Question

Which among the following are the squares of even numbers?

• A. $169$
• B. $256$
• C. $900$
• D. $144$

Answer 1

Answer: (B), (C), (D)

$144$

Let's take a look at each choice one-by-one:

$\bullet$ $\underset{–}{169}$
$169=13\times 13$
$\Rightarrow 169={13}^2$
$\Rightarrow {169}^{\frac{1}{2}}=({13}^2{)}^{\frac{1}{2}}$ (Taking square root both the sides)
$\Rightarrow {169}^{\frac{1}{2}}={13}^{2\times \frac{1}{2}}(\because (a^m{)}^n=a^{m\times n})$
$\Rightarrow \sqrt{169}=13$

Hence, $169$ is a perfect square, and $169$ is a square of $13$ (an odd number).
$\underset{–}{}$

$\bullet$ $\underset{–}{256}$
$256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$
$\Rightarrow 256=2^8$
$\Rightarrow 256=2^{4\times 2}$
$\Rightarrow 256=(2^4{)}^2$ $[\because a^{m\times n}=(a^m{)}^n]$
$\Rightarrow 256={16}^2$ $[\because 2^4=2\times 2\times 2\times 2=16]$
$\Rightarrow {256}^{\frac{1}{2}}=({16}^2{)}^{\frac{1}{2}}$ (Taking square root both the sides)
$\Rightarrow {256}^{\frac{1}{2}}={16}^{2\times \frac{1}{2}}(\because (a^m{)}^n=a^{m\times n})$
$\Rightarrow \sqrt{256}=16$

Hence, $256$ is a perfect square, and $256$ is a square of $16$ (an even number).
$\underset{–}{}$

$\bullet$ $\underset{–}{900}$
$900=2\times 2\times 3\times 3\times 5\times 5$
$\Rightarrow 900=2^2\times 3^2\times 5^2$
$\Rightarrow 900=(2\times 3\times 5{)}^2$ $(\because p^n\times q^n\times r^n=(p\times q\times r{)}^n)$

$\Rightarrow 900={30}^2$
$\Rightarrow {900}^{\frac{1}{2}}=({30}^2{)}^{\frac{1}{2}}$ (Taking square root both the sides)
$\Rightarrow {900}^{\frac{1}{2}}={30}^{2\times \frac{1}{2}}(\because (a^m{)}^n=a^{m\times n})$
$\Rightarrow \sqrt{900}=30$

Hence, $900$ is a perfect square, and $900$ is a square of $30$ (an even number).
$\underset{–}{}$

$\bullet$ $\underset{–}{144}$
$144=\underset{–}{2\times 2}\times \underset{–}{2\times 2}\times 3\times 3$
$\Rightarrow 144=4\times 4\times 3\times 3$
$\Rightarrow 144=4^2\times 3^2$
$\Rightarrow 144=(4\times 3{)}^2(\because a^n\times b^n=(a\times b{)}^n)$
$\Rightarrow 144={12}^2$
$\Rightarrow {144}^{\frac{1}{2}}=({12}^2{)}^{\frac{1}{2}}$ (Taking square root both the sides)
$\Rightarrow {144}^{\frac{1}{2}}={12}^{2\times \frac{1}{2}}(\because (a^m{)}^n=a^{m\times n})$
$\Rightarrow \sqrt{144}=12$

Hence, $144$ is a perfect square, and $144$ is a square of $12$ (an even number).
$\underset{–}{}$

Therefore, $256,900$ and $144$ are the squares of even numbers $16,30$ and $12,$ respectively.