Question

Using the given pattern, find the missing numbers.
12^2+ 22^2 + 22^2 = 32^2
22^2 + 32^2 + 62^2 = 72^2
32^2 + 42^2 + 122^2 =132^2
42^2 + 52^2 + —2^2 = 212^2
52^2 + —2^2 + 302^2 = 312 ^2
62^2 + 72^2 + —2^2 = —2^2
Note: To find pattern: Third number is related to first and second number. How? Fourth number is related to third number. How?

Answer 1

$\huge{\sf{Answer-}}$

$\sf{12^2 \: + \: 22^2 \: + \: 22^2 \: = \: 32^2 } \\ \sf{22^{2} + 32^{2} \: + 62^{2} + = 72^{2} } \\ \sf{32^{2} + 42^{2} \: + 122^{2} = 132^{2} } \\ \sf{42^{2} + 52^{2} \: + \red{202^{2}} = 212^{2} } \\ \sf{52^{2} + \red{62^{2}} \: + 302^{2} = 312^{2} } \\ \sf{62^{2} + 72^{2} \: + \red{422^{2} } = \red{432^{2} } }$

$\huge{\sf{Explanation-}}$

Here,

First is related to second. Let's see how?

★ 12² + 22²

From 1st and second number, taking bold ones :

1 - 2 [Consecutive numbers , so first is related to second]

★ 22² + 32²

2 - 3 [Consecutive numbers , so first is related to second]

★ 32² + 42²

3 - 4 [Consecutive numbers , so first is related to second]

★ 42² + 52²

4 - 5 [Consecutive numbers , so first is related to second]

★ 52² + 62²

5 - 6 [Consecutive numbers , so first is related to second]

★ 62² + 72²

6 - 7 [Consecutive numbers , so first is related to second]

$\rule{200}2$

First is related to third. Let's see how?

★ 12^2 + 22^2 + 22^2 = 32^2

1 × 2= 2

★ 22^2 + 32^2 + 62^2 = 72^2

2 × 3 = 6

★ 32^2 + 42^2 + 122^2 =132^2

3 × 4 = 12

★ 42^2 + 52^2 + 202^2 = 212^2

4 × 5 = 20

★ 52^2 + 62^2 + 302^2 = 312 ^2

5 × 6 = 30

★ 62^2 + 72^2 + 422^2 = 432^2

6 × 7 = 42

$\rule{200}2$

Now, fourth number is related to third number, let's see how?

★ 12^2+ 22^2 + 22^2 = 32^2

2 + 1 = 3

★ 22^2 + 32^2 + 62^2 = 72^2

6 + 1 = 7

★ 32^2 + 42^2 + 122^2 =132^2

12 + 1 = 13

★ 42^2 + 52^2 + 202^2 = 212^2

20 + 1 = 21

★ 52^2 + 622^2 + 302^2 = 312 ^2

30 + 1 = 31

★ 62^2 + 72^2 + 422^2 = 432^2

42 + 1 = 43

Answer 2

$\huge\star{\red{\underline{\mathfrak{Question :-}}}}$

12²+ 22² + 22² = 32²

22² + 32² + 62² = 72²

32² + 42² + 122² =132²

42² + 52² + _ _ 2² = 212²

52² + _2² + 302² = 312²

62² + 72² + _ _ 2² = _ _ 2²

$\rule{200}2$

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\sf{{12}^{2}+{22}^{2}+{22}^{2}={32}^{2}}$

$\sf{{22}^{2}+{32}^{2}+{62}^{2}={72}^{2}}$

$\sf{{32}^{2}+{42}^{2}+{122}^{2}={132}^{2}}$

$\sf{{42}^{2}+{52}^{2}+\blue{{202}^{2}}={212}^{2}}$

$\sf{{52}^{2}+\blue{{62}^{2}}+{302}^{2}={312}^{2}}$

$\sf{{62}^{2}+{72}^{2}+\blue{{422}^{2}}=\blue{{432}^{2}}}$

$\rule{200}2$

$\huge\star{\purple{\underline{\mathfrak{Explanation :-}}}}$

Here, we see that the tens digit of the second number is the consecutive of the first.

Like:-

• 12² and 22²
• 22² and 32²
• 32² and 42²
• 42² and 52²
• 52² and 62²
• 62² and 72²

$\rule{200}1$

Here, we see that tens digit of the first number and the tens digit of the third number are related in this way:-

(First number × Second number = Third number)

Note: We're only talking about tens digit.

• 12² and 22² (1×2=2)
• 22² and 32² (2×3=6)
• 32² and 42² (3×4=12)
• 42² and 52² (4×5=20)
• 52² and 62² (5×6=30)
• 62² and 72² (6×7=42)

$\rule{200}1$

Now, third number and fourth number are related in this way:-

(tens place of the third number + 1 =tenth place of the fourth number)

• 2 + 1 = 3
• 6 + 1 = 7
• 12 + 1 = 13
• 20 + 1 = 21
• 30 + 1 = 31
• 42 + 1 = 43

$\rule{200}4$