NCERT Solutions for Class 8 Maths Chapter 6: Squares and Square Roots - Exercise 6.2

Exercise 6.2

Refer to page 1-8 for exercise 6.2 in the PDF.

1. Find the square of the following numbers

i) The number is 32.

Ans: The number given is 32.

The number 32 can be written as the sum of 30 and 2.

$ \Rightarrow 32 = 30 + 2$

Thus, the square of the number is given as,

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

Thus, substituting the values we get,

$\Rightarrow {32^2} = {\left( {30 + 2} \right)^2}$ 

$\Rightarrow {32^2} = \left( {30 + 2} \right) \cdot \left( {30 + 2} \right)$

On evaluating further,

$\left( {30 + 2} \right) \cdot \left( {30 + 2} \right) = 30\left( {30 + 2} \right) + 2\left( {30 + 2} \right)$ 

${30^2} + 30 \times 2 + 2 \times 30 + {2^2}   = 900 + 60 + 60 + 4 $ 

${30^2} + 30 \times 2 + 2 \times 30 + {2^2}= 1024$

Therefore, the square of the number 32 is found to be 1024.

ii) The number is 35.

Ans: The number given is 35.

The number 35 can be written as the sum of 30 and 5.

$\Rightarrow 35=30+5$

Thus, the square of the number is given as,

$\Rightarrow {{35}^{2}}={{\left( 30+5 \right)}^{2}}$

$\Rightarrow {{35}^{2}}=\left( 30+5 \right)\cdot \left( 30+5 \right)$

On evaluating further,

$ \Rightarrow \left( 30+5 \right)\cdot \left( 30+5 \right)=30\left( 30+5 \right)+5\left( 30+5 \right) $

$ \Rightarrow \left( 30+5 \right)\cdot \left( 30+5 \right)={{30}^{2}}+30\times 5+5\times 30+{{5}^{2}}$

$\Rightarrow \left( 30+5 \right)\cdot \left( 30+5 \right)=900+150+150+25$

$\Rightarrow \left( 30+5 \right)\cdot \left( 30+5 \right)=1225$

Therefore, the square of the number 35 is found to be 1225.

iii) The number is 86.

Ans: The number given is 86.

The number 86 can be written as the sum of 80 and 6.

$ \Rightarrow 86 = 80 + 6$

Thus, the square of the number is given as,

$\Rightarrow {86^2} = {\left( {80 + 6} \right)^2} $ 

$\Rightarrow {86^2} = \left( {80 + 6} \right) \cdot \left( {80 + 6} \right)$

On evaluating further,

$\Rightarrow \left( 80+6 \right)\cdot \left( 80+6 \right)=80\left( 80+6 \right)+6\left( 80+6 \right)$ 

$\Rightarrow \left( 80+6 \right)\cdot \left( 80+6 \right)={{80}^{2}}+80\times 6+6\times 80+{{6}^{2}}$

$\Rightarrow \left( 80+6 \right)\cdot \left( 80+6 \right)=6400+480+480+36$

$\Rightarrow \left( 80+6 \right)\cdot \left( 80+6 \right)=7396$

Therefore, the square of the number 86 is found to be 7396.

iv) The number is 93.

Ans: The number given is 93.

The number 93 can be written as the sum of 90 and 3.

$ \Rightarrow 93 = 90 + 3$

Thus, the square of the number is given as,

$\Rightarrow {93^2} = {\left( {90 + 3} \right)^2}$ 

$\Rightarrow {93^2} = \left( {90 + 3} \right) \cdot \left( {90 + 3} \right)$

On evaluating further,

$\Rightarrow \left( 90+3 \right)\cdot \left( 90+3 \right)=90\left( 90+3 \right)+3\left( 90+3 \right)$

$\Rightarrow \left( 90+3 \right)\cdot \left( 90+3 \right)={{90}^{2}}+90\times 3+3\times 90+{{3}^{2}}$

$\Rightarrow \left( 90+3 \right)\cdot \left( 90+3 \right)=8100+270+270+9$

$\Rightarrow \left( 90+3 \right)\cdot \left( 90+3 \right)=8649$

Therefore, the square of the number 93 is found to be 8649.

v) The number is 71.

Ans: The number given is 71.

The number 71 can be written as the sum of  70 and 1.

$ \Rightarrow 71 = 70 + 1$

Thus, the square of the number is given as,

$\Rightarrow {71^2} = {\left( {70 + 1} \right)^2}$ 

$\Rightarrow {71^2} = \left( {70 + 1} \right) \cdot \left( {70 + 1} \right)$

On evaluating further,

$\Rightarrow \left( 70+1 \right)\cdot \left( 70+1 \right)=70\left( 70+1 \right)+1\left( 70+1 \right)$

$\Rightarrow \left( 70+1 \right)\cdot \left( 70+1 \right)={{70}^{2}}+70\times 1+1\times 70+{{1}^{2}}$

$\Rightarrow \left( 70+1 \right)\cdot \left( 70+1 \right)=4900+70+70+1$

$\Rightarrow \left( 70+1 \right)\cdot \left( 70+1 \right)=5041$

Therefore, the square of the number 71 is found to be 5041.

vi) The number is 46.

Ans: The number given is 46.

The number 46 can be written as the sum of  40 and 6.

