Question

$maths \: aryabhattas \: attention \: \\ \\ find \: the \: number \: of \: digits \: in \: the \: \\ square \: root \: of \: the \: following \: \\ \\ 1)81 \\ \\ 2)169 \\ \\ 3)6561 \\ \\ 4)273529 \\ \\ 5)1014049$
best will be marked as brainliest ​

Answer 1

1. 81. (9*9)
2. 169. (13*13)
3. 6561. (81*81)
4. 273529. (523*523)
5. 1014049. (1007*1007)

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Answer 2

Answer:-

To see the number of digits in the square root of a number, first let's check the maximum and minimum squares of numbers with differrent digits.

$1^2 = 1 \\ 10^2 = 100$

$\star$ So if the square root of a number has 1 digit, it will between 1 and 100

$10^2 = 100 \\ 100^2 = 10,000$

$\star$ If the square root of a number has 2 digits, it will lie between 100 and 10,000.

$100^2 = 10,000 \\ 1000^2 = 10,00,000$

$\star$ If the square root of a number has 3 digits, it will lie between 10,000 and 10,00,000.

$1000^2 = 10,00,000 \\ 10,000^2 = 10,00,00,000$

$\star$ If the square of a number lies between 10,00,000 and 10,00,00,000 , then the number has 4 digits.

Solution:-

81 lies between 10 and 100,

$\therefore \sqrt{81}$ will have 1 digit.

169 lies between 100 and 10,000

$\therefore \sqrt{169}$ will have 2 digits.

6561 lies between 100 and 10,000

$\therefore \sqrt{6,561}$ will have 2 digits.

2,73,529 lies between 10,000 and 10,00,000.

$\therefore \sqrt{2,73,529}$ will have 3 digits.

10,14,049 lies between 10,00,000 and 10,00,00,000

$\therefore \sqrt{10,14,049}$ will have 4 digits.