8th Maths 6.3

Chapter 6 Squares and Square Roots

NCERT Class 8th solution of Exercise 6.1

NCERT Class 8th solution of Exercise 6.2

NCERT Class 8th solution of Exercise 6.4

Exercise 6.3

$text {Square root you must learn.}$

$1^2=sqrt{1}=1$

$2^2=sqrt{4}=2$

$3^2=sqrt{9}=3$

$4^2=sqrt{16}=4$

$5^2=sqrt{25}=5$

$6^2=sqrt{36}=6$

$7^2=sqrt{49}=7$

$8^2=sqrt{64}=8$

$9^2=sqrt{81}=9$

$10^2=sqrt{100}=10$

$11^2=sqrt{121}=11$

$12^2=sqrt{144}=12$

$13^2=sqrt{169}=13$

$14^2=sqrt{196}=14$

$15^2=sqrt{225}=15$

$16^2=sqrt{256}=16$

$17^2=sqrt{289}=17$

$18^2=sqrt{324}=18$

$19^2=sqrt{361}=19$

$20^2=sqrt{400}=20$

$21^2=sqrt{441}=21$

$22^2=sqrt{484}=22$

$23^2=sqrt{529}=23$

$24^2=sqrt{756}=24$

$25^2=sqrt{625}=25$

$26^2=sqrt{676}=26$

$27^2=sqrt{729}=27$

$28^2=sqrt{224}=28$

$29^2=sqrt{841}=29$

$30^2=sqrt{900}=30$

$31^2=sqrt{961}=31$

$32^2=sqrt{1024}=32$

$33^2=sqrt{1089}=33$

$34^2=sqrt{1156}=34$

$35^2=sqrt{1225}=35$

$36^2=sqrt{1296}=36$

$37^2=sqrt{1369}=37$

$38^2=sqrt{1444}=38$

$39^2=sqrt{1521}=39$

$40^2=sqrt{1600}=40$

$41^2=sqrt{1681}=41$

$42^2=sqrt{1764}=42$

$43^2=sqrt{1849}=43$

$44^2=sqrt{1936}=44$

$45^2=sqrt{2025}=45$

$46^2=sqrt{2116}=46$

$47^2=sqrt{2209}=47$

$48^2=sqrt{2304}=48$

$49^2=sqrt{2401}=49$

$50^2=sqrt{2500}=50$

Q1. what will be the unit digit of the squares of the following nu

Q1. What could be the possible 'one's' digits of the square root of each of the following numbers?

$9801$

ii)

$99856$

iii)

$998001$

iv)

$657666025$

$text{Answer:}$

$text{One's digits of the square root of 9801 }$

$text{maybe 1 or 9.}$

ii)

$text{One's digits of the square root of 99856 }$

$text{maybe 4 or 6.}$

iii)

$text{One's digits of the square root of 998001 }$

$text{maybe 1 or 9.}$

iv)

$text{One's digits of the square root of 65666025 }$

$text{only 5.}$

Q2. Without doing any calculation, find the numbers which are surely not perfect squares.

$153$

ii)

$257$

iii)

$408$

iv)

$441$

$text{Answer:}$

$text{We know that}$

$text{Unit digits of all perfect square numbers}$

$text{ are 0, 1, 4, 5, and 9}$

$text{Unit digit of 153 is 3 so, it is not a perfect square.}$

ii)

$text{Unit digit of 257 is 7 so, it is not a perfect square.}$

iii)

$text{Unit digit of 408 is 8 so, it is not a perfect square.}$

iv)

$text{Unit digit of 441 is 1 so, it is a perfect square.}$

Q3. Find the square roots of

$100$ and

$169$ by the method of repeated subtraction.

