Answer:
[tex]\displaystyle 11\:units^2[/tex]
Step-by-step explanation:
Take the square root of both squares' areas first, then use the Pythagorean Theorem:
[tex]\displaystyle 2\sqrt{11} = \sqrt{44}; \sqrt{33} = \sqrt{33} \\ \\ -\sqrt{33}^2 + [2\sqrt{11}]^2 = b^2 → -33 + 44 = b^2 → 11 = b^2[/tex]
You keep 11 as is because this is the area of that tiny square.
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