Chapter 5 of Class 8 Mathematics, "Square and Square Roots," is an essential topic that helps students build a strong math foundation. This chapter covers the basics of square numbers and square roots, including their properties and some interesting patterns between them. Understanding square roots is important because they are the basis for many advanced calculations and algebraic expressions studied in higher classes.
The NCERT Solutions for Class 8 Maths Chapter 5 provides clear explanations and step-by-step solutions to textbook exercises, making it easier for students to understand and solve problems related to finding squares of big numbers by expansion techniques and other patterns. This blog is designed to help students master this chapter, boost their analytical skills, and gain confidence in math as they prepare for exams and future studies.
Downloading the NCERT Solutions of Class 8 Chapter 5 PDF will help you understand the curriculum's key concepts. Below is the link to download the NCERT solutions for Class 8 - Maths Chapter 5 PDF
NCERT Class 8 Maths Chapter 5 |
Exercise | Total Questions |
Exercise 5.1 | 9 Questions |
Exercise 5.2 | 2 Questions |
Exercise 5.3 | 10 Questions |
Exercise 5.4 | 9 Questions |
Sol. The unit digit of the squares of the given numbers is shown against the number in the following table:
S.No. | Number | Unit digit in the square of the number | Reason |
(i) | 81 | 1 | 1×1=1 |
(ii) | 272 | 4 | 2×2=4 |
(iii) | 799 | 1 | 9×9=81 |
(iv) | 3853 | 9 | 3×3=9 |
(v) | 1234 | 6 | 4×4=16 |
(vi) | 26387 | 9 | 7×7=49 |
(vii) | 52698 | 4 | 8×8=64 |
(viii) | 99880 | 0 | 0×0=0 |
(ix) | 12796 | 6 | 6×6=36 |
(x) | 55555 | 5 | 5×5=25 |
Sol. A number that ends either with 2,3,7 or 8 cannot be a perfect square. Also, a number that ends with odd number of zero(s) cannot be a perfect square. (i) Since the given number 1057 ends with 7, so it cannot be a perfect square. (ii) Since the given number 23453 ends with 3 , so it cannot be a perfect square. (iii) Since the given number 7928 ends with 8, so it cannot be a perfect square. (iv) Since the given number 222222 ends with 2 , so it cannot be a perfect square. (v) Since the given number 64000 ends with odd number of ' 0 ', so it cannot be a perfect square. (vi) Since the given number 89722 ends with 2, so it cannot be a perfect square. (vii) Since the given number 222000 ends with odd number of ' 0 ', so it cannot be a perfect square. (viii) Since the given number 505050 ends with odd number of ' 0 ', so it cannot be a perfect square.
3. The squares of which of the following would be odd numbers? (i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Sol. (i) The given number 431 is odd, so its square must be odd. (ii) The given number 2826 is even, so its square must be even. (iii) The given number 7779 is odd, so its square must be odd. (iv) The given number 82004 is even, so its square must be even.
4. Observe the following pattern and find the missing digits:
112=1211012=1020110012=10020011000012=1…….................. .
Sol. The missing digits are as under: 1000012=10000200001 100000012=100000020000001
5. Observe the following pattern and find the missing numbers:
112=1211012=10201101012=10203020110101012=…...........2=10203040504030201
Sol. 10101012=1020304030201 1010101012=10203040504030201
6. Using the given pattern, find the missing numbers: 12+22+22=32 22+32+62=72 32+42+122=132 42+52+ _ 2=212 52+…+302=312 62+72+2=2 Sol. The missing numbers are us under: 42+52+202=212 52+62+302=312 62+72+422=432
7. Without adding, find the sum: (i) 1+3+5+7+9 (ii) 1+3+5+7+9+11+13+15+17+19 (iii) 1+3+5+7+9+11+13+15+17+9+21+23
Sol. (i) 1+3+5+7+9= Sum of first 5 odd numbers =52=25 (ii) 1+3+5+7+9+11+13+15+17+19 = Sum of first 10 odd numbers =102=100 (iii) 1+3+5+7+9+11+13+15+17+19+21 +23= Sum of first 12 odd numbers =122= 144
8. (i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers.
