Fill in the blanks:
$6^{2} + 7^{2} +$ ___ $= 43^{2}$ is

84^{2}

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36^{2}

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49^{2}

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42^{2}

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Solution

The correct option is D $42^{2}$
6 and 7 are consecutive numbers

$6 \times 7 = 42 6^{2} + 7^{2} + 42^{2} = 43^{2}$
$∵ a^{2} + (a + 1)^{2} + (a \times (a + 1))^{2} = (a \times (a + 1) + 1)^{2}$

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Similar questions

Fill in the blanks:-

$6^{2}$ + $7^{2}$ + $($ __ $)^{2}$ = $43^{2}$

Fill in the blanks: $6^{2}$ + $7^{2}$ + $($ ___ $)^{2}$ = $43^{2}$

Choose the right option for the blank.

$6^{2}$ + $7^{2}$ + $($ __ $)^{2}$ = $43^{2}$

6^{2} + 7^{2} + 42^{2} = ()^{2}

$6^{2}$ + $7^{2}$ + $($ ___ $)^{2}$ = $43^{2}$

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