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

Question

https://gmatclub.com/forum/download/file.php?id=25932 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 . gpp-sgf_img1.png

littlegirl · Answer

Answer is E
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

Turkish · Answer

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

PareshGmat · Answer

Answer = E = 27 6^2 = a^2 + a^2 a = \sqrt{18} There are three equivalent right triangles as shown gpp-sgf_img1.png There total area = 3 * \frac{1}{2} * (\sqrt{18})^2 = 27

Bunuel · Answer

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)

BrushMyQuant · Answer

I think we can tag this under Geometry instead of Co-ordinate Geometry?
Thank you

bumpbot · Answer

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

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

Prep Toolkit

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