Sol. (i) 49=72=1+3+5+7+9+11+13 (ii) 121=112=1+3+5+7+9+11+13+15+17+19+21
9. How many numbers lie between squares of the following numbers? (i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Sol. (i) Between 122(=144) and 132(=169) there are twenty four (i.e., 2×12 ) numbers. (ii) Between 252(=625) and 262(=676), there are 50 (i.e. 2×25 ) numbers. (iii) Between 992(=9801) and 1002(= 10000), there are '198' (i.e. 2×99 ) numbers.
Sol. (i) 322=(30+2)2
=30(30+2)+2(30+2)=302+30×2+2×30+22=900+60+60+4=1024
(ii) 352=(30+5)2=30(30+5)+5(30+5) =302+30×5+5×30+52 =900+150+150+25=1225 (iii) 862=(80+6)2 =80(80+6)+6(80+6) =802+80×6+6×80+62 =6400+480+480+36=7396 (iv) 933=(90+3)2 =90(90+3)+3(90+3) =902+90×3+3×90+32 =8100+270+270+9=8649 (v) 712=(70+1)2 =70(70+1)+1(70+1) =702+70×1+1×70+12 =4900+70+70+1=5041 (vi) 462=(40+6)2 =40(40+6)+6(40+6) =402+40×6+6×40+62 =1600+240+240+36=2116
2. Write a Pythagorean triplet whose one member is (i) 6 (ii) 14 (iii) 16 (iv) 18
Sol. (i) 2 m=6⇒ m=3 Put m=3 in m2−1,m2+1 ∴2 m=6, m2−1=32−1=9−1=8 and m2+1=9+1=10 Thus, 6, 8 and 10 is Pythagorean triplets. (ii) 2 m=14⇒ m=7
Put m=7 in m2−1,m2+1 ∴2 m=14, m2−1=72−1=49−1=48 and m2+1=72+1=49+1=50 Thus, 14, 48 and 50 is Pythagorean triplets. (iii) 2 m=16⇒ m=8
Put m=8 in m2−1& m2+1 ∴2 m=16, m2−1=82−1=64−1=63 and m2+1=82+1=64+1=65 Thus 16, 63, and 65 is Pythagorean triplets. (iv) 2 m=18⇒ m=9
Put m=9 in m2−1 and m2+1 ∴2 m=18, m2−1=92−1=81−1=80 And m2+1=92+1=81+1=82 Thus 18, 80 and 82 is Pythagorean triplets.
Sol. The possible one's digits of the square root of the numbers:
S.no. | Numbers | Possible one's digits of square root |
(i) | 9801 | 1 or 9 |
(ii) | 99856 | 4 or 6 |
(iii) | 998001 | 1 or 9 |
(iv) | 657666025 | 5 |
Sol. We know that a number ending in 2,3,7 or 8 is never a perfect square. Therefore, (i) 153 is not a perfect square. (ii) 257 is not a perfect square. (iii) 408 is not a perfect square (iv) 441 may be a perfect square.
So, 153, 257 and 408 are surely not perfect squares.
3. Find the square roots of 100 and 169 by the method of repeated subtraction. Sol. From 100, we subtract successive odd numbers starting from 1 as under. 100-1 = 99 99−3=96 96−5=91 91−7=84 84−9=75 75−11=64 64−13=51 51−15=36 36−17=19 19-19 = 0 and obtain 0 at 10th step. ∴100=10 From 169, we subtract successive odd numbers starting from 1 as under.
