## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7

Other Exercises

Find the square root of the following numbers in decimal form :

Question 1.
84.8241
Solution:

Question 2.
0.7225
Solution:

Question 3.
0.813604
Solution:

Question 4.
0.00002025
Solution:

Question 5.
150.0625
Solution:

Question 6.
225.6004
Solution:

Question 7.
3600.720036
Solution:

Question 8.
236.144689
Solution:

Question 9.
0.00059049
Solution:

Question 10.
176.252176
Solution:

Question 11.
9998.0001
Solution:

Question 12.
0.00038809
Solution:

Question 13.
What is that fraction which when multiplied by itself gives 227.798649 ?
Solution:

Question 14.
square playground is 256.6404 square metres. Find the length of one side of the playground.
Solution:
Area of square playground = 256.6404 sq. m

Question 15.
What is the fraction which when multiplied by itself gives 0.00053361 ?
Solution:

Question 16.
Simplify :

Solution:

Question 17.
Evaluate $$\sqrt { 50625 }$$ and hence find the value of $$\sqrt { 506.25 } +\sqrt { 5.0625 }$$.
Solution:

Question 18.
Find the value of $$\sqrt { 103.0225 }$$ and hence And the value of

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6

Other Exercises

Question 1.
Find the square root of:

Solution:

Question 2.
Find the value of :

Solution:

Question 3.
The area of a square field is 80 $$\frac {244 }{ 729 }$$ square metres. Find the length of each side of the field.
Solution:

Question 4.
The area of a square field is 30 $$\frac {1 }{ 4}$$ m2. Calculate the length of the side of the squre.
Solution:

Question 5.
Find the length of a side of a square playground whose area is equal to the area of a rectangular field of dimensions 72 m and 338 m.
Solution:
Length of rectangular field (l) = 338 m
and breadth (b) = 72 m
∴ Area = 1 x 6= 338 x 72 m2
∴ Area of square = 338 x 72 m2 = 24336 m2
and length of the side of the square

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5

Other Exercises

Question 1.
Find the square root of each of the long division method.
(I) 12544
(ii) 97344
(iii) 286225
(iv) 390625
(v) 363609
(vi) 974169
(vii) 120409
(viii) 1471369
(ix) 291600
(x) 9653449
(xi) 1745041
(xii) 4008004
(xiii) 20657025
(xiv) 152547201
(jcv) 20421361
(xvi) 62504836
(xvii) 82264900
(xviii) 3226694416
(xix)6407522209
(xx) 3915380329
Solution:
<

Question 2.
Find the least number which must be subtracted from the following numbers to make them a perfect square :
(i) 2361
(ii) 194491
(iii) 26535
(iv) 16160
(v) 4401624
Solution:
(i) 2361
Finding the square root of 2361

We get 48 as quotient and remainder = 57
∴ To make it a perfect square, we have to subtract 57 from 2361
∴ Least number to be subtracted = 57
(ii) 194491
Finding the square root of 194491

We get 441 as quotient and remainder = 10
∴ To make it a perfect square, we have to subtract 10 from 194491
∴ Least number to be subtracted = 10
(iii) 26535
Finding the square root of 26535

We get 162 as quotient and 291 as remainder
∴ To make it a perfect square, we have to subtract 291 from 26535
∴ Least number to be subtracted = 291
(iv)16160
Finding the square root of 16160

We get 127 as quotient and 31 as remainder
∴ To make it a perfect square, we have to subtract 31 from 16160
∴ Least number to be subtracted = 31
(v) 4401624
Find the square root of 4401624

We get 2098 as quotient and 20 as remainder
∴ To make it a perfect square, we have to subtract 20 from 4401624
∴ Least number to be subtracted = 20

Question 3.
Find the least number which must be added to the following numbers to make them a perfect square :
(i) 5607
(ii) 4931
(iii) 4515600
(iv) 37460
(v) 506900
Solution:
(i) 5607

Finding the square root of 5607, we see that 742 = 5607- 131 =5476 and 752 = 5625
∴ 5476 < 5607 < 5625
∴ 5625 – 5607 = 18 is to be added to get a perfect square
∴ Least number to be added = 18
(ii) 4931

Finding the square root of 4931, we see that 702= 4900
∴ 712 = 5041 4900 <4931 <5041
∴ 5041 – 4931 = 110 is to be added to get a perfect square.
∴ Least number to be added =110
(iii) 4515600

Finding the square root of 4515600, we see
that 21242 = 4511376
and 2 1 252 = 45 1 56 25
∴ 4511376 <4515600 <4515625
∴ 4515625 – 4515600 = 25 is to be added to get a perfect square.
∴ Least number to be added = 25
(iv) 37460