$text{Sol.}$

$text{We use the method of repeated subtraction of}$

$text{consecutive odd numbers}$

$1) 100 - 1 = 99$

$2) 99 - 3 = 96$

$3) 96 - 5 = 91$

$4) 91 - 7 = 84$

$5) 84 - 9 = 75$

$6) 75 - 11 = 64$

$7) 64 - 13 = 51$

$8) 51 - 15 = 36$

$9) 36 - 17 = 19$

$10) 19 - 19 = 00$

$text{Ten times repeat}$

$text{Therefore }sqrt100 = 10$

$text{Answer:}$

$10$

ii)

$text{We use the method of repeated subtraction of}$

$text{consecutive odd numbers}$

$1) 169 - 1 = 168$

$2) 168 - 3 = 165$

$3) 165 - 5 = 160$

$4) 160 - 7 = 153$

$5) 153 - 9 = 144$

$6) 144 - 11 = 133$

$7) 133 - 13 = 120$

$8) 120 - 15 = 105$

$9) 105 - 17 = 88$

$10) 88 - 19 = 69$

$11) 69 - 21 = 48$

$12) 48 - 23 = 25$

$13) 25 - 25 = 00$

$text{Thirteen times repeat}$

$text{Therefore }sqrt169 = 13$

$text{Answer:}$

$13$

Q4. Find the square roots of the following numbers by the Prime Factorisation Method.

$729$

ii)

$400$

iii)

$1764$

iv)

$4096$

$7744$

vi)

$9604$

vii)

$5929$

viii)

$9216$

ix)

$529$

$8100$

$text{Sol.}$

$text{By Prime Factorisation Method}$

$729 = underline{3times3}times underline{3times3}times underline{3times3}$

$sqrt729 = 3times3times3$

$sqrt729 = 27$

$text{Answer:}$

$27$

ii)

$text{By Prime Factorisation Method}$

$400 = underline{2times2}times underline{2times2}times underline{5times5}$

$sqrt400 = 2times2times5$

$sqrt400 = 20$

$text{Answer:}$

$20$

iii)

$text{By Prime Factorisation Method}$

$1764 = underline{2times2}times underline{21times21}$

$sqrt1764 = 2times21$

$sqrt1764 = 42$

$text{Answer:}$

$42$

iv)

$text{By Prime Factorisation Method}$

$4096 = underline{2times2}times underline{2times2}$

$timesunderline{2times2}times$

$underline{2times2}$

$times underline{2times2}times underline{2times2}$

$sqrt4096 = 2times2times2times2times2times2$

$sqrt4096 = 64$

$text{Answer:}$

$64$

$text{By Prime Factorisation Method}$

$7744 = underline{2times2}times underline{2times2}$

$timesunderline{2times2}times$

$underline{11times11}$

$sqrt7744 = 2times2times2times11$

$sqrt7744 = 88$

$text{Answer:}$

$88$

vi)

$text{By Prime Factorisation Method}$

$9604 = underline{2times2}times underline{7times7}$

$timesunderline{7times7}$

$sqrt4096 = 2times7times7$

$sqrt4096 = 98$

$text{Answer:}$

$98$

vii)

$text{By Prime Factorisation Method}$

$5929 = underline{7times7}times underline{11times11}$

$sqrt5929 = 7times11$

$sqrt5929 = 77$

$text{Answer:}$

$77$

viii)

$text{By Prime Factorisation Method}$

$9216 = underline{2times2}times underline{2times2}$

$timesunderline{2times2}times$

$underline{2times2}$

$times underline{2times2}times underline{3times3}$

$sqrt9216 = 2times2times2times2times2times3$

$sqrt9216 = 96$

$text{Answer:}$

$96$

ix)

$text{By Prime Factorisation Method}$

$529 = underline{23times23}$

$sqrt529 = 23$

$text{Answer:}$

$23$

$text{By Prime Factorisation Method}$

$8100 = underline{2times2}times underline{3times3}$

$timesunderline{3times3}times$

$underline{5times5}$

$sqrt8100= 2times3times3times5$

$sqrt8100 = 90$

$text{Answer:}$

$90$

Q5. 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.

