Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

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Kerala State Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Square and Square Root Text Book Questions and Answers

Triangular numbers Textbook Page No. 80

See the dots arranged in triangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 1
How many dots are there in each?
1, 3, 6
How many dots would be there in the next triangle?
Such numbers as 1, 3, 6, 10, … are called triangular numbers.
The first triangular number is 1.
The second is 1 + 2 = 3.
The third is 1 + 2 + 3 = 6.
What is the tenth triangular number?
Answer:
From the above logic we need to add 10 numbers sum to obtain the 10th triangular number.
To find the 10th triangular number T10 we need to add 1+ 2+3+4+5+6+7+8+9+10 = 55
Therefore, 10th triangular number T10 is 55.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Squares and triangles Textbook Page No. 81

Look at these pictures:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 4
Each square is divided into two triangles.
Let’s translate this into numbers:
4 = 1 + 3
9 = 3 + 6
16 = 6 + 10
Check whether the same pattern continues. What do we see?
All perfect squares after 1 are the sums of two consecutive triangular numbers.
What is the sum of the seventh and eighth triangular numbers?
Answer:
Yes adding consecutive triangular numbers gives us perfect squares. So, we get the next pattern 9+16 = 25 which in turn is a perfect square.
7th Triangular number can be obtained as T7 = 1+2+3+4+5+6+7 = 28
8th Triangular number can be obtained as T8 = 1+2+3+4+5+6+7+8 = 36
Sum of Seventh and Eighth Triangular Numbers = T7+T8
= 28+36
= 64
Thus, the sum of 7th, 8th Triangular Numbers is 64

Increase and decrease Textbook Page No. 82

Look at this number pattern:
1 = 1
4 = 1 + 2 + 1
9 = 1 + 2 + 3 + 2 + 1
16 = 1 + 2 + 3 + 4 + 3 + 2 + 1
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 5
Can you split some more perfect squares like this?
Answer:
Writing some more perfect squares we have 25 = 1+2+3+4+5+4+3+2+1
36 = 1+2+3+4+5+6+5+4+3+2+1
49 = 1+2+3+4+5+6+7+6+5+4+3+2+1
64 = 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1

Square difference

We have see that
22 = 12 + (1 + 2)
32 = 22 + (2 + 3)
42 = 32 + (3 + 4) and so on.
We can write these in another manner also:
22 – 12 = 1 + 2
32 – 22 = 2 + 3
42 – 32 = 3 + 4
In general, the difference of the squares of two consecutive natural numbers is their sum. Now look at these:
32 – 12 = 9 – 1 = 8
42 – 22 = 16 – 4 = 12
52 – 32 = 25 – 9 = 16
What is the relation between the difference of the squares of alternative natural numbers and their sum?
Answer:
32 – 12 = (3+1) x 2 = 8
42 – 22 = (4+2) x 2 = 12
52 – 32 =  = (5+3) x 2 = 16

The difference between squares of two alternate natural numbers is always even i.e. twice the sum of two numbers that are squared.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Project Textbook Page No. 84

Last digit
Look at the last digit of squares of natural numbers from 1 to 10:
1, 4, 9, 6, 5, 6, 9, 4, 1, 0
Now, look at the last digits of squares of natural numbers from 11 to 20.
Do we have the same pattern?
Let’s look at another thing: Does any perfect square end in 2?
Which are the digits which do not occur at the end of perfect squares?
Is 2637 then a perfect square?
Answer:
Last digits of squares of natural numbers from 11 to 20 are 1, 4, 9, 6, 5, 6, 9, 4, 1, 0. Yes they have the same pattern
As we observe the digits 2, 3, 7, 8 doesn’t occur at the end of perfect squares.
As we know 2637 ends with digit 7 at the end it is not a perfect square.

To decide that a number is not a perfect square, we need only look at the last digit.
Can we decide that a number is a perfect square from its last digit alone?
Answer:
Yes, we can decide if a number is perfect square or not by seeing the last digit alone as you can see above.

