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PS: Geometry - Quad

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PS: Geometry - Quad [#permalink] New post 06 Sep 2004, 20:40
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Can you please explain your answer?
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Manager
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 [#permalink] New post 06 Sep 2004, 22:23
Since the four sides are equal, this quad is a rhombus.

We know that one angle is 45 degrees. So we can find the height between AD and BC.

Let us say that the perpendicular dist between AD and BC is x.

then sin 45 = x/ 5 sqrt(2).

sin45 = 1/ sqrt (2).

hence x = 5.

Now , we know that the area is base * height = 5 sqrt(2) * 5 = 25 sqrt(2).


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 [#permalink] New post 06 Sep 2004, 22:56
there is one more way to solve this sum

area is Base * Height
base is already provided

height can be found by drawing a perpendicular and then getting it by 45-45-90 triangle

height comes to 5
so area is 5 * 5 sqrt(2).
so 25 sqrt(2).


cheers
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 [#permalink] New post 07 Sep 2004, 16:14
Yes, OA is 25 sqrt(2)
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 [#permalink] New post 08 Sep 2004, 02:24
solved the same way as anuramm ,
trigonometry got 25sqrt2
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  [#permalink] 08 Sep 2004, 02:24
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