In the figure, KLMN is a square and angle KJN=45°...

Hello AAPL.

Let's take a look at your question.

We have a right triangle JKN, with hypotenuse 6 and angle KJN=45º. We have then $$\sin\left(45º\right)=\frac{KN}{6}\ \Leftrightarrow\ \ KN=\frac{6\sqrt{2}}{2}=3\sqrt{2}$$ and $$\cos\left(45º\right)=\frac{JN}{6}\ \Leftrightarrow\ \ JN=\frac{6\sqrt{2}}{2}=3\sqrt{2}.$$ Therefore, the height of the trapeze and JM are $$H=3\sqrt{2}\ and\ JM=JN+NM=3\sqrt{2}+3\sqrt{2}=6\sqrt{2}.$$ In conlusion, the area is $$A=\frac{\left(KL+JN\right)\cdot H}{2}=\frac{\left(3\sqrt{2}+6\sqrt{2}\right)\cdot3\sqrt{2}}{2}=\frac{54}{2}=27.$$ The correct answer is E.

I hope this explanation may help you.

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