Finding the square root of 37460
that 1932 = 37249, 1942 = =37636
∴ 37249 < 37460 < 37636
∴ 37636 – 37460 = 176 is to be added to get a perfect square.
∴ Least number to be added =176
(v) 506900

Finding the square root of 506900, we see that
7112 = 505521, 7122 = 506944
∴ 505521 < 506900 < 506944
∴ 506944 – 506900 = 44 is to be added to get a perfect square.
∴ Least number to be added = 44

Question 4.
Find the greatest number of 5 digits which is a perfect square.
Solution:
Greatest number of 5-digits = 99999 Finding square root, we see that 143 is left as remainder

∴ Perfect square = 99999 – 143 = 99856 If we add 1 to 99999, it will because a number of 6 digits
∴ Greatest square 5-digits perfect square = 99856

Question 5.
Find the least number of four digits which is a perfect square.
Solution:
Least number of 4-digits = 10000
Finding square root of 1000
We see that if we subtract 39

From 1000, we get three digit number
∴ We shall add 124 – 100 = 24 to 1000 to get a
perfect square of 4-digit number
∴ 1000 + 24 = 1024
∴ Least number of 4-digits which is a perfect square = 1024

Question 6.
Find the least number of six-digits which is a perfect square.
Solution:
Least number of 6-digits = 100000

Finding the square root of 100000, we see that if we subtract 544, we get a perfect square of 5-digits.
4389 – 3900 = 489
to 100000 to get a perfect square
Past perfect square of six digits= 100000 + 489 =100489

Question 7.
Find the greatest number of 4-digits which is a perfect square.
Solution:
Greatest number of 4-digits = 9999

Finding the square root, we see that 198 has been left as remainder
∴ 4-digit greatest perfect square = 9999 – 198 = 9801

Question 8.
A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row.
Solution:
Total number of soldiers = 8160 Soldiers left after arranging them in a square = 60
∴ Number of soldiers which are standing in a square = 8160 – 60 = 8100

Question 9.
The area of a square field is 60025 m2. A man cycle along its boundry at 18 km/hr. In how much time will be return at the starting point.
Solution:
Area of a square field = 60025 m2

Question 10.
The cost of levelling and turfing a square lawn at Rs. 250 per m2 is Rs. 13322.50. Find the cost of fencing it at Rs. 5 per metre ?
Solution:
Cost of levelling a square field = Rs. 13322.50
Rate of levelling = Rs. 2.50 per m2

and perimeter = 4a = 4 x 73 = 292 m Rate of fencing the field = Rs. 5 per m
∴ Total cost of fencing = Rs. 5 x 292 = Rs. 1460

Question 11.
Find the greatest number of three digits which is a perfect square.
Solution:
3-digits greatest number = 999

Finding the square root, we see that 38 has been left
∴ Perfect square = 999 – 38 = 961
∴ Greatest 3-digit perfect square = 961

Question 12.
Find the smallest number which must be added to 2300 so that it becomes a perfect square.
Solution:
Finding the square root of 2300

We see that we have to add 704 – 700 = 4 to 2300 in order to get a perfect square
∴ Smallest number to be added = 4

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4

Other Exercises

Question 1.
Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i)  In $$\sqrt { 9801 }$$ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9
(ii) In $$\sqrt { 799356 }$$ ∴ the units digit is 6
∴ The units digit of the square root can be 4 or 6
(iii) In $$\sqrt { 7998001 }$$ ∴ the units digit is 1
∴ The units digit of the square root can be 1 or 9
(iv) In 657666025
∴ The unit digit is 5
∴ The units digit of the square root can be 5

Question 2.
Find the square root of each of the following by prime factorization.
(i) 441
(ii) 196
(iii) 529
(iv) 1764
(v) 1156
(vi) 4096
(vii) 7056
(viii) 8281
(ix) 11664
(x) 47089
(xi) 24336
(xii) 190969
(xiii) 586756
(xiv) 27225
(xv) 3013696
Solution:

Question 3.
Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.
Solution:
Factorising 180,

180 = 2 x 2 x 3 x 3 x 5
Grouping the factors in pairs we see that factor 5 is left unpaired.
∴ Multiply 180 by 5, we get the product 180 x 5 = 900
Which is a perfect square
and square root of 900 = 2 x 3 x 5 = 30

Question 4.
Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorising 147,

147 = 3 x 7×7
Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired
∴ Multiplying 147 by 3, we get the product 147 x 3 = 441
Which is a perfect square
and its square root = 3×7 = 21

Question 5.
Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.
Solution:
Factorising 3645

3645 = 3 x 3 3 x 3 x 3 x 3 x 5
Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired
∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27

