8th class Square Roots and Cube Roots

       Navodaya

Skip to conten

Exercise 6.1

Question 1.

What will be the units digit of the square of the following numbers?
(i) 39
(ii) 297
(iii) 5125
(iv) 7286
(v) 8742
Solution:

Number	Square of the units digIt	Units digit of a squared number
i) 39	92 = 9 x 9 = 81	1
ii) 297	72 = 7 x 7 = 49	9
Iii) 5125	52 = 5 x 5 = 25	5
iv) 7286	62 = 6 x 6 = 36	6
v) 8742	22 = 2 x 2 = 4	4

Question 2.
Which of the following numbers are perfect squares?
(i) 121
(ii) 136
(iii) 256

(iv) 321

(v) 600
Solution:

Number	Prime factorizatlon	Perfect square numbers Yes/No
i) 121	121 = 11 x 11 = 112	yes
ii) 136	136 = 8 x 17 = 2 x 2 x 2 x 17	No
iii) 256	256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 28 = (24)2	yes
iv) 321	321 = 3 x 107	No
v) 600	600 = 120 x 5 = 12 x 10 x 5 = 2 x 2 x 2 x 3 x 5 x 5	No

Question 3.
The following numbers are not perfect squares. Give reasons?
(i) 257
(ii) 4592
(iii) 2433
(iv) 5050
(v) 6098
Solution:
i) 257 → The units digit of the number is 7. So it is not a perfect square number.
ii) 4592 → The units digit of the number is 2. So it is not a perfect square number.
iii) 2433 → The units digit of the number is 3. So, it is not a perfect square number.
Iv) 5050 → The last two digits of the number are not two zero’s. So, it is not a perfect square number.
v) 6098 → The units digit of the number is 8. So It is not a perfect square number

Question 4.
Find whether the square of the following numbers are even or odd?
(i) 431
(ii) 2826
(iii) 8204
(iv) 17779
(v) 99998
Solution:

Number	Units digit of square of a number	Even / Odd
(i) 431	12 = 1	1 , odd
(ii) 2826	62 = 36	6, odd
(iii) 8204	42 = 16	6 , even
(iv)17779	92 = 81	1 , odd
(v) 99998	82 = 64	4 , even

Question 5.
How many numbers lie between the square of the following numbers
(i) 25; 26
(ii) 56; 57
(iii) 107;108
Solution:
The numbers lie between the square of the numbers are:
1) 25,26 → 2 x 25=50
ii) 56, 57 → 2 x 56 = 112
iii) 107, 108 → 2 x 107 = 214

Question 6.
Without adding, find the sum of the following numbers
(i) 1 + 3 + 5 + 7 + 9 =
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 =
Solution:
(i) 1 + 3 + 5 + 7 + 9 = (5)2 = 5 x 5 = 25
[∵ Sum of ‘n’ consecutive odd number = n2]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 92 = 9 x 9 = 81
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 132 = 13 x 13 = 169

Exercise 6.2

Question 1.
Find the square roots of the following numbers by Prime factorization method.
(1) 441
(ii) 784
(iii) 4096
(iv) 7056
Solution:

Question 2.
Find the smallest number by which 3645 must be multiplied to get a perfect square.
Solution:

The prime factorization of 3645
= (3 × 3) × (3 × 3) (3 × 3) × 5
∴ Deficiency of one ‘5’ is appeared in the above product.
∴ 3645 is multiplied with 5 then we will get a perfect square.

Question 3.
Find the smallest number by which 2400 is to be multiplied to get a perfect square and also find the square root of the resulting number.
Solution:
The prime factorization of 2400
=(2 × 2) × (2 × 2) × 2 × (5 × 5) × 3
∴ 2,3 are needed to form a pair
∴ 2 × 3 = 6
∴ 6 should be multiplied with 2400 then we will get a perfect square number.
∴ 2400 × 6 = 14400
∴ 14400−−−−−√ = 120
Question 4.

Find the smallest number by which 7776 is to be divided to get a perfect square.
Solution:
The prime factorization of 7776
=(2 × 2) × (2 × 2) × 2 × (3 × 3) × (3 × 3) × 3
∴ 2, 3 are needed to form a pair
∴ 2 × 3 = 6
∴ 7776 should be divided by 6 then we will get a perfect square number.

Question 5.