169−1=168168−3=165165−5=160160−7=153153−9=144144−11=133133−13=120120−15=105105−17=8888−19=6969−21=4848−23=2525−25=0
and obtain 0 at 13th step. ∴169=13 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
Sol. (i) By prime factorisation, we get
3 | 729 |
3 | 243 |
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
1 |
729=3×3×3×3×3×3 ∴729=3×3×3=27 (ii) By prime factorisation, we get
2 | 400 |
2 | 200 |
2 | 100 |
2 | 50 |
5 | 25 |
5 | 5 |
1 |
400=2×2×2×2×5×5 ∴400=2×2×5=20 (iii) By prime factorisation, we get
2 | 1764 |
2 | 882 |
3 | 441 |
3 | 147 |
7 | 49 |
7 | 7 |
1 |
1764=2×2×3×3×7×7 ∴1764=2×3×7=42 (iv) By prime factorisation, we get
2 | 4096 |
2 | 2048 |
2 | 1024 |
2 | 512 |
2 | 256 |
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
1 |
4096=2×2×2×2×2×2×2×2×2×2×2×2 ∴4096=2×2×2×2×2×2 (vi) By prime factorisation, we get
2 | 9604 |
2 | 4802 |
7 | 2401 |
7 | 343 |
7 | 49 |
7 | 7 |
1 |
9604=2×2×7×7×7×7 ∴9604=2×7×7=98 (vii) By prime factorisation, we get
7 | 5929 |
7 | 847 |
11 | 121 |
11 | 11 |
1 |
5929=7×7×11×11 ∴5929=7×11=77 (viii) By prime factorisation, we get
2 | 9216 |
2 | 4608 |
2 | 2304 |
2 | 1152 |
2 | 576 |
2 | 288 |
2 | 144 |
2 | 72 |
2 | 36 |
2 | 18 |
3 | 9 |
3 | 3 |
1 |
9216=2×2×2×2×2×2×2×2×2×2×3×3 ∴9216=2×2×2×2×2×3 =96 (ix) By prime factorisation, we get
23 | 529 |
23 | 23 |
1 |
529=23×23 ∴529=23 (x) By prime factorisation, we get
2 | 8100 |
2 | 4050 |
3 | 2025 |
3 | 675 |
3 | 225 |
3 | 75 |
5 | 25 |
5 | 5 |
1 |
8100=2×2×3×3×3×3×5×5 ∴8100=2×3×3×5=90
2 | 252 |
2 | 126 |
3 | 63 |
3 | 21 |
7 | 7 |
1 |
252=2×2×3×3×7 It is clear that in order to get a perfect square, one more 7 is required. So, the given number should be multiplied by 7 to make the product a perfect square. ∴252×7=1764 is a perfect square. Thus, 1764=2×2×3×3×7×7 ∴1764=2×3×7=42 (ii) By prime factorisation, we get
2 | 180 |
2 | 90 |
3 | 45 |
3 | 15 |
5 | 5 |
1 |
180=2×2×3×3×5 It is clear that in order to get a perfect square, one more 5 is required. So, the given number should be multiplied by 5 to make the product a perfect square. ∴180×5=900 is a perfect square. Thus, 900=2×2×3×3×5×5 ∴900=2×3×5=30 (iii) By prime factorisation, we get
2 | 1008 |
2 | 504 |
2 | 252 |
2 | 126 |
3 | 63 |
3 | 21 |
7 | 7 |
1 |
1008=2×2×2×2×3×3×7 It is clear that in order to get a perfect square, one more 7 is required. So, the given number should be multiplied by 7 to make the product a perfect square. ∴1008×7=7056 is a perfect square. Thus, 7056=2×2×2×2×3×3×7×7 ∴7056=2×2×3×7=84 (iv) By prime factorisation, we get
2 | 2028 |
2 | 1014 |
3 | 507 |
3 | 169 |
13 | 13 |
1 |
2028=2×2×3×13×13 It is clear that the given number should be multiplied by 3 to make the product a perfect square ∴2028×3=6084 is a perfect square Thus 6084=2×2×3×3×13×13 ∴6084=2×3×13=78 (v) By prime factorisation, we get 1458=2×3×3×3×3×3×3 It is clear that the given number should be multiplied by 2 to make the product a perfect square. ∴1458×2=2916 is a perfect square. Thus 2916=2×2×3×3×3×3×3×3 ∴2916=2×3×3×3=54 (vi) By prime factorisation, we get
2 | 768 |
2 | 384 |
2 | 192 |
2 | 96 |
2 | 48 |
2 | 24 |
2 | 12 |
2 | 6 |
3 | 3 |
1 |
768=2×2×2×2×2×2×2×2×3 It is clear that the given number should be multiplied by 3 to make the product a perfect square. ∴768×3=2304 2304=2×2×2×2×3=48
Sol. (i) By prime factorisation, we get
2 | 252 |
2 | 126 |
3 | 63 |
3 | 21 |
7 | 7 |
1 |
252=2×2×3×3×7 Since the prime factor 7 cannot be paired. ∴ The given number should be divided by ∴7252=72×2×3×3×7=2×2×3×3 =36 is a perfect square and, 36=2×2×3×3 =2×3=6 (ii) By prime factorisation, we get
3 | 2925 |
3 | 975 |
5 | 325 |
5 | 65 |
13 | 13 |
1 |
2925=3×3×5×5×13 Since the prime factor 13 cannot be paired. ∴ The given number should be divided by 13 . ∴132925=133×3×5×5×13 =3×3×5×5=225 is a perfect square and, 225=3×3×5×5 =3×5=15 (iii) By prime factorisation, we get