Question 6.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorsing 1152,

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired.
∴ Dividing by 2, the quotient 576 is a perfect square .
∴ Square root of 576, it is 24

Question 7.
The product of two numbers is 1296. If one number is 16 times the others find the numbers.
Solution:
Product of two numbers = 1296
Let one number = x
Second number = 16x

∴ First number = 9
and second number = 16 x 9 = 144

Question 8.
A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents.
Solution:
Total donation collected = Rs. 202500
Let number of residents = x
Then donation given by each resident = Rs. x
∴ Total collection = Rs. x x x

Question 9.
A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?
Solution:
Total amount collected = Rs. 92.16 = 9216 paise
Let the number of members = x
Then amount collected by each member = x
paise

∴ Number of members = 96
and each member collected = 96 paise

Question 10.
A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ?
Solution:
Total fee collected = Rs. 2304
Let number of students = x
Then fee paid by each student = Rs. x
∴ x x x = 2304 => x2 = 2304
∴ x = $$\sqrt { 2304 }$$

Question 11.
The area of a square field is 5184 m2. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
The area of a square field = 5184 m2
Let side of the square = x

∴ side of square= 72 m
∴ Perimeter, of square field = 72 x 4 m = 288 m
Perimeter of rectangle = 288 m
Let breadth of rectangular field (b) = x
Then length (l) = 2x
∴ Perimeter = 2 (l + b)
= 2 (2x + x) = 2 x 3x = 6x
= 2 (2x + x) = 2 x 3x = 6x

∴ Length of rectangular field = 2x = 2 x 48 = 96 m
and area = l x b = 96 x 48 m2
= 4608 m2

Question 12.
Find the least square number, exactly divisible by each one of the numbers :
(i) 6, 9,15 and 20
(ii) 8,12,15 and 20

Solution:

LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180
=2 x 2 x 3 x 3 x 5
We see that after grouping the factors in pairs, 5 is left unpaired
∴ Least perfect square = 180 x 5 = 900

We see that after grouping the factors,
factors 2, 3, 5 are left unpaired
∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600

Question 13.
Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:

Question 14.
Write the prime factorization of the following numbers and hence find their square roots. ^
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056
Solution:
Factorization, we get:
(i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Grouping the factors in pairs of equal factors,

Question 15.
The students of class VIII of a school donated Rs. 2401 for PM’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.
Solution:
Total amount of donation = 2401
Let number of students in VIII = x
∴ Amount donoted by each student = Rs. x

Question 16.
A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.
Solution:
Number of students = 6000
Students left out = 71
∴ Students arranged in a field = 6000 – 71=5929

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3

Other Exercises

Question 1.
Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication :
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96
Solution:
(i) (25)2

Question 2.
Find the squares of the following numbers using diagonal method :
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171
Solution:

Question 3.
Find the squares of the following numbers :
(i) 127
(ii) 503
(iii) 451
(iv) 862
(v) 265
Solution:
(i) (127)2 = (120 + 7)2
{(a + b)2 = a2 + lab + b2}
= (120)2 + 2 x 120 x 7 + (7)2
= 14400+ 1680 + 49 = 16129

(ii) (503)2 = (500 + 3)2
{(a + b)2 = a2 + lab + b1}
= (500)2 + 2 x 500 x 3 + (3)2
= 250000 + 3000 + 9 = 353009

(iii) (451)2 = (400 + 51)2
{(a + b)2 = a2 + lab + b2}
= (400)2 + 2 x 400 x 51 + (5l)2
= 160000 + 40800 + 2601 = 203401

(iv) (451)2 = (800 + 62)2
{(a + b)2 = a2 + lab + b2}
= (800)2 + 2 x 800 x 62 + (62)2
= 640000 + 99200 + 3844 = 743044

(v) (265)2
{(a + b)2 = a2 + 2ab + b2}
(200 + 65)2 = (200)2 + 2 x 200 x 65 + (65)2
= 40000 + 26000 + 4225 = 70225

Question 4.
Find the squares of the following numbers
(i) 425
(ii) 575
(iii) 405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995
Solution:
(i) (425)2
Here n = 42
∴ n (n + 1) = 42 (42 + 1) = 42 x 43 = 1806
∴ (425)2 = 180625

(ii) (575)2
Here n = 57
∴ n (n + 1) = 57 (57 + 1) = 57 x 58 = 3306
∴ (575)2 = 330625

(iii) (405)2
Here n = 40
∴ n (n + 1) = 40 (40 + 1) -40 x 41 = 1640
∴ (405)2 = 164025

(iv) (205)2
Here n = 20
∴ n (n + 1) = 20 (20 + 1) = 20 x 21 = 420
∴ (205)2 = 42025

