Excercise 6.2 Squares and Square Roots- NCERT Solutions Class 8

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Chapter 6 Ex.6.2 Question 1

Find the squares of the following numbers.

   (i) \(32\)   

  (ii) \(35\) 

 (iii) \(86\)   

 (iv) \(93\)   

 (v) \(71\)   

(vi) \(46\)

Solution

Video Solution

What is known?

Numbers

What is unknown?

Squares of the numbers.

Reasoning 1:

There is a way to find this by without multiplication.

Use identity: 

\((a+b)^2=a^2+2ab+b^2\)

\((a+b)^2=a^2-2ab+b^2\)

Steps (i): 

\[\begin{align}32 &= 30 + 2\\{32^2} &= {(30 + 2)^2}\\ &= 30(30 + 2) + 2(30 + 2)\\ &= {30^2} + 30 \times 2 + 2 \times 30 + {2^2}\\ &= 900 + 60 + 60 + 4\\ &= 1024\end{align}\]

Reasoning 2:

If a number have its unit digit \(5\) i.e. \(a5\), its square number will be [\(a (a+1)\; \rm hundred+25\)].

Steps (ii): 

Hence \(a = 3\)

Square of the number \(35\)

\[\begin{align}&= [3(3+1) \text{hundreds}+25]\\&= [ (3\times 4) \;\text{hundreds}+25]\\&= 1200+25\\&= 1225 \end{align}\]

Reasoning:

There is a way to find this by without multiplication

Steps (iii): 

\[\begin{align}86 &= 80 + {86^2}\\ &= {(80 + 6)^2}{\rm{ }}\\ &= 80(80 + 6) + 6(80 + 6)\\ &= {80^2} + 80 \times 6 + 6 \times 80 + {6^2}\\ &= 6400 + 480 + 480 + 36\\&= 7396 \end{align}\]

Reasoning:

There is a way to find this by without multiplication.

Steps (iv):

\[\begin{align}93 &= 90 + 3\\{93^2} &= {(90 + 3)^2}\\ &= 90(90 + 3) + 3(90 + 3)\\ &= {90^2} + 90 \times 3 + 3 \times 90 + {3^2}\\ &= 8100 + 270 + 270 + 9\\ &= 8649 \end{align}\]

Reasoning

There is a way to find this by without multiplication.

Steps (v):

\[\begin{align}71 &= 70 + 1\\{71^2} &= {(70 + 1)^2}\\ &= 70(70 + 1) + 1(70 + 1)\\ &= {70^2} + 70 \times 1 + 1 \times 70 + {1^2}\\ &= 4900 + 70 + 70 + 1\\ &= 5041 \end{align}\]

Reasoning

There is a way to find this by without multiplication.

Steps (vi):

\[\begin{align} 46 &= 40 + 6\\{46^2} &= {(40 + 6)^2}\\ &= 40(40 + 6) + 6(40 + 6)\\ &= {40^2} + 40 \times 6 + 6 \times 40 + {6^2}\\ &= 1600 + 240 + 240 + 36\\ &= 2116 \end{align}\]

Chapter 6 Ex.6.2 Question 2

Write a Pythagorean triplet whose one member is.

(i) \(6\) 

(ii) \(14\) 

(iii) \(16 \)

(iv) \(18\)

Solution

Video Solution

What is known?

One of the members of Pythagorean triplet.

What is unknown?

Other two member of Pythagorean triplet.

Reasoning:

For any natural number \(''m''\) where \(m>1\), we have \(\begin{align} (2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2 \end{align}\) so, \(2m\), \(m^2 - 1 \,\,{\rm{and}}\,\,m^2 + 1\) forms a Pythagorean triplet 

Steps (i):

\(6\)

If we take

\(\begin{align} m^2 + 1 &= 6\\m^2 &= 5\end{align}\)

the value of \(m\) will not be an integer.

If we take 

\(\begin{align} m^2 - 1 &= 6\\m^2 &= 7 \end{align}\)

Again,the value of \(m\) will not be an integer.

Let ,\(2m=6\)

\[\begin{align}m &= \frac{6}{2} = 3\\\\m^2 - 1 &= (3)^2 - 1 \\&= 9 - 1 \\&= 8\\\\m^2 + 1 &= (3)^2 + 1 \\&= 9 + 1 \\&= 10 \end{align}\]

Therefore, pythagorean triplets are \(6, 8\) and \(10\)

Steps (ii):

\(14\)

If we take

\(\begin{align} m^2 + 1 &= 14\\m^2 &= 13 \end{align}\)

the value of the integer will not be an integer

If we take 

\(\begin{align} m^2 - 1 &= 14\\m^2& = 15 \end{align}\)

Again, the value of \(m\) will not be an integer.

Let \(2m=14\)

\[\begin{align} m&= \frac{14}{2} \\&= 7\\\\m^2 - 1 &= 7^2 - 1 \\&= 49 - 1 \\&= 48\\\\m^2 + 1 &= 7^2 + 1 \\&= 49 + 1 \\&= 50 \end{align}\]

Therefore ,\(14, 48, 50\) are pythagorean triplets.

Steps (iii):

\(16\)

If we take

\(\begin{align} m^2 + 1& = 16\\m^2& = 15 \end{align}\)

the value of \(m\) will be an integer.

If we take 

\(\begin{align} m^2 - 1& = 16\\m^2& = 17 \end{align}\)

The value of \(m\) will be an Integer.

Let ,\(2m=6\)

\[\begin{align} m &= \frac{16}{2} \\&= 8\\\\m^2 - 1 &= 8^2 - 1 \\&= 64 - 1 \\&= 63\\\\m^2 + 1 &= 8^2 + 1 \\&= 64 + 1 \\&= 65 \end{align}\]

Therefore, \(16, 63\) and \(65\) are pythagorean triplets.

Steps (iv):

\(18\)

If we take 

\(\begin{align} m^2 - 1 &= 18\\m^2& = 19 \end{align}\)

the value of \(m\) will not be an integer.

If we take

\(\begin{align} m^2 + 1& = 18\\m^2& = 17 \end{align}\)

again the value of \(m\) will not be an Integer.

Let \(2m=18\)

\[\begin{align}m &= \frac{18}{2} \\&= 9\\\\m^2 - 1 &= 9^2 - 1 \\&= 81 - 1 \\&= 80\\\\m^2 + 1 &= 9^2 + 1 \\&= 81 + 1 \\&= 82 \end{align}\]

Therefore, \(18, 80\) and \(82\) are pythagorean triplets.

  
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