Perfect square or square number - class-VIII

$(23)^2$
$a=2, b=3$

$\therefore (23)^2=529$

$(52)^2$
$a=5, b=2$

$\therefore (52)^2=2704$

$12 \times 12 =144, 12^2 = 144$. 

$144$ is a square of number $12$.

Therefore, option A is the correct answer.

Let the number be $x.$
As per the problem $\sqrt {x}=2\times \sqrt [ 3 ]{x  } $
Raising both sides by $6$ times
$=(x^{1/2})^6 = 2^6(x^{1/3})^6$
$= x^{1/2\times 6} = 2^6 x^{1/3\times 6}$
or $ x^3 = 64 x^2$
or $x=64$

The square of the number $78$ is $6084$

$Using\quad approximate\quad values,\quad we\quad will\quad calculate\quad the\quad values\quad the\quad given\quad problem\ \quad \quad =\quad { \left( 23.1 \right)  }^{ 2 }+{ \left( 48.6 \right)  }^{ 2 }-{ \left( 39.8 \right)  }^{ 2 }\ \quad \quad =\quad 533.61+2362.96-1584.04\ \quad \quad =\quad 1311.53\ \quad \quad =\quad approx...\quad { \left( 36.21 \right)  }^{ 2 }$

we know 

Take two numbers $3$ and $4$.

Square of a number means multiplying a number with the same number.

Square of $17\times 17 =289$.

Therefore, option B is the correct answer.

Square of  a number means multiply a number with the same number.

$22 \times 22= 22^2  = 484$

Thus the square of number $22$ is $484$.

Therefore, option C is the correct answer.

$3x\sqrt{2y}=\sqrt{2y\times3^2x^2}=\sqrt{18x^2y}$ 

$59^{2} = 59 \times 59 = 3481$

It can also be $(-13)^2$

$-13 \times -13 = 169$

Therefore, A is the correct answer.

Square of $6.3 = 6.3\times 6.3=\dfrac {63}{10}\times \dfrac {63}{10}=\dfrac {3936}{100}$
                        $= 39.69$

We have 2352 =  $ \displaystyle \underline{2\times 2}\times \underline{2\times 2}\times 3\times \underline{7\times 7} $
As the prime factor 3 has no pair 2352 is not a perfect square 

$112^2 = (100 + 12)^2$
$a = 100, b = 12$
$= a(a + b) + b(a + b)$
$= 100(100 + 12) + 12(100 + 12)$
$= 100^2 + 1200 + 1200 + 12^2$
$= 10000 + 2400 + 144$
$= 12544$
The missing number is 144.

Step 1: 

Step 1 :Multiply ten's digit with its next number.
$2 \times (2 + 1) = 2\times 3 = 6 ---(1)$
Step 2 : Find the square of unit's digit: 5
$5^2 = 25 ----(2)$
Combining (1) and (2), we get
$25 \times 25 = 625$
$6 $ is the missing number.

Square of rational numbers = $(\cfrac{7 \times 7\times 4}{28 \times 14})^2$
= $\cfrac{49 \times 49\times 16}{784 \times 196}$
= $\cfrac{7 \times 7\times 4}{28 \times 14}$
= $\cfrac{1}{4}$

First find the square $13$ and $26.$
$169 = 13 \times 13$
$676 = 26 \times 26$
So, the square of $\dfrac{13}{26}=\dfrac{169}{676}$

Multiply ten's digit with its next number.
$7 \times (7 + 1) = 7 \times 8 = 56 ---(1)$
Find the square of unit's digit: $5$
$5^2 = 25$ ----(2)
Combining (1) and (2), we get
$5625 = 75 \times75$
So, $6$ is the missing number.

$\dfrac{9}{10}$$\times$$\dfrac{9}{10}$ $=$ $\dfrac{81}{100}$

square of $\dfrac{36}{100} = \dfrac{1296}{10000}$

Square of rational number $ = (\dfrac{17 \times 18}{2 \times 9})^2$
                                            

Square of rational number = $(\cfrac{13 \times 12}{10 \times 13})^2$
= $\cfrac{169 \times 144}{100\times 169}$
= $\cfrac{36}{25}$

convert fraction into decimal first.

Multiply ten's digit with its next number.
$9 \times (9 + 1) = 9 \times 10 = 90---(1)$
Find the square of unit's digit: 5
$5^2 = 25$ ----(2)
Combining (1) and (2), we get
$9025 = 95\times95$
So, $9$ is the missing number.

$(\cfrac{9 \times 12}{4\times 3})^2 = (\cfrac{81 \times 144}{16\times 9})$ 
On simplifying, we get
$(\cfrac{9 \times 12}{4\times 3})^2 = 81$

$4\times12$ $=$ $48$

$30\times25$ $=$ $750$

Four digit number $accd$

$(\cfrac{16\times24}{48})^2$
On simplifying, we get

$(\cfrac{5\times25}{50})^2$
On simplifying, we get
$(\cfrac{5\times25}{50})^2 = (\cfrac{5}{2})^2$
= $\cfrac{25}{4}$