(v) (95)2
Here n = 9
∴ n (n + 1) = 9 (9 + 1) = 9 x 10 = 90
∴ (95)2 = 9025

(vi) (745)2
Here n = 74
∴ n (n + 1) = 74 (74 + 1) = 74 x 75 = 5550
∴ (745)2 = 555025

(vii) (512)2
Here a = 1, b = 2
∴ (5ab)2 = (250 + ab) x 1000 + (ab)2
∴ (512)2 = (250 + 12) x 1000 + (12)2
= 262 x 1000 + 144
= 262000 + 144 = 262144

(viii) (995)2
Here n = 99
∴ n (n + 1) = 99 (99 + 1) = 99 x 100 = 9900
∴ (995)2 = 990025

Question 5.
Find the squares of the following numbers using the identity (a + b)1 = a2 + lab + b1
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Solution:
a + b)2 = a2 + lab + b2

(i) (405)2 = (400 + 5)2
= (400)2 + 2 x 400 x 5 + (5)2
= 160000 + 4000 + 25 = 164025

(ii) (510)2 = (500 + 10)2
= (500)2 + 2 x 500 x 10 x (10)2
= 250000 + 10000 + 100
= 260100

(iii) (1001)2 = (1000+1)2
= (1000)2 + 2 X 1000 x 1 + (1)
= 1000000 + 2000 + 1
=1002001

(iv) (209)2 = (200 + 9)2
= (200)2 + 2 x 200 x 9 x (9)2
= 40000 + 3600 +81
= 43681

(v) (605)2 = (600 + 5)2
= (600)2 + 2 x 600 x 5 +(5)2
= 360000 + 6000       25
=366025

Question 6.
Find the squares of the following numbers using the identity (a – b)2 = a2 – 2ab + b2 :
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599
Solution:
a – b)2 = a2 – lab + b2

(i) (395)2 = (400 – 5)2
= (400)2 – 2 x 400 x 5 + (5)2
= 160000-4000 + 25
= 160025-4000
= 156025

(ii) (995)2 = (1000 – 5)2
= (1000)2 – 2 x 1000 x 5 + (5)2
= 1000000- 10000 + 25
= 1000025- 10000
= 990025

(iii) (495)2 = (500 – 5)2
= (500)2 – 2 x 500 x 5 + (5)2
= 250000 – 5000 + 25
= 250025 – 5000
= 245025

(iv) (498)2 = (500 – 2)2
= (500)2 – 2 x 500 x 2 + (2)2
= 250000 – 2000 + 4
= 250004 – 2000
= 248004

(v) (99)2 = (100 – l)2
= (100)2 – 2 x 100 x 1 + (1)2
= 10000 – 200 + 1
= 10001 – 200
= 9801

(vi) (999)2 = (1000- l)2
= (1000)2 – 2 x 1000 x 1+ (1)2
= 1000000-2000+1
= 10000001-2000=998001

(vii) (599)2 = (600 – 1)2
= (600)-2 x 600 X 1+ (1)2
= 360000 -1200+1
= 360001 – 1200 = 358801

Question 7.
Find the squares of the following numbers by visual method :
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99
Solution:
(a + b)2 = a2 – ab + ab + b2
(i) (52)2 = (50 + 2)2
= 2500 + 100 + 100 + 4
= 2704

(ii) (95)2 = (90 + 5)2
= 8100 + 450 + 450 + 25
= 9025

(iii) (505)2 = (500 + 5)2
= 250000 + 2500 + 2500 + 25
= 255025

(iv) (702)2 = (700 + 2)2
= 490000 + 1400+ 1400 + 4
= 492804

(v) (99)2 = (90 + 9)2
= 8100 + 810 + 810 + 81
= 9801

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2

Other Exercises

Question 1.
The following numbers are not perfect squares. Give reason :
(i) 1547
(ii) 45743
(iii) 8948
(iv) 333333

Solution:
We know that if the units digit is 2, 3, 7 or 8 of a number, then the number is not a perfect square.

(i) ∴ 1547 has 7 as units digit.
∴ It is not a perfect square.

(ii) 45743 has 3 as units digit
∴ It is not a perfect square.

(iii)  ∴ 8948 has 8 as units digit
∴ It is not a perfect square.

(iv)  ∴ 333333 has 3 as units digits
∴ It is not a perfect square.

Question 2.
Show that the following numbers are not perfect squares :
(i) 9327
(ii) 4058
(iii) 22453
(iv) 743522

Solution:
(i) 9327
∴ The units digit of 9327 is 7
∴ This number can’t be a perfect square.

(ii) 4058
∴ The units digit of 4058 is 8
∴ This number can’t be a perfect square.

