Exercise 5.2 (Revised) - Chapter 6 - Square & Square Roots - Ncert Solutions class 8 - Maths

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Chapter 5 - Square & Square Roots - NCERT Solutions Class 8 Maths | Expert Solutions

Ex 5.2 Question 1.

Find the squares of the following numbers:
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46

Answer.

$
\begin{aligned}
& \text {(i) }(32)^2=(30+2)^2=(30)^2+2 \times 30 \times 2+(2)^2 \\
& {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} \\
& =900+120+4=1024
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) }(35)^2=(30+5)^2=(30)^2+2 \times 30 \times 5+(5)^2 \\
& {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} \\
& =900+300+25=1225
\end{aligned}
$
$
\begin{aligned}
& \text { (iii) }(86)^2=(80+6)^2=(80)^2+2 \times 80 \times 6+(6)^2 \\
& {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} \\
& =6400+960+36=7396
\end{aligned}
$
(iv) $(93)^2=(90+3)^2=(90)^2+2 \times 90 \times 3+(3)^2$

$\begin{aligned}
& {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} \\
& \quad=8100+540+9=8649
\end{aligned}$

$
\begin{aligned}
& \text { (v) }(71)^2=(70+1)^2=(70)^2+2 \times 70 \times 1+(1)^2 \\
& {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} \\
& =4900+140+1=5041
\end{aligned}
$
(vi) $(46)^2=(40+6)^2=(40)^2+2 \times 40 \times 6+(6)^2$
$
\begin{aligned}
& {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} \\
& \quad=1600+480+36=2116
\end{aligned}
$
Ex 5.2 Question 2.

Write a Pythagoras triplet whose one member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18

Answer.

(i) There are three numbers $2 m, m^2-1$ and $m^2+1$ in a Pythagorean Triplet. Here, $2 m=6 \Rightarrow m=\frac{6}{2}=3$ Therefore, Second number $\left(m^2-1\right)=(3)^2-1=9-1=8$ Third number $m^2+1=(3)^2+1=9+1=10$

Hence, Pythagorean triplet is $(6,8,10)$.
(ii) There are three numbers $2 m=m^2-1$ and $m^2+1$ in a Pythagorean Triplet. Here, $2 m=14 \Rightarrow m=\frac{14}{2}=7$ Therefore, Second number $\left(m^2-1\right)=(7)^2-1=49-1=48$

Third number $m^2+1=(7)^2+1=49+1=50$
Hence, Pythagorean triplet is $(14,48,50)$.
(iii) There are three numbers $2 m, m^2-1$ and $m^2+1$ in a Pythagorean Triplet.

Here, $2 m=16 \Rightarrow m=\frac{16}{2}=8$
Therefore, Second number $\left(m^2-1\right)=(8)^2-1=64-1=63$
Third number $m^2+1=(8)^2+1=64+1=65$
Hence, Pythagorean triplet is $(16,63,65)$.
(iv) There are three numbers $2 m_{=} m^2-1$ and $m^2+1$ in a Pythagorean Triplet.

Here, $2 m=18 \Rightarrow m=\frac{18}{2}=9$
Therefore, Second number $\left(m^2-1\right)=(9)^2-1=81-1=80$
Third number $m^2+1=(9)^2+1=81+1=82$

$\text { Hence, Pythagorean triplet is }(18,80,82) \text {. }$