question_answer 26)

                Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.        (i) 525                                    (ii) 1750                (iii) 252                                  (iv) 1825 (v) 6412

Answer:

                (i) 525                          We have

	22
2	\[\overline{5}\]          \[\overline{25}\] ?4
42	1            25 ?             84
	41

  This shows that \[{{22}^{2}}<525\]. Next perfect square is \[{{23}^{2}}=529\]. Hence, the number to be added is \[{{23}^{2}}-525=529-525=4\]. Therefore, the perfect square so obtained is \[525+4=529\]. Hence,  \[\sqrt{529}=23\]. (ii) 1750 We have

	41
4	\[\overline{17}\]          \[\overline{50}\] ?16
81	1            50 ?             81
	69

  This shows that \[{{41}^{2}}<1750\]. Next perfect square is \[{{42}^{2}}=1764\]. Hence, the number to be added is \[{{42}^{2}}-1750=1764-1750=14\] Therefore, the perfect square so obtained is \[1750+14=1764\]. Hence, \[\sqrt{1764}=42\]. (iii) 252 We have   

	15
1	\[\overline{2}\]          \[\overline{52}\] ?1
25	1            52 ?1           25
	27

This shows that     \[{{15}^{2}}<252\]. Next perfect square is \[{{16}^{2}}=256\]. Hence, the number to be added is \[{{16}^{2}}-252=256-252=4\]. Therefore, the perfect square so obtained is \[252+4=256\]. Hence, \[\sqrt{256}=16\]. (iv) 1825 We have  

	42
4	\[\overline{18}\]          \[\overline{25}\] ?16
82	2            25 ?1           64
	61

Thus shows that     \[{{42}^{2}}<1825\]. Next perfect square is \[{{43}^{2}}=1849\]. Hence, the number to be added is \[{{43}^{2}}-1825=1849-1825=24\]. Therefore, the perfect square so obtained is \[1825+24=1849\]. Hence, \[\sqrt{1849}=43\]. (v) 6412 We have

	80
8	\[\overline{64}\]          \[\overline{12}\] ?64
160	12                 ?0
	12

This shows that  \[{{80}^{2}}<6412\]. Next perfect square is \[{{81}^{2}}=6561\]. Hence, the number to be added is \[{{81}^{2}}-6412=6561-6412=149\]. Therefore, the perfect square so obtained is \[6412+149=6561\]. Hence, \[\sqrt{6561}\,=81\].