(iii) 22453
∴ The units digit of 22453 is 3
.∴ This number can’t be a perfect square.

(iv) 743522
∴ The units digit of 743522 is 2
∴ This number can’t be a perfect square.

Question 3.
The square of which of the following numbers would be an odd number ?
(i) 731
(ii) 3456
(iii) 5559
(iv) 42008
Solution:
We know that the square of an odd number is odd and of even number is even. Therefore
(i) Square of 731 would be odd as it is an odd number.
(ii) Square of 3456 should be even as it is an even number.
(iii) Square of 5559 would be odd as it is an odd number.
(iv) The square of 42008 would be an even number as it is an even number.
Therefore suqares of (i) 731 and (ii) 5559 will be odd numbers.

Question 4.
What will be the units digit of the squares of the following numbers ?
(i) 52
(ii) 977
(iii) 4583
(iv) 78367
(v) 52698
(vi) 99880
(vii) 12796
(viii) 55555
(ix) 53924

Solution:

(i) Square of 52 will be 2704 or (2)2 = 4
∴ Its units digit is 4.

(ii) Square of 977 will be 954529 or (7)2 = 49 .
∴ Its units digit is 9

(iii) Square of 4583 will be 21003889 or (3)2 = 9
∴ Its units digit is 9

(iv) IS 78367, square of 7 = 72 = 49
∴ Its units digit is 9

(v) In 52698, square of 8 = (8)2 = 64
∴ Its units digit is 4

(vi) In 99880, square of 0 = 02 = 0
∴ Its units digit is 0

(vii) In 12796, square of 6 = 62 = 36
∴ Its units digit is 6

(viii) In In 55555, square of 5 = 52 = 25
∴ Its units digit is 5

(ix) In 53924, square pf 4 = 42 = 16
∴ Its units digit is 6

Question 5.
Observe the following pattern
1 + 3 = 22
1 + 3 + 5 = 32
1+34-5 + 7 = 42
and write the value of 1 + 3 + 5 + 7 + 9 +…………upto n terms.
Solution:
The given pattern is
1 + 3 = 22
1 + 3 + 5 = 32
1+3 + 5 + 7 = 42
1+3 + 5 + 7 + 9 +……………… upto n terms (number of terms)2 = n2

Question 6.
Observe the following pattern :
22 – 12 = 2 + 1
32 – 22 = 3 + 2
42 – 32 = 4 + 3
52 – 42 = 5 + 4
Find the value of
(i) 1002 – 992
(ii) 1112 – 1092
(iii) 992 – 962
Solution:
From the given pattern,
22 – 12 = 2 + 1
32 – 22 = 3 + 2
42 – 32 = 4 + 3
52 – 42 = 5 + 4
Therefore
(i) 1002-99° = 100 + 99

(ii) 1112 – 1092 = 1112 – 1102– 1092
= (1112 – 1102) + (1102 – 1092)
= (111 + 110) + (110+ 109)
= 221 + 219 = 440

(iii) 992 – 962 = 992 – 982 + 982 – 972 + 972 – 962
= (992 – 982) + (982 – 972) + (972 – 962)
= (99 + 98) + (98 + 97) + (97 + 96)
= 197 + 195 + 193 = 585

Question 7.
Which of the following triplets are Pythagorean ?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(vi) (16, 63, 65)
(vii) (12, 35, 38)
Solution:
A pythagorean triplet is possible if (greatest number)2 = (sum of the two smaller numbers)

(i) 8, 15, 17
Here, greatest number =17
∴ (17)2 = 289
and (8)2 + (15)2 = 64 + 225 = 289
∴ 82 + 152 = 172
∴ 8, 15, 17 is a pythagorean triplet

(ii) 18, 80, 82
Greatest number = 82
∴ (82)2 = 6724
and 182 + 802 = 324 + 6400 = 6724
∴ 182 + 802 = 822
∴ 18, 80, 82 is a pythagorean triplet

(iii) 14, 48, 51
Greatest number = 51
∴ (51)2 = 2601
and 142 + 482 = 196 + 2304 = 25 00
∴ 512≠ 142 + 482
∴ 14, 48, 51 is not a pythagorean triplet

(iv) 10, 24, 26
Greatest number is 26
∴ 262 = 676
and 102 + 242 = 100 + 576 = 676
∴ 262 = 102 + 242
∴ 10, 24, 26 is a pythagorean triplet

(vi) 16, 63, 65
Greatest number = 65
∴ 652 = 4225
and 162 + 632 = 256 + 3969 = 4225
∴ 652 = 162 + 632
∴ 16, 63, 65 is a pythagorean triplet

(vii) 12, 35, 38
Greatest number = 38
∴ 382 = 1444
and 122 + 352 = 144 + 1225 = 1369
∴ 382 ≠122 + 352
∴ 12, 35, 38 is not a pythagorean triplet.