$ \Rightarrow 46 = 40 + 6$

Thus, the square of the number is given as,

$\Rightarrow {46^2} = {\left( {40 + 6} \right)^2}$ 

$\Rightarrow {46^2} = \left( {40 + 6} \right) \cdot \left( {40 + 6} \right)$

On evaluating further,

$\Rightarrow \left( 40+6 \right)\cdot \left( 40+6 \right)=40\left( 40+6 \right)+6\left( 40+6 \right)$

$\Rightarrow \left( 40+6 \right)\cdot \left( 40+6 \right)={{40}^{2}}+40\times 6+6\times 40+{{6}^{2}}$

$\Rightarrow \left( 40+6 \right)\cdot \left( 40+6 \right)=1600+240+240+36$

$\Rightarrow \left( 40+6 \right)\cdot \left( 40+6 \right)=2116$

Therefore, the square of the number 46 is found to be 2116.

2. Write a Pythagorean triplet whose one member is,

i) The number 6.

Ans: It is known that for any natural number $n$ greater than 1,$2n,{\text{ }}{n^2} - 1,{\text{ }}{n^2} + 1$ forms a Pythagorean triplet.

Now, the member of the triplet is given to be 6.

Consider substituting 6 to find $n$.

If, 6 is substituted in ${n^2} + 1$ then its value will be,

$\Rightarrow {n^2} + 1 = 6$

$\Rightarrow {n^2} = 6 - 1$

$\Rightarrow {n^2} = 5$

On evaluation, the value of $n$ will not be an integer, hence it is neglected.

If, 6 is substituted in ${n^2} - 1$ then its value will be,

$\Rightarrow {n^2} - 1 = 6$ 

$\Rightarrow {n^2} = 6 + 1$

$\Rightarrow {n^2} = 7$

Again on evaluation, the value of $n$ will not be an integer, hence it is neglected.

Now, substitute 6 in $2n$.

$\Rightarrow 2n = 6$

$\Rightarrow n = 3$

Therefore, the required triplet is $2 \times 3,{\text{ }}{3^2} - 1,{\text{ }}{3^2} + 1{\text{ or }}6,{\text{ 8, and }}10.$

ii) The number 14.

Ans: It is known that for any natural number $n$ greater than 1,$2n,{\text{ }}{n^2} - 1,{\text{ }}{n^2} + 1$ forms a Pythagorean triplet.

Now, the member of the triplet is given to be 14.

Consider substituting 14 to find $n$.

If, 14 is substituted in ${n^2} + 1$ then its value will be,

$\Rightarrow {n^2} + 1 = 14$

$\Rightarrow {n^2} = 14 - 1$

$\Rightarrow {n^2} = 13$

On evaluation, the value of $n$ will not be an integer, hence it is neglected.

If, 14 is substituted in ${n^2} - 1$ then its value will be,

$\Rightarrow {n^2} - 1 = 14$

$\Rightarrow {n^2} = 14 + 1$

$\Rightarrow {n^2} = 15$

Again on evaluation, the value of $n$ will not be an integer, hence it is neglected.

Now, substitute 14 in $2n$.

$\Rightarrow 2n = 14$

$\Rightarrow n = 7$

Therefore, the required triplet is $2 \times 7,{\text{ }}{{\text{7}}^2} - 1,{\text{ }}{{\text{7}}^2} + 1{\text{ or 14}},{\text{ 48, and 50}}.$

iii) The number 16.

Ans: It is known that for any natural number $n$ greater than 1,$2n,{\text{ }}{n^2} - 1,{\text{ }}{n^2} + 1$ forms a Pythagorean triplet.

Now, the member of the triplet is given to be 16.

Consider substituting 16 to find $n$.

If, 16 is substituted in ${n^2} + 1$ then its value will be,

$\Rightarrow{n^2} + 1 = 16$

$\Rightarrow{n^2} = 16 - 1$

$\Rightarrow{n^2} = 15$

On evaluation, the value of $n$ will not be an integer, hence it is neglected.

If, 16 is substituted in ${n^2} - 1$ then its value will be,

$\Rightarrow{n^2} - 1 = 16$

$\Rightarrow{n^2} = 16 + 1$

$\Rightarrow{n^2} = 17$

Again on evaluation, the value of $n$ will not be an integer, hence it is neglected.

Now, substitute 16 in $2n$.

$\Rightarrow 2n = 16$

$\Rightarrow n = 8$

Therefore, the required triplet is $2 \times 8,{\text{ }}{{\text{8}}^2} - 1,{\text{ }}{{\text{8}}^2} + 1{\text{ or 16}},{\text{ 63, and 65}}$.

iv) The number 18.

Ans: It is known that for any natural number $n$ greater than 1,$2n,{\text{ }}{n^2} - 1,{\text{ }}{n^2} + 1$ forms a Pythagorean triplet.

Now, the member of the triplet is given to be 18.

Consider substituting 18 to find $n$.

If, 18 is substituted in ${n^2} + 1$ then its value will be,

$\Rightarrow{n^2} + 1 = 18$

$\Rightarrow{n^2} = 18 - 1$

$\Rightarrow{n^2} = 17 $

On evaluation, the value of $n$ will not be an integer, hence it is neglected.

If, 18 is substituted in ${n^2} - 1$ then its value will be,

$\Rightarrow{n^2} - 1 = 18$

$\Rightarrow{n^2} = 18 + 1$

$\Rightarrow{n^2} = 19$

Again on evaluation, the value of $n$ will not be an integer, hence it is neglected.

Now, substitute 18 in $2n$.

$\Rightarrow 2n = 18$

$\Rightarrow n = 9$

Therefore, the required triplet is $2 \times 9,{\text{ }}{{\text{9}}^2} - 1,{\text{ }}{{\text{9}}^2} + 1{\text{ or 18}},{\text{ 80, and 82}}$.

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.2

Opting for the NCERT solutions for Ex 6.2 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.2 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 6 Exercise 6.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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