Rectangle and square

Look at this picture.
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 6
Dots in a rectangle.
Can you rearrange the dots to make another rectangle?
Can you rearrange the dots to make a square? Start like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 7
How many more are needed to make a square?
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 8
How many dots were there in the original rectangle? How many in this square?
What do we see here?
42 = (3 × 5) + 1
Can we do this for all rectangular arrangements?
The numbers here are 3, 4, 5.
So, for this trick to work, what should be the relation between the number of dots in each row and column of the rectangle?
We can write this in numbers as
22 = (1 × 3) + 1
32 = (2 × 4) + 1
42 = (3 × 5) + 1
Try to continue this
Answer:
52 = (4 x 6) +1
62 = (5 x 7) +1
72 = (6 x 8) +1
82 = (7 x 9) +1
92 = (8 x 10)+1

Square root of a perfect square

784 is a perfect square. What is its square root? 784 is between the perfect squares 400 and 900; and we know that their square roots are 20 and 30. So \(\sqrt{784}\) is between 20 and 30. Since last digit of 784 is 4, its square root should have 2 or 8 as the last digit. So \(\sqrt{784}\) is either 22 or 28.
784 is near to 900 than 400. So \(\sqrt{784}\) must be 28. Now calculate 282 and check.
Given that 1369, 2116, 2209 are perfect squares, find their square roots like this.

Project Textbook Page No. 87

Digit sum

16 is a perfect square and the sum of its digits is 7.
The next perfect square 25 also has digit sum 7.
The digit sum of 36 is 9.
The sum of the digits of the next perfect square 49 is 13. If we add the digits again, the sum is 4. Find the sum of the digit sums (reduced to a single digit number) of perfect squares starting from 1.
Do you see any pattern?
Is 3324 is perfect square?
Answer:
Given number is 3324
Now Split the number and add each number 3 + 3 + 2 + 4  = 12
As the result is more than 1 number we should add it again 1 + 2 = 3
All possible numbers that are perfect square have roots of either 1, 4, 7, 9
As 3 is not in the list 3324 is not a Perfect Square.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Rows and columns Textbook Page No. 80

Look this picture:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 2
Dots in rows and columns make a rectangle.
How many dots in all?
Did you count the dots one by one?
Can you make other rectangles with 24 dots?
Is any one of these a square?
How many more dots do we need to make a square? Can you remove some dots and make a square? How many?
Can you remove some dots and make a square? How many?
Numbers which can be arranged in squares are called square numbers.
Do you see anything special about of the number of dots making a square?
Answer:
There are total 4×6=24 dots in all.
No, the dots were counted by multiplying the number of dots in rows and number of dots in column.
None of these is a square.
We need more 12 dots to make a square, 24+12=36 or 62.
Yes, we can remove some dots and make a square.
We need to remove 8 dots to make a square, 24-8=16 or 42.
The number of dots making a square is the sqaure of that number.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root img_1

Squares

What are the ways in which we can write 36 as the product of two numbers?
2 × 18, 3 × 12, 4 × 9
We can also write
36 = 6 × 6
And we have seen that it can also be shortened as 36 = 62.
36 is 6 multiplied by 6 itself; that is, the second power of 6.
There is another name for this:
36 is the square of 6.
Then what is the square of 5?
What is the square of \(\frac{1}{2}\)?
Answer:
Square of 5 = 52 = 25
\(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

Perfect squares

1, 4, 9, 16,… are the squares of the natural numbers. They are called perfect squares.
What is the perfect square after 16?
Why is 20 not a perfect square?
Answer:
Prime Factorization of 20 can be written as 22 x 51
As 5 is not in pair 20 is not a perfect square.
The next perfect square after 16 is 25.

Let us look at the succession of perfect squares in another way.
To reach 4 from 1, we must add 3.
To reach 9 from 4?
We can state these as
4 – 1 = 3
9 – 4 = 5
16 – 9 = 7
All these differences are odd numbers, right?
So, the difference of two consecutive perfect squares is an odd number.
Let’s write this as,
4=1 + 3
9 = 4 + 5 = 1 + 3 + 5
16 = 9 + 7 = 1 + 3 + 5 + 7
What do we see here?
When we add consecutive odd numbers starting from 1, we get the perfect squares.
This can be seen from these pictures also.
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 3
Can you write down the squares of natural numbers upto 20, by adding odd numbers? You can proceed like this
12 = 1
22 = 1 + 3 = 4
32 = 4 + 5 = 9
42 = 9 + 7 = 16

What is the relation between the number of consecutive odd numbers from 1 and their sum?
What is the sum of 30 consecutive odd numbers starting from 1?
Answer:
Let us assume the arithmetic series 1, 3, 5, 7, 9, 11, 13…..
we know the formula for sum of arithmetic series Sn = n/2[2a+(n-1)d]
Here d = 2
first term a = 1
Substituting the inputs we have Sn= 30/2[2×1+(30-1)2]
= 30/2[2+58]
= 30/2[60]
= 900
Therefore, the sum of 30 consecutive odd numbers starting from 1 is 900.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Tricks with ten Textbook Page No. 82