Question 8.
Observe the following pattern

Solution:
From the given pattern

Question 9.
Observe the following pattern

and find the values of each of the following :
(i) 1 + 2 + 3 + 4 + 5 +….. + 50
(ii) 31 + 32 +… + 50
Solution:
From the given pattern,

Question 10.
Observe the following pattern

and find the values of each of the following :
(i) 12 + 22 + 32 + 42 +…………… + 102
(ii) 52 + 62 + 72 + 82 + 92 + 102 + 12
2
Solution:
From the given pattern,

Question 11.
Which of the following numbers are squares of even numbers ?
121,225,256,324,1296,6561,5476,4489, 373758
Solution:
We know that squares of even numbers is also are even number. Therefore numbers 256, 324,1296, 5476 and 373758 have their units digit an even number.
∴ These are the squares of even numbers.

Question 12.
By just examining the units digits, can you tell which of the following cannot be whole squares ?

1. 1026
2. 1028
3. 1024
4. 1022
5. 1023
6. 1027

Solution:
We know that a perfect square cam at ends with the digit 2, 3, 7, or 8
∴ By examining the given number, we can say that 1028, 1022, 1023, 1027 can not be perfect squares.

Question 13.
Write five numbers for which you cannot decide whether they are squares.
Solution:
A number which ends with 1,4, 5, 6, 9 or 0
can’t be a perfect square
2036, 4225, 4881, 5764, 3349, 6400

Question 14.
Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit.
Solution:
A number which does not end with 2, 3, 7 or 8 can be a perfect square
∴ The five numbers can be 2024, 3036, 4069, 3021, 4900

Question 15.
Write true (T) or false (F) for the following statements.
(i) The number of digits in a square number is even.
(ii) The square of a prime number is prime.
(iii) The sum of two square numbers is a square number.
(iv) The difference of two square numbers is a square number.
(vi) The product of two square numbers is a square number.
(vii) No square number is negative.
(viii) There is not square number between 50 and 60.
(ix) There are fourteen square number upto 200.

Solution:
(i) False : In a square number, there is no condition of even or odd digits.
(ii) False : A square of a prime is not a prime.
(iii) False : It is not necessarily.
(iv) False : It is not necessarily.
(vi) True.
(vii) True : A square is always positive.
(viii) True : As 72 = 49, and 82 = 64.
(ix) True : As squares upto 200 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 which are fourteen in numbers.

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1

Other Exercises

Question 1.
Which of the following numbers are perfect squares ?
(i)484
(ii) 625
(iii) 576
(iv) 941
(v) 961
(vi) 2500
Solution:

Grouping the factors in pairs, we have left no factor unpaired
∴ 484 is a perfect square of 22

∴ Grouping the factors in pairs, we have left no factor unpaired
∴ 625 is a perfect square of 25.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 576 is a perfect square of 24
(iv) 941 has no prime factors

∴ 941 is not a perfect square.
(v) 961 =31 x 31
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 961 is a perfect square of 31

Grouping the factors in pairs, we see that no factor is left impaired
∴ 2500 is a perfect square of 50 .

Question 2.
Show that each of the following* numbers is a perfect square. Also find the number whose square is the given number in each case :
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761
Solution:

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 1156 is a perfect square of 2 x 17 = 34

Grouping the factors in pairs, we see that no factor is left unpaired
2025 is a perfect square of 3 x 3 x 5 =45

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 14641 is a perfect square of 11×11 = 121

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 4761 is a perfect square of 3 x 23 = 69

Question 3.
Find the smallest number by which the given number must be multiplied so that the product is a perfect square.
(i) 23805
(ii) 12150
(iii) 7688
Solution:

Grouping the factors in pairs of equal factors, we see that 5 is left unpaird
∴ In order to complete the pairs, we have to multiply 23805 by 5, then the product will be the perfect square.
Requid smallest number = 5
(ii) 12150 = 2 x 3 x 3×3 x 3×3 x 5×5

Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired
∴ In order to complete the pairs, we have to multiply 12150 by 2 x 3 =6 i.e., then the product will be the complete square.
∴ Required smallest number = 6

Grouping the factors in pairs of equal factors, we see that factor 2 is left unpaired
∴ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square
∴ Required smallest number = 2

Question 4.
Find the smallest number by which the given number must be divided so that the resulting number is a perfect square.
(i) 14283
(ii) 1800
(iii) 2904
Solution:

Grouping the factors in pairs of equal factors, we see that factors we see that 3 is left unpaired
Deviding by 3, the quotient will the perfect square.