$252$

ii)

$180$

iii)

$1008$

iv)

$2028$

$1458$

vi)

$768$

$text{Sol.}$

$text{By Prime Factorisation Method}$

$252 = underline{2times2}times underline{3times3}$

$times7$

$text{As the prime factor 7 has no pair}$

$text{So, we multiply 252 by 7}$

$252times 7= underline{2times2}times underline{3times3}$

$timesunderline{7times7}$

$sqrt1764 = 2times3times7$

$sqrt1764 = 42$

$text{Answer:}$

$7; 42$

ii)

$text{By Prime Factorisation Method}$

$180 = underline{2times2}times underline{3times3}$

$times5$

$text{As the prime factor 5 has no pair}$

$text{So, we multiply 180 by 5}$

$180times5 = underline{2times2}times underline{3times3}$

$timesunderline{5times5}$

$sqrt900 = 2times3times5$

$sqrt900 = 30$

$text{Answer:}$

$5; 30$

iii)

$text{By Prime Factorisation Method}$

$1008 = underline{2times2}times underline{2times2}times underline{3times3}$

$times7$

$text{As the prime factor 7 has no pair}$

$text{So, we multiply 1008 by 7}$

$1008times 7= underline{2times2}timesunderline{2times2}times$

$underline{3times3}$

$timesunderline{7times7}$

$sqrt7056 = 2times2times3times7$

$sqrt7056 = 84$

$text{Answer:}$

$7; 84$

iv)

$text{By Prime Factorisation Method}$

$2028 = underline{2times2}times underline{3times3}$

$times$

$3timesunderline{13times13}$

$text{As the prime factor 3 has no pair}$

$text{So, we multiply 2028 by 3}$

$2028times3 = underline{2times2}timesunderline{3times3}$

$timesunderline{13times13}$

$sqrt6084 = 2times3times13$

$sqrt6084 = 78$

$text{Answer:}$

$3; 78$

$text{By Prime Factorisation Method}$

$1458 = 2times underline{3times3}times underline{3times3}$

$timesunderline{3times3}$

$text{As the prime factor 2 has no pair}$

$text{So, we multiply 1458 by 2}$

$1458times2 = underline{2times2}timesunderline{3times3}times$

$underline{3times3}$

$timesunderline{3times3}$

$sqrt2916 = 2times3times3times3$

$sqrt2916 = 54$

$text{Answer:}$

$2; 54$

vi)

$text{By Prime Factorisation Method}$

$768 = underline{2times2}times underline{2times2}times$

$underline{2times2}times3$

$text{As the prime factor 3 has no pair}$

$text{So, we multiply 768 by 3}$

$768times3 = underline{2times2}timesunderline{2times2}times$

$underline{2times2}$

$timesunderline{2times2}timesunderline{3times3}$

$sqrt2304 = 2times2times2times2times3$

$sqrt2304 = 48$

$text{Answer:}$

$3; 48$

Q6. 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.

$252$

ii)

$2925$

iii)

$396$

iv)

$2645$

$2800$

vi)

$1620$

$text{Sol.}$

$text{By Prime Factorisation Method}$

$252 = underline{2times2}times underline{3times3}$

$times7$

$text{As the prime factor 7 has no pair}$

$text{So, we divide 252 by 7}$

$252÷7 = underline{2times2}times underline{3times3}times7÷7$

$sqrt36 = underline{2times2}timesunderline{3times3}$

$sqrt36 = 2times3$

$sqrt36 = 6$

$text{Answer:}$

$7; 6$

ii)

$text{By Prime Factorisation Method}$

$2925 = underline{3times3}times underline{5times5}$

$times13$

$text{As the prime factor 13 has no pair}$

$text{So, we divide 2925 by 13}$

$2925÷13= underline{3times3}times underline{5times5}times13÷13$

$sqrt225 = underline{3times3}timesunderline{5times5}$

$sqrt225 = 3times5$

$sqrt225 = 15$

$text{Answer:}$

$13; 15$

iii)

