# NCERT Solutions Class 8 Maths Chapter 6 Exercise 6.2

The Central Board of Secondary Education (CBSE) is one of India’s main national school education boards. Thousands of students are enrolled in the CBSE. Mathematics is one of the compulsory subjects taught to Class 8 CBSE students. It is an essential component of the school curriculum and is essential for achieving success on the yearly exams. In Class 8, students are introduced to a variety of fresh concepts. Some of these are squares of numbers as well as their square roots. Students also need to use these concepts in their higher classes. As a result, Class 8 students must fully comprehend the concepts to do well in their annual examinations as well as in their future classes. Class 8 students may find the yearly exam to be very difficult. Students are frequently overwhelmed by the extensive syllabus in Class 8. They become extremely stressed out and anxious about finishing their coursework on time. Subsequently, they are less effective in exams and are unable to reach their full potential. They receive less than expected marks in the exams because they are unable to complete the question paper. However, they can significantly improve their performance with adequate examination preparation beforehand. They can manage their stress and perform well for the annual exam if given the right support and resources.

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## NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.2) Exercise 6.2

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

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### NCERT Solutions for Class 8

It is crucial for Class 8 students to carefully study all the subjects. However, that is insufficient for students to achieve high test scores. To avoid mistakes and errors during the exam, students should practice numerous questions. Practising questions repeatedly also improves students’ ability to remember information. Students can easily do so with the help of Extramarks’ NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.2 and other exercises. Not only the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.2, but NCERT solutions also are available for Class 8 students for all subjects and topics.

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Q.1 Find the square of the following numbers.

(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{}{32}^{2}={\left(30\text{}+\text{}2\right)}^{2}\\ =\left(30\text{}+\text{}2\right)\left(30\text{}+\text{}2\right)\\ =30\left(30\text{}+\text{}2\right)+2\left(30\text{}+\text{}2\right)\\ =30×30+2×30+2×30+\text{}2×2\\ =900+120+4\\ =\text{}1024\\ \\ \left(\mathrm{ii}\right){35}^{2}={\left(30+5\right)}^{2}\\ =\left(30+5\right)\left(30+5\right)\\ =30\left(30+5\right)+5\left(30+5\right)\\ =30×30+5×30+5×30+\text{}5×5\\ =900+150+150+25\\ =\text{}1225\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{}{86}^{2}={\left(80+6\right)}^{2}\\ =\left(80+6\right)\left(80+6\right)\\ =80\left(80+6\right)+6\left(80+6\right)\\ =80×80+6×80+6×80+\text{}6×6\\ =6400+480+480+36\\ =7396\\ \text{}\\ \left(\mathrm{iv}\right)\text{}{93}^{2}={\left(90+3\right)}^{2}\\ =\left(90+3\right)\left(90+3\right)\\ =90\left(90+3\right)+3\left(90+3\right)\\ =90×90+3×90+3×90+3×3\\ =8100+270+270+9\\ =8649\\ \\ \left(\mathrm{v}\right)\text{}{71}^{2}\text{}={\left(70+1\right)}^{2}\\ =\left(70+1\right)\left(70+1\right)\\ =70\left(70+1\right)+1\left(70+1\right)\\ =70×70+1×70+1×70+1×1\\ =4900+70+70+1\\ =5041\end{array}$

$\begin{array}{l}\left(\mathrm{vi}\right)\text{}{46}^{2}={\left(40+6\right)}^{2}\\ =\left(40+6\right)\left(40+6\right)\\ =40\left(40+6\right)+6\left(40+6\right)\\ =40×40+6×40+6×40+6×6\\ =1600+240+240+36\\ =\text{}2116\end{array}$

Q.2 Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14 (iii) 16 (iv) 18

Ans

$\begin{array}{l}\text{For any natural number}\mathrm{m}>\text{1},\\ \text{2}\mathrm{m},{\mathrm{m}}^{\text{2}}-\text{1},{\mathrm{m}}^{\text{2}}+\text{1 forms a Pythagorean triplet}.\\ \left(\text{i}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{6},\text{then}{\mathrm{m}}^{\text{2}}=\text{5}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\end{array}$

$\begin{array}{l}\text{If we take}{m}^{\text{2}}-\text{1}=\text{6},\text{then}{m}^{\text{2}}=\text{7}\\ \text{Again the value of}m\text{is not an integer}.\\ {\text{So, we try to take m}}^{\text{2}}\text{+ 1 = 6}\text{.}\\ {\text{Again m}}^{\text{2}}\text{= 5 will not give an integer value for m}\text{.}\\ \text{Let 2}m=\text{6}⇒m=\text{3}\\ \text{Therefore},\text{the Pythagorean triplets are}\\ \text{2}×\text{3},{\text{3}}^{\text{2}}-\text{1},{\text{3}}^{\text{2}}+\text{1 or 6},\text{8},\text{and 1}0.\end{array}$

$\begin{array}{l}\left(\text{ii}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{14},\text{then}{\mathrm{m}}^{\text{2}}=\text{13}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{14},\text{then}{\mathrm{m}}^{\text{2}}=\text{15.}\\ \text{Again the value of}\mathrm{m}\text{is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{14}⇒\mathrm{m}=\text{7}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{49}-\text{1}=\text{48 and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{49}+\text{1}=\text{5}0\\ \text{Therefore},\text{the required triplet is 14},\text{48},\text{and 5}0.\end{array}$

$\begin{array}{l}\left(\text{iii}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{16},\text{then}{\mathrm{m}}^{\text{2}}=\text{15}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{16},\text{then}{\mathrm{m}}^{\text{2}}=\text{17}\\ \text{Again the value of}\mathrm{m}\text{\hspace{0.17em}is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{16}⇒\mathrm{m}=\text{8}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{64}-\text{1}=\text{63 and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{64}+\text{1}=\text{65}\\ \text{Therefore},\text{the Pythagorean triplet is 16},\text{63},\text{and 65}.\end{array}$

$\begin{array}{l}\left(\text{iv}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{18},{\mathrm{m}}^{\text{2}}=\text{17}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{18},\text{then}{\mathrm{m}}^{\text{2}}=\text{19}\\ \text{Again the value of}\mathrm{m}\text{\hspace{0.17em}\hspace{0.17em}is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{18}⇒\mathrm{m}=\text{9}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{81}-\text{1}=\text{8}0\text{and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{81}+\text{1}=\text{82}\\ \text{Therefore},\text{the Pythagorean triplet is 18},\text{8}0,\text{and 82}.\end{array}$