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

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

by AAPL » Fri Dec 22, 2017 2:55 pm
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In the figure, KLMN is a square and angle KJN=45°. Find the area of figure JKLMN.

$$A.\ 9+9\sqrt{2}$$
$$B.\ 9+18\sqrt{2}$$
$$C.\ 18$$
$$D.\ 18+9\sqrt{2}$$
$$E.\ \ 27$$

The OA is E.

I know that I can get the value of the KN side of sin(45)=(KN)/6, and then with it I can determine the are of JKN=1/2*B*H and KLMN=(side)^2, finally I just need to make the sum of the two areas and get the total area, right? I appreciate if any expert explain it for me. Thank you so much.
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by EconomistGMATTutor » Sat Dec 23, 2017 3:17 am
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.

I'm available if you'd like a follow-up.

Regards.
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