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Angle $eab$ is a right angle, and $be = 9$ units. what is the number of square units in the sum of the areas of the two squares $abcd$ and $aefg$?

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Let the measure of side AB be x, then, the measue of side AE is given by


AE=√(9^2-x^2).

Now, ABCD is a square of size x, thus the area of square ABCD is given by


Area=x^2

Also, AEFG is a square of size
√(9^2-x^2), thus, the area of square AEFG is given by


Area=\left(√(9^2-x^2)\right)^2=9^2-x^2=81-x^2

The sum of the areas of the two squares ABCD and AEFG is given by


x^2+81-x^2=81

Therefore, the number of square units in the sum of the areas of the two squares ABCD and AEFG is 81 square units.
User Roman Glass
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