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Director
Joined: 20 Jul 2004
Posts: 593
Followers: 1

Kudos [?]: 47 [0], given: 0

00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Manager
Joined: 31 Dec 2003
Posts: 214
Location: US
Followers: 0

Kudos [?]: 8 [0], given: 0

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).

Thanks.
Manager
Joined: 28 Jul 2004
Posts: 54
Followers: 1

Kudos [?]: 0 [0], given: 0

there is one more way to solve this sum

area is Base * Height

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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Jim

Director
Joined: 20 Jul 2004
Posts: 593
Followers: 1

Kudos [?]: 47 [0], given: 0

Yes, OA is 25 sqrt(2)
Manager
Joined: 18 Jun 2004
Posts: 105
Location: san jose , CA
Followers: 1

Kudos [?]: 27 [0], given: 0

solved the same way as anuramm ,
trigonometry got 25sqrt2
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