
(ii) \(2925 = 3^2 \times 5^2 \times 13\). Multiplier = 13. New number = 38025, Square root = 195.
(iii) \(396 = 2^2 \times 3^2 \times 11\). Multiplier = 11. New number = 4356, Square root = 66.
(iv) \(2028 = 2^2 \times 3 \times 13^2\). Multiplier = 3. New number = 6084, Square root = 78.
(v) \(1458 = 2 \times 3^6\). Multiplier = 2. New number = 2916, Square root = 54.
(vi) \(768 = 2^8 \times 3\). Multiplier = 3. New number = 2304, Square root = 48.
In simple words: Multiply by the single unpaired prime factor.

Exam Tip: Always verify that the new product has even powers for all prime factors.

Question 8. For each of the following number, 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.) 180 iii.) 1008 iv.) 2028 v.) 1458 vi.) 768
Answer:
(i) \(252 = 2^2 \times 3^2 \times 7\). Divisor = 7. Quotient = 36, Square root = 6.
(ii) \(180 = 2^2 \times 3^2 \times 5\). Divisor = 5. Quotient = 36, Square root = 6.
(iii) \(1008 = 2^4 \times 3^2 \times 7\). Divisor = 7. Quotient = 144, Square root = 12.
(iv) \(2028 = 2^2 \times 13^2 \times 3\). Divisor = 3. Quotient = 676, Square root = 26.
(v) \(1458 = 2 \times 3^6\). Divisor = 2. Quotient = 729, Square root = 27.
(vi) \(768 = 2^8 \times 3\). Divisor = 3. Quotient = 256, Square root = 16.
In simple words: Divide by the unpaired prime factor to eliminate it.

Exam Tip: The divisor is identical to the multiplier for making a number a perfect square.

Question 9. The students of class viii of a school donated Rs 2401 in all, for prime minister's national relief fund. each student donated as many rupees as the no: of students in the class. Find the no: of students in the class.
Answer:
Let the number of students be \(x\).
Donation per student = Rs \(x\).
Total collection = \(x \times x = x^2\)
\(x^2 = 2401\)
\(x = \sqrt{2401} = 49\)
Number of students = 49.
In simple words: When things are arranged in rows and columns equally, or contribution equals total members, take the square root of the total.

Exam_Tip: Look for key phrases like "each gave as many as the total number" to immediately recognize square root word problems.

Question 10. 2116 plants are to be planted in a garden in such a way that each row contains as many plants as the no: of rows. Find the no: of rows and the no: of plants in each row.
Answer:
Let the number of rows be \(x\).
Number of plants in each row = \(x\).
Total plants = \(x^2 = 2116\)
\(x = \sqrt{2116} = 46\)
Number of rows = 46, and number of plants in each row = 46.
In simple words: Square root of total plants gives the number of rows and plants per row.

Exam Tip: Use the long division method to find the square root of 4-digit numbers quickly.

Question 11. Find the smallest square no: that is divisible by each of the no: 4, 9 and 10.
Answer:
First, find the LCM of 4, 9, and 10.
Prime factors: \(4 = 2^2\), \(9 = 3^2\), \(10 = 2 \times 5\).
\(\text{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\)
Prime factorization of 180 = \(2^2 \times 3^2 \times 5\).
The factor 5 is unpaired. To make it a square, multiply by 5.
Smallest square number = \(180 \times 5 = 900\)
In simple words: Find the LCM first, then multiply by the missing factor to make it a perfect square.

Exam Tip: "Divisible by each" always starts with LCM, and "smallest square" means make the LCM prime factors complete pairs.

Question 12. Find the smallest square no: that is divisible by each of the no: 8, 15 and 20.
Answer:
Prime factors: \(8 = 2^3\), \(15 = 3 \times 5\), \(20 = 2^2 \times 5\).
\(\text{LCM} = 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120\)
Prime factorization of 120 = \(2^3 \times 3 \times 5 = 2^2 \times 2 \times 3 \times 5\).
Unpaired factors are 2, 3, and 5. We must multiply by \(2 \times 3 \times 5 = 30\).
Smallest square number = \(120 \times 30 = 3600\)
In simple words: Find LCM, find which prime factors lack pairs, and multiply by them.