The square of 10 is 100. What is the square of 100?
In the square of 1000, how many zeros are there after 1?
What about the square of 10000?
What happens to the number of zeros on squaring?
So how do we spot the perfect squares among 10, 100, 1000, 10000 and so on?
Is one lakh a perfect square?
What about ten lakhs (million)?
Now find out the squares of 20, 200 and 2000.
Is 400000000 a perfect square?
What if we put in one more zero?
Answer:
Square of a number containing x zeros will become 2 times number of zeros.
1 lakh is not a perfect square as the number of zeros is not even.
Ten Lakhs is a perfect square as it comes with 6 zeros that are even.
20 = 20 x 20 = 400
200 = 200 x 200 = 40000
2000 = 2000 x 2000 = 4000000
400000000 is a perfect square as the number of zeros 8 is even.

Now some problems. Do them all in your head.

• Find out the squares of these numbers.

• 30
Answer:
302 = 30 x 30
= 900
The square of 30 is 900.

• 400
Answer:
4002 = 400 x 400
= 160000
The square of 400 is 160000

• 7000
Answer:
70002 = 7000 x 7000
= 49000000
The Square of 7000 is 49000000

• 6 × 1025
Answer:
(6 × 1025)= (6 × 1025) x (6 × 1025)
= 36 x (1025)2
= 36 x 1050

• Find out the perfect squares among these numbers.

• 2500
Answer:
The given number 2500 can be written as 502
Hence the 2500 is a perfect square number.

• 36000
Answer:
We cannot write the given number 36000 as square of two equal numbers. Hence it is not a perfect square.

• 1500
Answer:
We cannot write the given number 1500 as square of two equal numbers. Hence it is not a perfect square.

• 9 × 107
Answer:
We cannot write the given number 9 × 107 as square of two equal numbers. Hence it is not a perfect square.

• 16 × 1024
Answer:
The given number 2500 can be written as (4x 1012)2
Hence the given number 16 × 1024 is a perfect square.

Next square

What is the square of 21?
Wait a bit before you start multiplying.
The square of 20 is 400, isn’t it? So to get the square of 21, we need only add an odd number.
Which odd number?
Let’s start from the beginning. We can write
22 = 12 + 3 = 12 + (1 + 2)
32 = 22 + 5 = 22 + (2 + 3)
42 = 32 + 7 = 32 + (3 + 4)
52 = 42 + 9 = 42 + (4 + 5)
and so on. Continuing like this, how do we write 212?
212 = 202 + (20 + 21)
That is,
212 = 400 + 41 = 441
Now we can continue as before with
222 = 441 +43 = 484
and so on.
How do we find out the square of 101?
1002 = 10000
What more should we add?
100 + 101 = 201
So, 1012 = 10000 + 201 = 10201

• Find out the squares of these numbers using the above idea.
• 51
Answer:
Using the above process we write the 512 = 502+(50+51)
= 2500+(50+51)
= 2500+101
= 2601

• 61
Answer:
It can be written as 612 = 602+(60+61)
= 3600+(60+61)
= 3600+121
= 3721

• 121
Answer:
The given number is written as 1212 = 1202+(120+121)
= 14400+(120+121)
= 14400+241
= 14641

• 1001
Answer:
It is written as 10012 = 10002+(1000+1001)
= 10002+(1000+1001)
= 1000000+2001
= 1002001

• Compute the squares of natural numbers from 90 to 100.
Answer:
902
= 90 x 90
=8100
912
= 91 x 91
= 8281
922
= 92 x 92
= 8464
932
= 93 x 93
= 8649
942
= 94 x 94
= 8836
952
= 95 x 95
= 8025
962
= 96 x96
= 8928
972
= 97 x 97
= 9409
982
= 98 x 98
= 9310
992
= 99 x 99
= 9801
1002
= 100 x 100
= 10,000

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Fraction squares Textbook Page No. 83

A fraction multiplied by itself is also a square.
What is the square of \(\frac{3}{4}\) ?
(\(\frac{3}{4}\))2 = \(\frac{3}{4}\) × \(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 4}\) = \(\frac{9}{16}\)
That is,
(\(\frac{3}{4}\))2 = \(\frac{9}{16}\) = \(\frac{3^{2}}{4^{2}}\)
So to square a fraction, we need only square the numerator and denominator separately.

