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Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o [#permalink]
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Answer = E = 27

\(6^2 = a^2 + a^2\)

\(a = \sqrt{18}\)

There are three equivalent right triangles as shown

Attachment:
gpp-sgf_img1.png
gpp-sgf_img1.png [ 3.65 KiB | Viewed 5439 times ]


There total area \(= 3 * \frac{1}{2} * (\sqrt{18})^2 = 27\)
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Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o [#permalink]
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Bunuel wrote:

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

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

Since KN must be perpendicular to JM, we know that JNK must be a 45-45-90 triangle. The legs are equal, and the hypotenuse is the square root of 2 times larger than either leg. Let JN = KN = x. Then

\(\frac{x}{6}=\frac{1}{\sqrt{2}}\) --> \(x=3\sqrt{2}\)

In that work, we rationalized the denominator when we divided by the radical. This number for x is the side of the square, so square KLMN has an area which is this number squared.

Area of KLMN = \((3\sqrt{2})^2=18\)

That’s part of the area. Now, notice that triangle JKN would be equivalent to a square of the same size cut in half along the diagonal. This triangle must have exactly half the area of the square. Well, the square is 18, so the triangle must be 9, and together, they must have an area of 27.

Answer = (E)
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Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o [#permalink]
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Expert Reply
I think we can tag this under Geometry instead of Co-ordinate Geometry?
Thank you
Bunuel wrote:
Attachment:
gpp-sgf_img1.png
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

Kudos for a correct solution.
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Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o [#permalink]
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Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o [#permalink]
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