Exam Tip: Check that every prime factor in your LCM has an even exponent.

Question 13. Find the least no: that must be subtracted from 5607 so as to get a perfect square. Also find the square root of the perfect square.
Answer:
Using long division for 5607:
\(70^2 = 4900\)
\(75^2 = 5625\)
Let us test near 74:
\(74 \times 74 = 5476\)
Let us test 75:
\(75 \times 75 = 5625\)
By division method for 5607:
\(74^2 = 5476\) with a remainder of \(5607 - 5476 = 131\).
Therefore, 131 must be subtracted from 5607 to get the perfect square 5476.
Square root = 74.
In simple words: Find the remainder when doing long division; subtracting that remainder gives the square.

Exam Tip: For "what must be subtracted", always look at the remainder of the long division process.

Question 14. Find the greatest 4- digit no: which is a perfect square.
Answer:
The greatest 4-digit number is 9999.
Using long division on 9999, we find \(99^2 = 9801\) with a remainder of \(9999 - 9801 = 198\).
Greatest 4-digit perfect square = \(9999 - 198 = 9801\).
In simple words: Subtract the remainder of 9999 from itself.

Exam Tip: The greatest 4-digit square is always \(99^2\).

Question 15. Find the least no: that must be added to 1300 so as to get a perfect square. Also finds the square root of the perfect square.
Answer:
\(36^2 = 1296\) (less than 1300)
Next number is 37:
\(37^2 = 1369\)
Number to be added = \(1369 - 1300 = 69\).
Square root of the new number = 37.
In simple words: Find the next integer's square and subtract the given number from it.

Exam Tip: For "what must be added", find the square of (quotient + 1) and subtract the original number.

Question 16. Find the square root of 12.25.
Answer:
\(\sqrt{12.25} = \sqrt{\frac{1225}{100}} = \frac{\sqrt{1225}}{\sqrt{100}} = \frac{35}{10} = 3.5\)
In simple words: Convert the decimal to a fraction over 100, then take the square root of numerator and denominator.

Exam Tip: Count decimal places; the square root will have half as many decimal places.

Question 17. Area of a square plot is \(2304\,\text{m}^2\). find the side of the square plot. also find the square root of the perfect square.
Answer:
Area = \(\text{side}^2 = 2304\,\text{m}^2\)
Side = \(\sqrt{2304} = 48\,\text{m}\)
Square root of 2304 is 48.
In simple words: Side length is the square root of the area.

Exam Tip: Always include proper units like meters for side length.

Question 18. find the square root of each of the following no: by division method.
i.) 2304 ii.) 4489 iii.) 3481 iv.) 529 v.) 3249 vi.) 1369 vii.) 5776 viii.) 7921 ix.) 576 x.) 1024 xi.) 3136 xii.) 900
Answer:
(i) \(\sqrt{2304} = 48\)
(ii) \(\sqrt{4489} = 67\)
(iii) \(\sqrt{3481} = 59\)
(iv) \(\sqrt{529} = 23\)
(v) \(\sqrt{3249} = 57\)
(vi) \(\sqrt{1369} = 37\)
(vii) \(\sqrt{5776} = 76\)
(viii) \(\sqrt{7921} = 89\)
(ix) \(\sqrt{576} = 24\)
(x) \(\sqrt{1024} = 32\)
(xi) \(\sqrt{3136} = 56\)
(xii) \(\sqrt{900} = 30\)
In simple words: Pair digits from right to left and use long division.

Exam Tip: Group digits in pairs from the decimal point before starting long division.

Question 19. Find the square root of the following decimal no:
2.56 ii 7.29 iii. 51.84 iv 42.25 v 31.36
Answer:
(i) \(\sqrt{2.56} = 1.6\)
(ii) \(\sqrt{7.29} = 2.7\)
(iii) \(\sqrt{51.84} = 7.2\)
(iv) \(\sqrt{42.25} = 6.5\)
(v) \(\sqrt{31.36} = 5.6\)
In simple words: Ignore the decimal point, find the square root, then place the decimal point after one digit.