Now do these problems without pen and paper.
• Find out the squares of these numbers.
• \(\frac{2}{3}\)
Answer:
\(\frac{2}{3}\)2  = \(\frac{2}{3}\)*\(\frac{2}{3}\)
= \(\frac{2×2}{3×3}\)
= \(\frac{4}{9}\)

• \(\frac{1}{5}\) 
Answer:
\(\frac{1}{5}\)2  = \(\frac{1}{5}\)*\(\frac{1}{5}\)
= \(\frac{1×1}{5×5}\)
= \(\frac{1}{25}\)

• \(\frac{7}{3}\)
Answer:
\(\frac{7}{3}\)2  = \(\frac{7}{3}\)*\(\frac{7}{3}\)
= \(\frac{7×7}{3×3}\)
= \(\frac{49}{9}\)

• 1\(\frac{1}{2}\)
Answer:
1\(\frac{1}{2}\)2  = \(\frac{1}{2}\)*\(\frac{1}{2}\)
=1 \(\frac{1×1}{2×2}\)
=1 \(\frac{1}{4}\)
= \(\frac{5}{4}\)

• Which of the fractions below are squares?

• \(\frac{4}{15}\)
Answer:
Numerator is 4 and Denominator is 15 .
Both are not perfect Square numbers so it is not possible to write the given fraction as squares.

• \(\frac{8}{9}\)
Answer:
Numerator is 8 and Denominator is 9.
Here 8 is not a perfect Square number and Denominator is a perfect square number. so it is not possible to write the given fraction as squares.

• \(\frac{16}{25}\)
Answer:
In a given fraction, numerator is 16 and Denominator is 25 which were square numbers of 4 and 5.Hence we can Write as \(\frac{4}{5}\)2

• 2\(\frac{1}{4}\)
Answer:
When the given 2\(\frac{1}{4}\) is converted to improper fraction, we get \(\frac{9}{4}\).
Here the Numerator is 9 and Denominator is 4
We can write the numerator as perfect square i.e. (3)2
Denominator 4 can be written as (2)2
Therefore, 2\(\frac{1}{4}\) can be written as \(\frac{3}{2}\)2

• 4\(\frac{1}{9}\)
Answer:
When the given 4\(\frac{1}{9}\) is converted to improper fraction, we get \(\frac{1}{9}\).
Here the Numerator is 1 and but the Denominator is 9 as a square of number (3)2
So, we cannot write the given fraction as Squares.

• \(\frac{8}{18}\)
Answer:
Numerator is 8 and Denominator is 18.
Here numerator and Denominator are not perfect square numbers so there is no chance to write the given fraction as square of numbers.

Decimal squares

What is the square of 0.5?
We know that 52 = 25. How many decimal places would be there in the product 0.5 × 0.5?
Why?
0.5 = \(\frac{5}{10}\), right?
Can you find out the square of 0.05?
Answer:
0.052 =0.05×0.05
= 0.0025 Hence four decimal places are there.
0.0025 = \(\frac{25}{1000}\)

You have computed the squares of many natural numbers. Using that table, can you find out the square of 0.15?
Do these problems also in your head.

• Find out the squares of these numbers.
Answer:
0.152 = 0.15 x 0.15
= 0.0225

• 1.2
Answer:
1.22 = 1.2 x 1.2
= 1.44

• 0.12
Answer:
0.122 = 0.12 x 0.12
= 0.0144

• 0.013
Answer:
0.0132 = 0.013 x 0.013
= 0.000169

• Which of the following numbers are squares?

• 2.5
Answer:
We cannot write the number 2.5 as a square.

• 0.25
Answer:
0.25 = 0.5 x 0.5.
Hence the number 0.25 is written as 0.52
• 0.0016
Answer:
0.0016 = 0.04 x 0.04.
Hence the number 0.0016 is written as 0.042

• 14.4
Answer:
We cannot write the number 14.4 as square.

• 1.44
Answer:
1.44= 1.2×1.2
Hence the number 1.44 is written as 1.22

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Square product Textbook Page No. 84

What is 52 × 42?
52 × 42 = 25 × 16 = ……..
There is an easier way:
52 × 42 = 5 × 5 × 4 × 4
= (5 × 4) × (5 × 4)
= 20 × 20
= 400
= (5 x 4) x (5 x 4)
= 20 x 20
= 400

Can you find out the products below like this, without pen and paper?