1521 trees are planted in a garden in such a way that there are as many trees in each row as there are rows in the garden. Find the number of rows and number of trees in each row.
Solution:
Let the no. of trees planted in a garden for each row = x say.
No. of rows in the garden = x
∴ Total no. of trees in the garden = x × x = x2
According to the sum x2 = 1521
x = 1521−−−−√=39×39−−−−−−√ = 39
∴ No. of trees for each row = 39
No. of rows in the garden = 39

Question 6.
A school collected ₹ 2601 as fees from its students. If fee paid by each student and number students in the school were equal, how many students were there in the school?
Solution:
Let the no. of students in a school = x say
The (amount) fee paid by each student = ₹ x
Amount collected by all the students
= x × x = x2
According to the sum
∴ x2 = 2601
x = 2601−−−−√=51×51−−−−−−√ = 51
∴ x = 51
∴ No. of students in the school = 51

Question 7.
The product of two numbers is 1296. If one number is 16 times the other, find the two numbers?
Solution:
Given that the product of two numbers = 1296.
Let the second number = x say
Then first number = 16 × x = 16x
∴ The product of two numbers
= x × 16x= 16x2
According to the sum
16x2 = 1296
⇒ x2 = 129616 = 81
⇒ x2 = 81
⇒ x = 8–√1¯¯¯=9×9−−−−√ = 9
⇒ x = 9
∴ The first number = 16x
= 16 × 9
=144
The second number = x = 9

Question 8.
7921 soldiers sat in an auditorium in such a way that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium’?
Solution:
Let the number of soldiers sat in an auditorium for each row = x say
∴ No. of rows in an auditorium = x
∴ Total no. of soldiers = x × x = x2
According to the sum,
x2 = 7921
x = 7921−−−−√=89×89−−−−−−√ = 89
∴ No. of rowS in an auditorium = 89

Question 9.
The area of a square field is 5184 m2. Find the area of a rectangular field, whose perimeter is equal to the perimeter of the square field and whose length is twice of its breadth.
Solution:
Area of a square field = 5184 m2
A = s2 = 5184
:. s = 5184−−−−√=72×72−−−−−−√ = 72
∴ s = 72
∴ Perimeter of the square field = 4 × s
= 4 × 72
= 288 m
According to the sum,
Perimeter of a rectangular field
= Perimeter of a square field = 288 m
Let the breadth of a rectangular field
= x m say
∴ Length = 2 × x = 2 × m
∴ Perimeter of the rectangular field
= 2 (1 + b)
= 2 (2x + x)
= 2 × 3x
= 6x
∴ 6x = 288 .
x = 2886
x = 48
∴ Breadth of the rectangular field
= x = 48 m
Length of the rectangular field = 2x
= 2 × 48
=96m
∴ Area of the rectangular field
= l × b
= 96 × 48
= 4608 m2

Exercise 6.3

Question 1.
Find the square roots of the following numbers by division method.
(i) 1089
(ii) 2304
(iii) 7744
(iv) 6084
(v) 9025
Solution:
(i) 1089

(ii) 2304

(iii) 7744

(iv) 6084

(v) 9025

Question 2.
Find the square roots of the following decimal numbers.
(i) 2.56
(ii) 18.49
(iii) 68.89
(iv) 84.64
Solution:
(i) 2.56

(ii) 18.49

(iii) 68.89

(iv) 84.64

Question 3.
Find the least number that is to be subtracted from 4000 to make it perfect square
Solution:
Square root of 4000 by
Division Method:

∴ The least number 31 should be subtracted from 4000 we will get a perfect square number4
∴ 4000 – 31 = 3969
= 3969−−−−√=63×63−−−−−−√ = 63

Question 4.
Find the length of the side of a square whose area is 4489 sq.cm.
Solution:
Area of a square (A) = 4489 sq.cms
A = s2
s2 = 4489
s = 4489−−−−√=67×67−−−−−−√ = 67cms.
∴ The side of a square (s) = 67cms.

Question 5.A gardener wishes to plant 8289 plants in the form of a square and found that there were 8 plants left. How many plants were planted in each row?