Question 20. Find at least no: which must be subtracted from each of the following no: 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
Answer:
(i) For 402: \(20^2 = 400\). Remainder = 2. Subtract 2, square root = 20.
(ii) For 1989: \(44^2 = 1936\), \(45^2 = 2025\). Using \(44^2 = 1936\), remainder = \(1989 - 1936 = 53\). Subtract 53, square root = 44.
(iii) For 3250: \(57^2 = 3249\). Remainder = \(3250 - 3249 = 1\). Subtract 1, square root = 57.
(iv) For 825: \(28^2 = 784\), \(29^2 = 841\). Using \(28^2 = 784\), remainder = \(825 - 784 = 41\). Subtract 41, square root = 28.
(v) For 4000: \(63^2 = 3969\). Remainder = \(4000 - 3969 = 31\). Subtract 31, square root = 63.
In simple words: Subtract the remainder obtained from long division.

Exam Tip: Remainder subtraction always yields the greatest square less than or equal to the number.

Question 21. Find the least no: which must be added to each of the following no: 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
Answer:
(i) For 525: \(22^2 = 484\), \(23^2 = 529\). Add \(529 - 525 = 4\). Square root = 23.
(ii) For 1750: \(41^2 = 1681\), \(42^2 = 1764\). Add \(1764 - 1750 = 14\). Square root = 42.
(iii) For 252: \(15^2 = 225\), \(16^2 = 256\). Add \(256 - 252 = 4\). Square root = 16.
(iv) For 1825: \(42^2 = 1764\), \(43^2 = 1849\). Add \(1849 - 1825 = 24\). Square root = 43.
(v) For 6412: \(80^2 = 6400\), \(81^2 = 6561\). Add \(6561 - 6412 = 149\). Square root = 81.
In simple words: Find the next larger square and subtract the original number.

Exam Tip: Addition problems always require checking \((\text{root} + 1)^2\).

Question 22. Find the length of the side of a square whose area is \(441\,\text{m}^2\).
Answer:
Area = \(\text{side}^2 = 441\,\text{m}^2\)
Side = \(\sqrt{441} = 21\,\text{m}\)
In simple words: Side = \(\sqrt{\text{Area}}\).

Exam Tip: Remember units in final answers.

Question 23. In a right triangle ABC angle B = \(90^\circ\)
Answer:
In right triangle ABC with \(\angle B = 90^\circ\), the side opposite to angle B is AC, which is the hypotenuse. By Pythagoras theorem, \(AC^2 = AB^2 + BC^2\).
In simple words: AC is the hypotenuse because it faces the right angle at B.

Exam Tip: Always identify the hypotenuse first before applying Pythagoras theorem.

Question 24. a. If AB = 6 cm, BC = 8 cm, find AC      b. If AC = 13 cm, BC = 5 cm, find AB
Answer:
(a) In right triangle ABC, \(AC^2 = AB^2 + BC^2\)
\(AC^2 = 6^2 + 8^2 = 36 + 64 = 100\)
\(AC = \sqrt{100} = 10\,\text{cm}\)
(b) \(AC^2 = AB^2 + BC^2\)
\(13^2 = AB^2 + 5^2\)
\(169 = AB^2 + 25\)
\(AB^2 = 169 - 25 = 144\)
\(AB = \sqrt{144} = 12\,\text{cm}\)
In simple words: Use Pythagoras theorem: sum of squares of legs equals square of hypotenuse.

Exam Tip: Recognize Pythagorean triples like (6, 8, 10) and (5, 12, 13) to save time.

Question 25. A gardener has 1000 plants. He wants to plant this in such a way that the no: of rows and the no: of columns remain same. Find the minimum no: of plants he needs more for this.
Answer:
\(31^2 = 961\) (less than 1000)
\(32^2 = 1024\)
Minimum plants needed more = \(1024 - 1000 = 24\).
In simple words: Find the next complete square greater than 1000 and find the difference.

Exam Tip: "Minimum plants needed more" is an addition-type square root problem.

Question 26. There are 500 children in a school for a P.Ed. drill. They have to stand in such a manner that the no. of rows is equal to the no: of columns. How many children would be left out in this arrangement? [least no :]
Answer:
\(22^2 = 484\) (less than 500)
\(23^2 = 529\)
Using \(22^2 = 484\), the number of children accommodated is 484.
Children left out = \(500 - 484 = 16\).
In simple words: Find the remainder by subtracting the largest square less than 500.

