Ex.6.1 Q6 Squares and Square Roots Solutions - NCERT Maths Class 8

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Question

Using the given pattern, find the missing numbers.

\[\begin{align}{1^2} + {2^2} + {2^2} &= {3^2}\\{2^2} + {3^2} + {6^2} &= {7^2}\\{3^2} + {4^2} + {{12}^2} &= {{13}^2}\\{4^2} + {5^2} + { - ^2} &= {{21}^2}\\{5^2} + { - ^2} + {{30}^2} &= {{31}^2}\\{6^2} + {7^2} + \_\_{\_^2} &= \_\_\_{\_^2}\end{align}\]

Text Solution

What is known?

Pattern

What is uknown?

Missing Number in the pattern

Reasoning:

The third number is the product of the first two numbers and the fourth number is obtained by adding \(1\) to the third number

Steps:

\[\begin{align}{1^2} + {2^2} + {2^2} &= {3^2}\\{2^2} + {3^2} + {6^2} &= {7^2}\\{3^2} + {4^2} + {12}^2 &= {13}^2\\{4^2} + {5^2} + {20}^2 &= {21}^2\\{5^2} + {6^2} + {30}^2 &= {31}^2\\{6^2} + {7^2} + {42}^2 &= {43}^2\end{align}\]