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Square and Square Roots Class 8 Questions

Square and Square Roots Class 8 Questions, deals with various types of questions which are as under:-

  • Finding the perfect square
  • Finding the number to be multiplied to make perfect square
  • Finding the square of two-digit number
  • Finding the value of expression using identity
  • Finding the value of square root term
  • Finding the value of unknown term using identity
  • Word problems on Square root

Square and Square Roots Class 8 Questions – Finding the perfect square

Question 1:

Which one of the following is perfect square ?

488 , 732 , 121,  3482

Solution:

The numbers which end with 2 , 3, 7 or 8 cannot be the complete squares.

Since, 488 , 732 and 3482 end with 8 , 2, 2 , therefore, they cannot be perfect square.

But, 121 can be written as

121 = 11 x 11

Hence, 121 is a perfect square.

Square and Square Roots Class 8 Questions – Finding the number to be multiplied to make perfect square

Question 2:

What is the least number by which 450 should be multiplied, so that the product of the two numbers would be a perfect square?

Solution:

If we find the factors of 450 , we would get 450 = 3 x 3 x 5 x 5 x 2

If we draw out the factors in pairs, we would notice that we have pairs of 3 and 5 , but 2 is the only single number, which is not in pair.

Therefore if 450 is multiplied by 2 then the product would be a complete square.

 

Square and Square Roots Class 8 Questions – Finding the square of two digit number

Question 3:

What is the value of \left(91 \right) ^{ 2 }

Solution:

We know that { (a-b) }^{ 2 }\quad =\quad { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad -\quad 2ab

When finding the square of a larger number, we can use this equation to break the larger number into parts, whose squares can be easily calculated

We can write 91 as ( 100 – 9 )

So

91^{ 2 } =\left(100-9 \right) ^{2 }

Applying { (a-b) }^{ 2 }\quad =\quad { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad -\quad 2ab and substituting the value of a and b, we get

= \left(100 \right) ^{ 2 }+ 9^{2}– 2 x 100 x 9

= 10000 + 81 – 1800

= 10081 – 1800

= 8281

 

Square and Square Roots Class 8 Questions – Finding the value of expression using identity

Question 4:

\cfrac {22\times 22-2\times 20\times 22+20\times 20 }{ \left( 2 \right) ^{ 2 } } = ?

Solution:

We know that { (a-b) }^{ 2 }\quad =\quad { a }^{ 2 }\quad -\quad 2ab\quad +\quad { b }^{ 2 }\quad

If we compare the above equation with the Numerator, we would find that the equation can be re-written as

= 22^{ 2 }-2\times 20\times 22+20^{ 2 }

Or we can say that the given equation has value of

a = 22 and b = 20

So we can replace the numeration

22 x 22 – 2 x 20 x 22 + 20 x 20 with \left( 22-20 \right) ^{ 2 }=\left( 2 \right) ^{ 2 }

Now putting these values in Numerator, we would get

\cfrac {2^{ 2 } }{ 2^{ 2 } } = \cfrac {4}{4} = 1

 

Question 5:

Find the value of \left(44 \right) ^{ 2 }-\left(43 \right) ^{ 2 }+2^{2}

Solution:

\left(44 \right) ^{ 2 }-\left(43 \right) ^{ 2 }+2^{2}

We know that\left( a^{ 2 }-b^{ 2 } \right) =\left( a-b \right) \left( a+b \right)

a = 44 and b = 43

= ( 44 + 43 ) ( 44 – 43 ) + 4

= 87 x 1 + 4

= 87 + 4

= 91

 

Question 6:

The value of \left(555 \right) ^{ 2 }– \left(554 \right) ^{ 2 }?

Solution:

We know that \left( a \right) ^{ 2 }-\left( b \right) ^{ 2 }=\left( a-b \right) \left( a+b \right)

If we compare the above equation to the problem , we find that

a = 555 and b = 554

So we can rewrite the above equation as

555^{2 }-554^{2 } = ( 555 + 554 ) ( 555 – 554 )

= 1109 x 1

= 1109

 

Square and Square Roots Class 8 Questions – Finding the value of square root term

Question 7:

If\sqrt {18 } = 4.24 , what is the value of\sqrt { \cfrac {2 }{ 9} } ?

Solution:

In order to solve these type of problems, we first need to equate the value which has to be solved \sqrt { \cfrac {2 }{ 9} } of the given equation in the given equation

We can write \sqrt {18} = 4.24

Replacing 18 by ( 2 x 9 ) in the given equation, and dividing both the sides by 9 , we would get

\sqrt { \cfrac { 2\times 9 }{ 9\times 9 }}

= \sqrt { \cfrac {18}{81} }

= \cfrac { \sqrt {18 } }{9}

= \cfrac {4.24}{9}

= 0.47

 

Square and Square Roots Class 8 Questions – Finding the value of unknown term using identity

Question 8:

If 42^{2} – 40^{2} = {\left(12 \right) ^{ 2 } – \left(10 \right) ^{ 2 }} x Y , then the value of Y is ?

Solution:

Since,

\left( a^{ 2 }-b^{ 2 } \right) =\left( a-b \right) \left( a+b \right)

We can rewrite both the equation as

42^{2} – 40^{2} = {\left(12 \right) ^{ 2 } – \left(10 \right) ^{ 2 }} x Y

( 42 + 40 ) x ( 42 – 40 ) = ( 12 + 10 ) x ( 12 – 10 ) x Y

82 x 2 = 22 x 2 x Y

164 = 44 x Y

Or,

Y = \cfrac {164}{44}

Y = \cfrac {41}{11}

 

Square and Square Roots Class 8 Questions – Word problems on Square root

Question 9:

In a conference hall of a school, 900 students were sitting in such a way that there were as many rows as there were students in a row. How many rows were there?

Solution:

Let the number of rows = Number of students in a row = x

Total number of students = Number of rows x Number of students in a row

Then

x\times x=900,

x^{ 2 }=900

x=√900

= √ 3 x 3 x 5 x 5 x 2 x 2

= 3 x 5 x 2

= 30

Hence, the number of rows = Number of students in a row = 30

 

Question 10:

A group of People decided to collect as many paisa from each member of the group as is the number of members. If the total collection amounts to Rs. 11.56 , the number of members in the group is ?

Solution:

Funds collected in paisa= 11.56 x 100

= 1156

Funds collected= Numbers of the members of the group x Paisa collected from each memeber

Or

1156= Numbers of the members of the group x Paisa collected from each memeber

Let the Number of members of the group= x

Then Paisa collected from each member= x

Substituting these values in the given equation

1156= x x x,

x^{ 2 }= 1156

Or,

x = \sqrt {1156},

x = 34

Numbers of the members of the group = 34

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