• 52 × 82
Answer:
52 × 82 = 5 x 5 x 8 x 8
= (5 x 8) x (5 x 8)
= 40 x 40
= 1600

• 2.52 × 42
Answer:
2.52 × 42 = 2.5 x 2.5 x 4 x 4
= (2.5 x 4) x (2.5 x 4)
= 10 x 10
= 100

• (1.5)2 × (0.2)2
Answer:
(1.5)2 × (0.2)2 = 1.5 x 1.5 x 0.2 x 0.2
= (1.5 x 0.2) x (1.5 x 0.2)
= 0.3 x 0.3
= 0.009

What general rule did we use in all these?
The product of the squares of two numbers is equal to the square of their product.
How do we say this in algebra?
x2y2 = (xy)2, for any numbers x, y
What about for three numbers?
Answer:
let the three numbers be x, y, and z
x2y2z2 = (xyz)2, for any numbers x, y, and z.

Square factors

How do we write 30 as a product of prime numbers?
30 = 2 × 3 × 5
So how do we factorize 900?
900 = 302 = (2 × 3 × 5)2 = 22 × 32 × 52
Similarly, using the facts that 24 = 23 × 3 and 242 = 576, we get
576 = 242 = (23 × 3)2 = (23)2 × 32 = 26 × 32

Can you write each number below and its square as a product of prime powers?
• 35
Answer:
35 = 5 x 7
Thus 35 can be written as product of prime powers 51 x 71

• 45
Answer:
45 = 5 x 9
= 5 x 3 x 3
= 51 x 32
Thus, 45 can be written as 51 x 32

• 72
Answer:
72 = 24 x 3
= 8 x 3 x 3
= 2 x 2 x 2 x 3 x 3
= 23  x 32
Thus, 72 can be written as 23  x 32

• 36
Answer:
36 = 9 x 4
= 9 x 2 x 2
= 3 x 3 x 2 x 2
= 32 x 22
Thus, 36 can be written as 32 x 22

• 49
Answer:
49 = 7 x 7
= 72
Thus, 49 can be written as 72

Did you note any peculiarity of the exponents of the factors of the squares?
Answer:

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Reverse computation

We have to draw a square; and its area must be 9 square centimetres.
How do we do it?
The area of a square is the square of the side.

So if the area is to be 9 square centimetres, what should be the side?
To draw a square of area 169 square centimetres, what should be the length of a side?
For that, we must find out which number squared gives 169. Looking up our table of squares, we find 132 = 169. So we must draw a square of side 13 centimetres.

Here, given a number we found out which number it is the square of. This operation is called extracting the square root.
That is, instead of saying the square of 13 is 169, we can say in reverse that the square root of 169 is 13.
Just as we write
132 = 169
as shorthand for statement “the square of 13 is 169”, we write the statement “the square root of 169 is 13” in shorthand form as
\(\sqrt{169}\) = 13
(the extraction of square root is indicated by the symbol Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 9)
Similarly, the fact that the square of 5 is 25 can also be stated, the square root of 25 is 5. In short hand form,
52 = 25
\(\sqrt{25}\) = 5
In general
For numbers x and y, if x2 = y, then \(\sqrt{y}\) = x

Now find out the square root of these numbers:

• 100
Answer:
10 = 100
\(\sqrt{100}\) = 10
Therefore, square root of 100 is 10

• 256
Answer:
162 = 256
\(\sqrt{256}\) = 16
Thus, the square root of 256 is 16.

• \(\frac{1}{4}\)
Answer:
\(\frac{1}{4}\) = \(\frac{1}{2}\)2
Thus, the Square root of \(\frac{1}{4}\) = \(\frac{1}{2}\)

• \(\frac{16}{25}\)
Answer:
\(\frac{16}{25}\) = \(\frac{4}{5}\)2
Thus, the square root of \(\frac{16}{25}\) is \(\frac{4}{5}\).

• 1.44
Answer:
1.44 = (1.2)2
Thus, the square root of 1.44 is 1.2

• 0.01
Answer:
0.01 = (0.1)2
Thus, the square root of 0.01 is 0.1

Square root factors Textbook Page No. 87

How do we find the square root of 1225?
Since a product of squares is the square of the product, we need only write 1225 as a product of squares.
First factorize 1225 into primes:
1225 = 52 × 72
And we can write
52 × 72 = (5 × 7)2 = 352
So, 1225 = 352
From this, we get \(\sqrt{1225}\) = 35
Let’s take another example. What is the \(\sqrt{3969}\) ?
As before, we first factorize 3969 into primes.
3969 = 32 × 32 × 72
= (3 × 3 × 7)2
From this, we get \(\sqrt{3969}\) = 3 × 3 × 7 = 63