2 | 396 |
2 | 198 |
3 | 99 |
3 | 33 |
11 | 11 |
1 |
396=2×2×3×3×11 Since the prime factor 11 cannot be paired so 396 should be divided by 11 . 11396=36 is a perfect square. and 36=2×2×3×3=2×3=6 (iv) By prime factorisation, we get
5 | 2645 |
23 | 529 |
23 | 23 |
1 |
2645=5×23×23 Since the prime factor 5 cannot be paired, so 2645 should be divided by 5 . 52645=529 is a perfect square and 529=23×23=23 (v) By prime factorisation, we get
2 | 2800 |
2 | 1400 |
2 | 700 |
2 | 350 |
5 | 175 |
5 | 35 |
7 | 7 |
1 |
2800=2×2×2×2×5×5×7 Since the prime factor 7 cannot be paired, so 2800 should be divided by 7 . =72800 =400 is a perfect square and 400=20 (vi) By prime factorisation, we get
2 | 1620 |
2 | 810 |
3 | 405 |
3 | 135 |
3 | 45 |
3 | 15 |
5 | 5 |
1 |
1620=2×2×3×3×3×3×5 Since the prime factor 5 cannot be paired, so 1620 should be divided by 5 . 51620 =324 is a perfect square. and 324=18
7 | 2401 |
7 | 343 |
7 | 49 |
7 | 7 |
1 |
=7×7×7×7 =7×7=49 ⇒x=3×3×3×3×5×5 ⇒x=3×3×5 ⇒x=45 number of rows as well as number of plants.
2 | 8 | 15 | 20 |
2 | 4 | 15 | 10 |
2 | 2 | 15 | 5 |
3 | 1 | 15 | 5 |
5 | 1 | 5 | 5 |
1 | 1 | 1 |
=2×2×2×3×5=120 To make 120 a perfect square, we will multiply this number by 2×3×5. So the number will be 2×2×2×2×3×3×5×5=3600 Hence, 3600 is the smallest square number, which is divisible by 8,15 and 20.
NCERT Solutions for Class 8 Maths Other Chapters:- |
Chapter 5: Squares and Square Roots |
NCERT Solutions for Class 8 Maths Chapter 5 provides detailed and step-by-step solutions to all of the questions. When taking a board exam, students must submit a clear, well-structured paper to receive the highest possible score. These answers provide students with a solid grasp of the subject at hand and serve as guidelines for them to follow.
Yes, practising all the questions is beneficial as they cover various algebraic concepts related to Square and Square Roots. Start by solving the examples to understand the steps involved in solving square root-related problems.
The properties of square numbers, Pythagorean triplets, and methods of finding square numbers through various techniques are crucial concepts in class 8, Chapter 5, squares and square roots. Exam questions covering all three of these topics carry a lot of weight, so students should prepare thoroughly. To fully understand this topic, they can also try the optional exercises and solved examples.
NCERT Class 8 Chapter 5 Squares and Square Roots contains 30 questions in total. In addition to the exercise questions, students can practice the solved examples to gain a conceptual understanding.
(Session 2025 - 26)
Linear Equations in One Variable is the second chapter of Class 8 Maths. It covers various topics, including Introduction, Solving Equations having the Variable on both Sides, Reducing Equations to Simpler Form.
In class 8 quadrilaterals, students are exposed to different types of polygons mostly concentrated on four-sided figures.
Mensuration is the ninth chapter of Class 8 Maths. It covers various topics, including an introduction, area of polygon, solid shapes, surface area of cube, cuboid and cylinder, volume of cube, cuboid and cylinder, volume and capacity.
This chapter introduces students to the principles of algebraic expressions, their operations, and numerous standard identities.
Chapter 10 of Class 8 Maths, Exponents and Powers, provides a comprehensive understanding of key concepts such as an introduction to exponents, handling powers with negative exponents, and the important laws of exponents.
Friction chapter class 8, explains friction and examines the reasons for it and its effects on various surfaces.