Solution:
No. of plants are planted = 8289 If 8289 plants are planted in a square shape, 8 plants are left.
Then remaining plants = 8289 – 8 = 8281

∴ No. of plants for each row = 91
∴ 8281 plants are planted in a square shape then no. of plants are planted for each row = 91

Question 6.
Find the least perfect square with four digits.
Solution:
The smallest number of 4 digits = 1000

∴ 24 should be added tö 1000 then 1000 + 24 = 1024
∴ The smallest 4 digited perfect square number is 1024.
[∵ 1024−−−−√ = 32]

Question 7.
Find the least number which must be added to 6412 to make it a perfect square?
Solution:

∴ The least number 149 should be added to 6412 then we will get a perfect square number.
∴ 6412 + 149 = 6561
∴ 6561−−−−√=81×81−−−−−−√ = 81

Question 8.
Estimate the value of the following numbers to the nearest whole number
(i) 97−−√
(ii) 250−−−√
(iii) 780−−−√
Solution:
ï) 97−−√ , 97 lie between the perfect
square numbers 81 and 100.
∴ 81 <97< 100
92 < 97 < 102
=9 < 97−−√ < 10
∴ 97−−√ Is nearest to 10.
[∵ 97 is nearest to 100]

ii) 250−−−√, 250 lie between the perfect square numbers 225 and 256.
∴ 225 < 250 < 256
152 < 250 < 162
= 15 < 250−−−√ <16
∴ 250−−−√ is nearest to 16.
[ ∵ 250 is nearest to 256]

ii) 780−−−√, 780 lie between the perfect
square numbers 729 and 784.
∴ 729 < 780 < 784
272 < 780 < 282
= 27< 780−−−√ <28
∴ 780−−−√ is nearest to 28.
[ ∵ 780 is nearest to 784]

Exercise 6.4

Question 1.
Find the cubes of the following numbers
(1) 8
(ii) 16
(iii) 21
(iv) 30
Solution:

Number	Cube Of a Number
i) 8	83 =  8 × 8 × 8 = 512
ii) 16	163 = 16 × 16 × 16 = 4096
iii) 21	213 = 21 × 21 × 21 = 9261
iv) 30	303 = 30 × 30 × 30 = 27000

Question 2.
Test whether the given numbers are perfect cubes or not.
(i) 243
(ii) 516
(iii) 729
(iv) 8000
(v)2700
Solution:

Number	Cube Of a Number	Yes / No
i) 243	3 × 3 × 3 × 3 × 3 = 35	No
ii) 516	2 × 2 × 3 × 43	No
iii) 729	9 × 9 × 9 = 93	Yes
iv) 8000	20 × 20 × 20 = (20)3	Yes
v) 2700	(30) × (30) × 3	No

Question 3.
Find the smallest number by which 8788 must be multiplied to obtain a perfect cube?
Solution:
The prime factorisation of 8788
= (2 × 2) × (13 × 13 × 13)

∴ From the above product 2 ¡s left in the triplet.
∴ 2 should be multiplied with 8788 we will get a perfect cube number.

Question 4.
What smallest number should 7803 be multiplied with so that the product becomes a perfect cube?
Solution:
The prime factorisation of 7803
= (3 × 3 × 3) × (17 × 17)

∴ From the above product 17 is left in
the triplet.
∴ 17 should be multiplied to 7803 then we will get a perfect cube number.

Question 5.
Find the smallest number by which 8640 must be divided so that the quotient is a perfect cube’?
Solution:
The prime factorisation of 8640
= (2 × 2 × 2) × (2 × 2 × 2) × 5 × (3 × 3 × 3)
= (2)3 × (2)3 × 5 × (3)3

Question 6.
Ravi made a cuboid of plasticine ofdimensions 12cm, 8cm and 3cm. How many minimum number of such cuboids will be needed to form a cube’?
Solution:
The volume of a plasticine cuboid
= l × b × h
= 12 × 8 × 3
= 288 cm3
If the minimum no. of such cuboids will be needed to form a cube then its volume be less than 288 i.e., 216 cm3
∴ s3 = 216
s = 216−−−√3=6×6×6−−−−−−−√3=63−−√3 = 6
∴ The side of the cube 6 cm

Question 7.
Find the smallest prime number dividing the sum 311 +513.
Solution:
The units digit in 311 is 7

∴ The units digit in 311 is 7
The units digit in 513 is 5
7 + 5 = 12 is divided by a smallest prime number 2.
∴ The smallest prime number that divide the sum 311 + 513 = 2