Exam Tip: "Left out" means find the remainder using subtraction from the number.

Worksheet - 2 Chapter 2 - Squares and Square root

Question 1. A natural number is called a ____________ or ___________ if it is the square of some natural number.
Answer: square number or perfect square.
In simple words: A square number is the product of a natural number multiplied by itself.

Exam Tip: Memorize basic definitions for fill-in-the-blank questions.

Question 2. \(\sqrt{0.09} =\) _________ .
Answer: \(0.3\)
In simple words: \(0.3 \times 0.3 = 0.09\).

Exam Tip: Remember that \(\sqrt{0.09}\) is 0.3, not 0.03.

Question 3. \(\frac{1}{\sqrt{0.09}} \times \sqrt{5.76} =\) ________ .
Answer:
\(\sqrt{0.09} = 0.3\)
\(\sqrt{5.76} = 2.4\)
\(\frac{1}{0.3} \times 2.4 = \frac{2.4}{0.3} = \frac{24}{3} = 8\)
In simple words: Simplify each square root first, then perform division and multiplication.

Exam Tip: Convert decimals to whole number ratios to divide easily.

Question 4. \(121^2 - 120^2 =\) ___________ .
Answer:
Using identity \(a^2 - b^2 = (a - b)(a + b)\):
\(121^2 - 120^2 = (121 - 120)(121 + 120) = 1 \times 241 = 241\)
In simple words: Difference of consecutive squares is equal to the sum of the numbers.

Exam Tip: Shortcut: \((n+1)^2 - n^2 = 2n + 1\), or \(121 + 120 = 241\).

Question 5. Is ( 1, 2, 3 ) a Pythagorean triplet ?
Answer:
Check: \(1^2 + 2^2 = 1 + 4 = 5\)
\(3^2 = 9\)
Since \(5 \neq 9\), (1, 2, 3) is not a Pythagorean triplet.
In simple words: Sum of squares of the two smaller numbers must equal the square of the largest number.

Exam Tip: Always test \(a^2 + b^2 = c^2\) where c is the largest number.

Question 6. Is 2352 a perfect square? If not, find the smallest number by which 2352 must be multiplied so that the product is a perfect square. Find the square root of the perfect square obtained.
Answer:
Prime factorization of \(2352 = 2^4 \times 3 \times 7^2\).
Since 3 is unpaired, it is not a perfect square.
Multiplier = 3.
New number = \(2352 \times 3 = 7056\).
Square root = 84.
In simple words: Multiply by 3 to pair the lonely factor 3.

Exam Tip: This is a repeated question from earlier in the document; consistency is key.

Question 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 perfect square obtained.
Answer:
Prime factorization of \(9408 = 2^6 \times 3 \times 7^2\).
Divisor = 3.
Quotient = 3136.
Square root = 56.
In simple words: Divide by 3 to eliminate the unpaired factor.

Exam Tip: Match divisors with the unpaired prime factor.

Question 8. The student of class VIII of a school donated Rs. 2401 for prime ministers 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.
Answer:
Number of students = \(\sqrt{2401} = 49\).
In simple words: Take the square root of 2401.

Exam Tip: Equate total amount to \(x^2\).

Question 9. Find the least number which must be subtracted from 18265 so as to get a perfect square.
Answer:
Using long division for 18265:
\(135^2 = 18225\)
\(136^2 = 18496\)
Let us check \(135\): \(135 \times 135 = 18225\).
Remainder = \(18265 - 18225 = 40\).
Therefore, 40 must be subtracted.
In simple words: Subtract the remainder to get the nearest smaller square.

Exam Tip: Long division remainder is your answer for subtraction problems.

Question 10. Find the least number which must be added to 893304 to obtain a perfect square.
Answer:
Using long division for 893304:
\(945^2 = 893025\)
\(946^2 = 894916\)
Number to be added = \(894916 - 893304 = 1612\).
In simple words: Find the next integer's square and subtract 893304 from it.

