Question

PLEASE HELP!!!! WILL MARK BRAINLIEST!!!! 15points

Answer 1

Answer:

$\displaystyle 11\:units^2$

Step-by-step explanation:

Take the square root of both squares' areas first, then use the Pythagorean Theorem:

$\displaystyle 2\sqrt{11} = \sqrt{44}; \sqrt{33} = \sqrt{33} \\ \\ -\sqrt{33}^2 + [2\sqrt{11}]^2 = b^2 → -33 + 44 = b^2 → 11 = b^2$

You keep 11 as is because this is the area of that tiny square.

I am joyous to assist you anytime.