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# In the figure, KLMN is a square, and angle KJN = 45°. Find the area o

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Joined: 02 Sep 2009
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In the figure, KLMN is a square, and angle KJN = 45°. Find the area o  [#permalink]

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15 Mar 2015, 23:13
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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

Kudos for a correct solution.

Attachment:

gpp-sgf_img1.png [ 3.61 KiB | Viewed 2179 times ]

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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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16 Mar 2015, 02:41
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Since KJN =45* then JN=KN = 6/ sqrt(2)=3* sqrt(2)
Area of JLKM = area of JNK + area of KLMN
=1/2*KH^2+KH^2
=3/2*KH^2=(3/2)*18=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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16 Mar 2015, 03:02
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KJN is a 45:45:90 triangle or x:x:x√2 one side is given as KJ (Hypotenuse) =6. Therefore x√2=6 or x=3√2. Area of a triangle is 1/2*3√2*3√2= 9
Area of a square = S^2. We know KN is 3√2. Therefore (3√2)^2 = 18. Add both the areas = 18+9=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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16 Mar 2015, 03:05
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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 [ 3.65 KiB | Viewed 1991 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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20 Jul 2015, 05:28
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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.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 60555
Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o  [#permalink]

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23 Mar 2015, 03:40
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.

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Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o   [#permalink] 23 Mar 2015, 03:40
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# In the figure, KLMN is a square, and angle KJN = 45°. Find the area o

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