Grouping the factors in pair of equal factors, we see that 2 is left unpaired.
∴ Dividing by 2, the quotient will be the perfect square.

Grouping the factors in pairs of equal factors, we see that 2 x 3 we left unpaired
∴ Dividing by 2 x 3 = 6, the quotient will be the perfect square.

Question 5.
Which of the following numbers are perfect squares ?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121
Solution:
11 is not a perfect square as 11 = 1 x 11
12 is not a perfect square as 12 = 2×2 x 3
16 is a perfect square as 16 = 2×2 x 2×2
32 is not a perfect square as 32 = 2×2 x 2×2 x 2
36 is a perfect square as 36 = 2×2 x 3×3
50 is not a perfect square as 50 = 2 x 5×5
64 is a perfect square as 64 = 2×2 x 2×2 x 2×2
79 is not a perfect square as 79 = 1 x 79
81 is a perfect square as 81 = 3×3 x 3×3
111 is not a perfect square as 111 = 3 x 37
121 is a perfect square as 121 = 11 x 11
Hence 16, 36, 64, 81 and 121 are perfect squares.

Question 6.
Using prime factorization method, find which of the following numbers are perfect squares ?
∴ 189,225,2048,343,441,2916,11025,3549
Solution:

Grouping the factors in pairs, we see that are 3 and 7 are left unpaired
∴ 189 is not a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 225 is a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 2048 is a perfect square

Grouping the factors in pairs, we see that one 7 is left unpaired
∴ 343 is not a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 441 is a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 2916 is a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 11025 is a perfect square.

Grouping the factors in pairs, we see that 3, no factor 7 are left unpaired
∴ 3549 is a perfect square.

Question 7.
By what number should each of the following numbers be multiplied to get a perfect square in each case ? Also, find the number whose square is the new number.
(i) 8820
(ii) 3675
(iii) 605
(iv) 2880
(v) 4056
(vi) 3468
Solution:

Grouping the factors in pairs, we see that 5 is left unpaired
∴ By multiplying 8820 by 5, we get the perfect square and square root of product will be
= 2 x 3 x 5 x 7 = 210

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3675 by 3, we get a perfect square and square of the product will be
= 3 x 5 x 7 = 105

Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 605 by 5, we get a perfect square and square root of the product will be
= 5 x 11 =55

Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 2880 by 5, we get the perfect square.
Square rooi of product will be = 2 x 2 * 2 – 3 x 5 = 120

Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 4056 by 2 x 3 i.e., 6, we get the perfect square.
and square root of the product will be
= 2 x 2 x 3 x 13 = 156

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3468 by 3 we get a perfect square, and square root of the product will be 2 x 3 x 17 = 102

Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 7776 by 2 x 3 or 6 We get a perfect square and square root of the product will be
= 2 x 2 x 2 x 3 x 3 x 3 = 216

Question 8.
By what numbers should each of the following be .divided to get a perfect square in each case ? Also find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575
Solution:

Grouping the factors in pairs, we see that 2 is left unpaired
∴ Dividing by 2, we get the perfect square and square root of the quotient will be 7 x 13 = 91

Grouping the factors in pairs, we see that 2 is left unpaired,
∴ Dividing 3698 by 2, the quotient is a perfect square
and square of quotient will be = 43

Grouping the factors in pairs, we see that 7 is left unpaired
∴ Dividing 5103 by 7, we get the quotient a perfect square.
and square root of the quotient will be 3 x 3 x 3 = 27

Grouping the factors iq pairs, we see that 2 and 3 are left unpaired
∴ Dividing 3174 by 2 x 3 i.e. 6, the quotient will be a perfect square and square root of the quotient will be = 23

Grouping the factors in pairs, we find that 7 is left unpaired i
∴ Dividing 1575 by 7, the quotient is a perfect square
and square root of the quotient will be = 3 x 5 = 15

Question 9.
Find the greatest number of two digits which is a perfect square.
Solution:
The greatest two digit number = 99 We know, 92 = 81 and 102 = 100 But 99 is in between 81 and 100
∴ 81 is the greatest two digit number which is a perfect square.

Question 10.
Find the least number of three digits which is perfect square.
Solution:
The smallest three digit number =100
We know that 92 = 81, 102 = 100, ll2 = 121
We see that 100 is the least three digit number which is a perfect square.

Question 11.
Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.
Solution:
By factorization:

Grouping the factors in pairs, we see that 11 is left unpaired
∴ The least number is 11 by which multiplying 4851, we get a perfect square.

Question 12.
Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.
Solution:
By factorization,

Grouping the factors in pairs, we see that 13 is left unpaired
∴ Dividing 28812 by 3, the quotient will be a perfect square.