$text{By Prime Factorisation Method}$

$396 = underline{2times2}times underline{3times3}$

$times11$

$text{As the prime factor 11 has no pair}$

$text{So, we divide 396 by 11}$

$396÷11= underline{2times2}times underline{3times3}times11÷11$

$sqrt36 = underline{2times2}timesunderline{3times3}$

$sqrt36 = 2times3$

$sqrt36 = 6$

$text{Answer:}$

$11; 6$

iv)

$text{By Prime Factorisation Method}$

$2645 = underline{2times2}times underline{3times3}$

$times5$

$text{As the prime factor 5 has no pair}$

$text{So, we divide 2645 by 5}$

$2645÷5= underline{23times23}times5÷5$

$sqrt529 = underline{23times23}$

$sqrt529 = 23$

$text{Answer:}$

$5; 23$

$text{By Prime Factorisation Method}$

$2800 = underline{2times2}timesunderline{2times2}timesunderline{5times5}$

$times7$

$text{As the prime factor 7 has no pair}$

$text{So, we divide 2800 by 7}$

$2800÷7= underline{2times2}times underline{2times2}timesunderline{5times5}times7÷7$

$sqrt400 = underline{2times2}timesunderline{2times2}timesunderline{5times5}$

$sqrt400 = 2times2times5$

$sqrt400 = 20$

$text{Answer:}$

$7; 20$

vi)

$text{By Prime Factorisation Method}$

$1620 = underline{2times2}times underline{3times3}times underline{3times3}$

$times5$

$text{As the prime factor 5 has no pair}$

$text{So, we divide 1620 by 5}$

$1620÷5= underline{2times2}times underline{3times3}timesunderline{3times3}times5÷5$

$sqrt324 = underline{2times2}timesunderline{2times2}timesunderline{3times3}$

$sqrt324 = 2times3times3$

$sqrt324 = 18$

$text{Answer:}$

$5; 18$

Q7. 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.

$text{Sol.}$

$text{All students of Class VIII of a school donated}$

$text{total money ₹ 2401}.$

$text{The number of students in class VIII }$

$= sqrt2401$

$text{By Prime Factorisation Method}$

$2401 = underline{7times7}timesunderline{7times7}$

$sqrt2401 = 7times7$

$sqrt2401 = 49$

$text{Answer:}$

$text{The number of students in class VIII is 49.}$

Q8.

$2025$ plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

$text{Sol.}$

$text{Total number of plants 2025}.$

$text{The number of rows and plants in each row }$

$= sqrt2025$ ,

$text{By Prime Factorisation Method}$

$2025 = underline{3times3}timesunderline{3times3}timesunderline{5times5}$

$sqrt2025 = 3times3times5$

$sqrt2025 = 45$

$text{Answer:}$

$text{The number of rows and plants in each row}$

$=45$ .

Q9. Find the smallest square number that is divisible by each of the numbers

$4, 9$ and

$10$ .

$text{Sol.}$

$text{LCM of 4, 9 and 10 is 180}$

$text{By Prime Factorisation Method}$

$180 = 2times2times3times3times5$

$text{As the prime factor 5 has no pair}$ .

$text{So, we multiply 180 by 5},$

$180times5 = 900$

$text{Answer:}$

$text{The smallest number is 900}$ .

Q10. Find the smallest square number that is divisible by each of the numbers

$8, 15$ and

$20$ .

$text{Sol.}$

$text{LCM of 8, 15 and 20 is 120}$

$text{By Prime Factorisation Method}$

$120 = 2times2times2times3times5$

$text{As the prime factor 2, 3, and 5 has no pair}$

$text{So, we multiply 120 by }2times3times5 = 30$

$120times30 = 3600$

$text{Answer:}$

$text{The smallest number is 3600}$ .

Search This Blog

8th Maths 6.3

Chapter 6

Squares and Square Roots

NCERT Class 8th solution of Exercise 6.1

NCERT Class 8th solution of Exercise 6.2

NCERT Class 8th solution of Exercise 6.4

Exercise 6.3

Comments

Popular posts from this blog

10th Maths Chapter 1

CBSE 10th and 12th Result

RSKMP 5th & 8th Result