Exam Tip: For addition, use \((\text{root} + 1)^2\).

Question 11. Find the square root of 2.9 correct to two places of decimal.
Answer:
Using long division method for \(2.90 \, 00 \, 00\):
1. \(1 \times 1 \le 2 \implies\) root starts with 1, remainder 1.
2. Bring down 90 to get 190. Double 1 to get 2_ . \(27 \times 7 = 189\), remainder 1.
3. Bring down 00 to get 100. Double 17 to get 34_ . \(340 \times 0 = 0\), remainder 100.
4. Bring down another 00 to get 10000. Double 170 to get 340_ . \(3402 \times 2 = 6804\).
Continuing gives approximately 1.702.
Correct to two decimal places, \(\sqrt{2.9} \approx 1.70\).
In simple words: Calculate up to three decimal places and round off to two.

Exam Tip: Add pairs of trailing zeros to the decimal number before performing long division.

Question 12. Find the square root of \(11\frac{2}{3}\) correct to two places of decimal.
Answer:
Convert mixed fraction to improper fraction: \(11\frac{2}{3} = \frac{35}{3} = 11.6667\)
Find \(\sqrt{11.6667}\) using long division:
\(3.415 \dots\)
Rounded to two decimal places, the square root is 3.42.
In simple words: Convert to decimal first, then find the square root.

Exam Tip: Convert mixed fractions to decimals up to 4 decimal places before extracting the square root.

Question 13. Find the square root of \(21\frac{51}{169}\).
Answer:
Convert mixed fraction: \(21\frac{51}{169} = \frac{21 \times 169 + 51}{169} = \frac{3549 + 51}{169} = \frac{3600}{169}\)
Take square root: \(\sqrt{\frac{3600}{169}} = \frac{\sqrt{3600}}{\sqrt{169}} = \frac{60}{13} = 4\frac{8}{13}\)
In simple words: Convert to an improper fraction and take the square root of numerator and denominator separately.

Exam Tip: Always look for perfect squares in both numerator and denominator of mixed fractions.

Question 14. Find the least perfect square which is divisible by 5 , 6 and 8.
Answer:
Prime factors: \(5 = 5\), \(6 = 2 \times 3\), \(8 = 2^3\).
\(\text{LCM} = 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120\)
Prime factorization of 120 = \(2^3 \times 3 \times 5 = 2^2 \times 2 \times 3 \times 5\).
Unpaired factors are 2, 3, and 5. Multiply by \(2 \times 3 \times 5 = 30\).
Least perfect square = \(120 \times 30 = 3600\).
In simple words: Find LCM and multiply by missing prime factors to form pairs.

Exam Tip: Similar to Question 12 on page 2; LCM approach is standard.

Question 15. Find the smallest five digit number which is a perfect square.
Answer:
Smallest 5-digit number is 10000.
Notice that \(10000 = 100^2\).
Therefore, 10000 is already a perfect square and is the smallest 5-digit perfect square.
In simple words: 10000 is \(100 \times 100\), so it is the smallest 5-digit square.

Exam Tip: Recognize powers of 10 instantly.

Question 16. Find the other two members of a Pythagorean triplet, one of the number of which is 16.
Answer:
General form of Pythagorean triplet is \(2m\), \(m^2 - 1\), \(m^2 + 1\).
Let \(2m = 16 \implies m = 8\).
Second member = \(m^2 - 1 = 8^2 - 1 = 64 - 1 = 63\).
Third member = \(m^2 + 1 = 8^2 + 1 = 64 + 1 = 65\).
The other two members are 63 and 65.
In simple words: Set \(2m = 16\) to find m, then calculate \(m^2 - 1\) and \(m^2 + 1\).

Exam Tip: Always use \(2m\) as the even number given in the problem.

Question 17. Check, is ( 12 , 35, 37 ) a Pythagorean triplet?
Answer:
Check sum of squares of smaller numbers:
\(12^2 + 35^2 = 144 + 1225 = 1369\)
Square of the largest number:
\(37^2 = 1369\)
Since \(12^2 + 35^2 = 37^2\), (12, 35, 37) is a Pythagorean triplet.
In simple words: Verify that \(a^2 + b^2 = c^2\) holds true.