Question 13.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.
Solution:
By factorization,

Grouping the factors in pairs, we see that one 2 is left unpaired.
∴ Dividing 1152 by 2, we get the perfect square and square root of the resulting number 576, will be 2 x 2 x 2 x 3 = 24

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 2 Powers MCQS

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 2 Powers MCQS

Other Exercises

Choose the correct alternative in each of the following :

Question 1.
Square of $$\left( \frac { -2 }{ 3 } \right)$$

Solution:

Question 2.
Cube of $$\frac { -1 }{ 2 }$$ is

Solution:

Question 3.
Which of the following is not equal to

Solution:

Question 4.
Which of the following in not reciprocal of

Solution:

Question 5.
Which of the following numbers is not equal to $$\frac { -8 }{ 27 }$$ ?

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.
For any two non-zero rational numbers a and b,a4+b4 is equal to
(a) (a + b)1
(b) (a + b)0
(c) (a + b)4
(d) (a + b)8
Solution:
(c) {∵ a4 + b4 = (a + b)4}

Question 16.
For any two rational numbers a and b, a5 x b5 is equal to
(a) (a x b)0
(b) (a x b)10

(c) (a x b)5
(d) (a x b)25

Solution:
(c) {∵ a5 x b5 = (a x b)5}

Question 17.
For a non-zero rational number a, a7 + a12 is equal to
(a) a5
(b) a-19

(c) a-5
(d) a19

Solution:
(c) {a5 a12 = a7-12 =a-5}

Question 18.
For a non-zero rational number a, (a3)-2 is equal to
(a) a6
(b) a-6
(c) a-9
(d) a1

Solution:
(b) {(a3)-2 = a3 x (-2)= a6}

Hope given RD Sharma Class 8 Solutions Chapter 2 Powers MCQS are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.3

Other Exercises

Question 1.
Express the following numbers in standard form :
(i) 6020000000000000
(ii) 0.00000000000942
(iii) 0.00000000085
(iv) 846 X 107
(v) 3759 x 10-4
(vi) 0.00072984
(vii) 0.000437 x 104
(Viii) 4 + 100000
Solution:

Question 2.
Write the following numbers in the usual form :
(i) 4.83 x 107
(ii) 3.02 x 10-6
(iii) 4.5 x 104

(iv) 3 x 10-8
(v) 1.0001 x 109
(vi) 5.8 x 102
(vii) 3.61492 x 106
(viii) 3.25 x 10-7
Solution:

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## RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2

Other Exercises

Question 1.
Write each of the following in exponential form :

Solution:

Question 2.
Evaluate :

Solution:

Question 3.
Express each of the following as a rational number in the form $$\frac { p }{ q } :$$

Solution:

Question 4.
Simplify :

Solution:

Question 5.
Express each of the following rational numbers with a negative exponent :

Solution:

Question 6.
Express each of the following rational numbers with a positive exponent :

Solution:

Question 7.
Simplify :

Solution:

Question 8.
By what number should 5-1 be multiplied so that the product may be equal to (-7)-1 ?
Solution:

Question 9.
By what number should $${ \left( \frac { 1 }{ 2 } \right) }^{ -1 }$$ be multiplied so that the product may be equal to $${ \left( \frac { -4 }{ 7 } \right) }^{ -1 }$$ ?
Solution:

Question 10.
By what number should (-15)-1 be divided so that the quotient may be equal to (-5)-1 ?
Solution:

Question 11.
By what number should $${ \left( \frac { 5 }{ 3 } \right) }^{ -2 }$$ be multiplied so that the product may be $${ \left( \frac { 7 }{ 3 } \right) }^{ -1 }$$ ?
Solution:

Question 12.
Find x, if

Solution:

Question 13.

Solution:

Question 14.
Find the value of x for which 52x + 5-3 = 55.
Solution:

Hope given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.1

Other Exercises

Question 1.
Express each of the following as a rational number of the form $$\frac { p }{ q } ,$$ where p and q are integers and q≠ 0 :

Solution:

Question 2.
Find the values of each of the following

Solution:

Question 3.
Find the values of each of the following :

Solution:

Question 4.
Simplify :

Solution:

Question 5.
Simplify :

Solution:

Question 6.
By what number should 5-1 be multiplied so that the product may be equal to(-7)-1 ?
Solution:

Question 7.
By what number should $${ \left( \frac { 1 }{ 2 } \right) }^{ -1 }$$ multiplied so that the product many be equal to $${ \left( \frac { -4 }{ 7 } \right) }^{ -1 }$$
Solution:

Question 8.
By what number should (-15)-1 be divided so that the quotient may be equal to (-5)-1 ?
Solution:

Hope given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.1 are helpful to complete your math homework.

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