Squares and Square Roots, Cubes and Cube Roots — Exercise 3(C)

Find the square root of 576 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{576} = 24}$.

Find the square root of 961 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{961} = 31}$.

Find the square root of 1444 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{1444} = 38}$.

Find the square root of 4489 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{4489} = 67}$.

Find the square root of 6241 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{6241} = 79}$.

Find the square root of 5476 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{5476} = 74}$.

Find the square root of 9025 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{9025} = 95}$.

Find the square root of 7569 by division method.

Answer

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{7569} = 87}$.

The area of a square field is 7744 sq. metres. Find its perimeter.

Answer

Let the side of the square field be $x$ metres.

Area of the square field = $x^2$

Area of the square field = 7744 sq. metres (given)

$\Rightarrow x^2 = 7744\\[1em] \Rightarrow x = \sqrt{7744}$

Marking periods and using the division method, we have :

⇒ $x = 88$

Side of the square field = 88 metres

Perimeter of the square field = 4 × side = 4 × 88 = 352 metres

Hence, the perimeter of the square field is 352 metres.

Find the least number which must be subtracted from 1104 to obtain a perfect square. Find this perfect square and its square root.

Answer

Let us try to find the square root of 1104.

Clearly, (33)² is less than 1104 by 15.

So, the least number to be subtracted from 1104 to get a perfect square = 15.

Resulting perfect square number = 1104 - 15 = 1089 and $\sqrt{1089} = 33$

Hence, 15 should be subtracted from 1104, the perfect square so obtained is 1089 and $\mathbf{\sqrt{1089} = 33}$.

Find the least number which must be subtracted from 7956 to obtain a perfect square. Find this perfect square and its square root.

Answer

Let us try to find the square root of 7956.

Clearly, (89)² is less than 7956 by 35.

So, the least number to be subtracted from 7956 to get a perfect square = 35.

Resulting perfect square number = 7956 - 35 = 7921 and $\sqrt{7921} = 89$

Hence, 35 should be subtracted from 7956, the perfect square so obtained is 7921 and $\mathbf{\sqrt{7921} = 89}$.

Find the least number which must be added to 6203 to obtain a perfect square. Find the perfect square and its square root.

Answer

Let us try to find the square root of 6203.

Clearly, (78)² < 6203 < (79)².

∴ Required number to be added = (79)² - 6203 = 6241 - 6203 = 38.

Resulting number = 6203 + 38 = 6241 = (79)² and $\sqrt{6241} = 79$

Hence, 38 should be added to 6203, the perfect square so obtained is 6241 and $\mathbf{\sqrt{6241} = 79}$.

Find the least number which must be added to 7348 to obtain a perfect square. Find the perfect square and its square root.

Answer

Let us try to find the square root of 7348.

Clearly, (85)² < 7348 < (86)².

∴ Required number to be added = (86)² - 7348 = 7396 - 7348 = 48.

Resulting number = 7348 + 48 = 7396 = (86)².

and, $\sqrt{7396} = 86$

Hence, 48 should be added to 7348, the perfect square so obtained is 7396 and $\mathbf{\sqrt{7396} = 86}$.

Find the greatest number of six digits, which is a perfect square. Find the square root of this number.

Answer

Greatest number of six digits = 999999

Let us try to find the square root of 999999

Clearly, (999)² is less than 999999 by 1998.

So, the least number to be subtracted is 1998.

Required number = 999999 - 1998 = 998001 and $\sqrt{998001} = 999$

Hence, the greatest number of six digits which is a perfect square is 998001 and $\mathbf{\sqrt{998001} = 999}$.

Find the least number of four digits which is a perfect square.

Answer

Least number of four digits = 1000.

Let us find its square root.

Clearly, (31)² < 1000 < (32)².

∴ Required number to be added = (32)² - 1000 = 1024 - 1000 = 24.

Required number = 1000 + 24 = 1024 = (32)².

Hence, the least number of four digits which is a perfect square is 1024.

Find the least number by which 69192 must be decreased, to make it a perfect square.

Answer

Let us try to find the square root of 69192.

Clearly, (263)² is less than 69192 by 23.

So, the least number to be subtracted from 69192 to get a perfect square = 23.

Resulting perfect square number = 69192 - 23 = 69169 = (263)².

Hence, 69192 must be decreased by 23 to make it a perfect square.

Find the least number by which 69192 must be increased, to make it a perfect square.

Answer

Let us try to find the square root of 69192.

Clearly, (263)² < 69192 < (264)².

∴ Required number to be added = (264)² - 69192 = 69696 - 69192 = 504.

Resulting number = 69192 + 504 = 69696 = (264)².

Hence, 69192 must be increased by 504 to make it a perfect square.

Find the least number by which 69192 must be multiplied, to make it a perfect square.

Answer

Resolving 69192 into prime factors, we get :

$\begin{array}{l|r} 2 & 69192 \\ \hline 2 & 34596 \\ \hline 2 & 17298 \\ \hline 3 & 8649 \\ \hline 3 & 2883 \\ \hline 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$

69192 = (2 × 2) × (3 × 3) × (31 × 31) × 2

Thus, 2, 2 and 3, 3 and 31, 31 exist in pairs while one 2 is alone.

So, we should multiply the given number by 2 to get a perfect square number.

$\Rightarrow 69192 \times 2 = 138384\\[1em] \Rightarrow 138384 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (31 \times 31)\\[1em] \Rightarrow \sqrt{138384} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (31 \times 31)} \\[1em] \Rightarrow \sqrt{138384} = 2 \times 2 \times 3 \times 31\\[1em] \Rightarrow \sqrt{138384} = 372$

Hence, 69192 must be multiplied by 2 to make it a perfect square.

Find the least number by which 69192 must be divided, to make it a perfect square.

Answer

Resolving 69192 into prime factors, we get :

$\begin{array}{l|r} 2 & 69192 \\ \hline 2 & 34596 \\ \hline 2 & 17298 \\ \hline 3 & 8649 \\ \hline 3 & 2883 \\ \hline 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$

69192 = (2 × 2) × (3 × 3) × (31 × 31) × 2

Clearly, in the product of prime factors, the factor 2 does not exist in pair.

So, we must divide the given number by 2.

$\Rightarrow 69192 \div 2 = 34596\\[1em] \Rightarrow 34596 = (2 \times 2) \times (3 \times 3) \times (31 \times 31)\\[1em] \Rightarrow \sqrt{34596} = \sqrt{(2 \times 2) \times (3 \times 3) \times (31 \times 31)} \\[1em] \Rightarrow \sqrt{34596} = 2 \times 3 \times 31\\[1em] \Rightarrow \sqrt{34596} = 186$

Hence, 69192 must be divided by 2 to make it a perfect square.