Now compute the square roots of these.
• 256
Answer:
Given number is 256
Firstly factorizing it into primes we have 256 = 2 x 128
= 2 x 2 x 64
= 2 x 2 x 2 x 32
= 2 x 2 x 2 x 2 x 16
= 2 x 2 x 2 x 2 x 2 x 8
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= (2 x 2) x (2 x 2)  x (2 x 2) x (2 x 2)
= 4 x 4 x 4 x 4
= 16 x 16
= (16)2
From this, we get \(\sqrt{256}\) = 16

• 2025
Answer:
Given number is 2025
Writing the factorization of 2025 we have
= 3 x 3 x 3 x 3 x 5 x 5
= 9 x 9 x 5 x 5
= (9 x 5)2
= (45)2
Therefore, the square root of \(\sqrt{2025}\) is 45

• 441
Answer:
Given number is 441
Writing the factorization of 441 we have
= 3 x 3 x 7 x 7
= 32 x 72
= (3 x 7)2
= (21)2

Therefore, the square root of 441 is 21.

• 921
Answer:
Writing the factorization of 921 we have 3 x 307
Thus, 921 can’t be written as perfect square and the square root of 921 isn’t a natural number.

• 1089
Answer:
Given number is 1089
Writing the factorization of it we have 1089 = 3 x 3 x 11 x 11
= 32 x 112
= (3 x 11)2
= (33)2
Therefore, square root of 1089 is 33.

• 15625
Answer:
Given number is 15625
Writing the factorization of it we have 15625 = 5 x 5 x 5 x 5 x 5 x 5
= 52 x 52 x 52
= (5 x 5 x 5)2
= (125)2
Therefore, the square root of 15625 is 125

• 1936
Answer:
Given number is 1936
Writing the factorization of 1936 we have 2 x 2 x 2 x 2 x 11 x 11
= (22 x 22 x 112)
= (2 x 2 x 11)2
= 442
Therefore, the square root of 1936 is 44.

• 3025
Answer:
Given number is 3025
Writing the factorization of 3025 we have 5 x 5 x 11 x 11
= 52 x 112
= (5 x 11)2
= 552
Therefore, the square root of 3025 is 55.

• 12544
Answer:
Given number is 12544
Writing the factorization of 12544 we have 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7
= 28 x 72
= (112)2
Therefore, Square Root of 12544 is 112.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Let’s do it!

Question 1.
The area of a square plot is 1024 square meters. What is the length of its sides?
Answer:
Given area of Square plot = 1024 Square Meters
Imagine ‘S’ is side of a Square. As we know Area of Square = Side x Side
1024 = Side x Side
1024 = Side2
Side = √1024
Side = 2 5
Hence Length of Square Plot = 32 meters

Question 2.
In a hall, 625 chairs are arranged in rows and columns, with the number of rows equal to the number of columns. The chairs in one row and one column are removed. How many chairs remain?
Answer:
Total number of chairs = 625
Number of Rows and Columns present are Equal i.e. Rows =  Columns , Let it be y.
Rows .Columns=625
y. y =625
y 2= 625
y = √625
y = √(25)2
y = 25
so Number of Rows and Number of Columns =25
When 1 Row and 1 Column is removed as below
Number of Rows =25-1=24
Number of Rows =25-1=24
When the 24 Rows and 24 Columns are arranged = 24 x 24
= 576

Question 3.
The sum of a certain number of consecutive odd numbers, starting with 1, is 5184. How many odd numbers are added?
Answer:
We know the formula of sum of arithmetic series Sn = n/2[2a+(n-1)d]
The arithmetic series 1, 3, 5, 7, 9, 11, 13 ……
Here the first term a = 1
Number of odd numbers to be added = n
Common difference d = 2
Substituting in the formula above we have
5184 = n/2[2×1+(n-1)2]
5184 = n/2[2+2n-2]
5184 = n/2[2n]
5184 = n2
n = 72
Thus, the number of odd numbers to be added is 72.

Question 4.
The sum of two consecutive natural numbers and the square of the first is 5329. What are the numbers?
Answer:
Suppose x and x+1 are two Consecutive natural numbers
As per the Given data, Square of First number x2  = 5329
When they are added it gives as  x+(x+1) +x2 = 5329
x + x+1+x2 = 5329
x + x+1+x = 5329
2x +1+x = 5329
(x+1)= 5329
(x+1) = √5329
x+1 = √(73)2
x+1 = 73
x = 73-1
x= 72
Therefore, the numbers are 72 and 73.

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