Question

ans the given question​

Answer 1

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:\begin{tabular}{|c|c|c|}\cline{1-3}121&52&225\\\cline{1-3}256&60&196\\\cline{1-3}289&?&144\\\cline{1-3}\end{tabular}$

To Find:

• The value that replace ‘?’

Solution:

We are given that,

$\implies\begin{tabular}{|c|c|c|}\cline{1-3}121&52&225\\\cline{1-3}256&60&196\\\cline{1-3}289&?&144\\\cline{1-3}\end{tabular}$

Let us assume the value of the ? as x.
$\implies\begin{tabular}{|c|c|c|}\cline{1-3}121&52&225\\\cline{1-3}256&60&196\\\cline{1-3}289&x&144\\\cline{1-3}\end{tabular}$

We can see that the values in column 1 and 3, are perfect squares, i.e.,

⇒ 121 = 11²

⇒ 256 = 16²

⇒ 289 = 17²

⇒ 225 = 15²

⇒ 196 = 14²

⇒ 144 = 12²

Replacing these values in the table, we get,

$\implies\begin{tabular}{|c|c|c|}\cline{1-3}11^2&52&15^2\\\cline{1-3}16^2&60&14^2\\\cline{1-3}17^2&x&12^2\\\cline{1-3}\end{tabular}$

Now, we can see that, the values in 2nd column can be expressed as twice the sum of the values whose squares are present in column 1 and 3.

That is,

⇒ 52 = 2(11 + 15) [2*26 = 52]

⇒ 60 = 2(16 + 14) [2*30 = 60]

Similarly we can express x as,

⇒ x = 2(17 + 12)

⇒ x = 2(29)

⇒ x = 58

Hence, the value that replaces ‘?’ in the table is 58.

Therefore, Option 4) 58 is correct.
FINAL TABLE:

$\bf\implies\begin{tabular}{|c|c|c|}\cline{1-3}121&52&225\\\cline{1-3}256&60&196\\\cline{1-3}289&58&144\\\cline{1-3}\end{tabular}$