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Using cosine formula of area,

10*10*sin(120)=\(100*\sqrt{3}/2\)=\(50\sqrt{3}\)

Ans. (E)
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Bunuel

Note: Figure not drawn to scale

If AD is the same length as DC, what is the area of the parallelogram shown?


A. \(\frac{25\sqrt{3}}{2}\)

B. 25

C. \(25\sqrt{3}\)

D. \(\frac{75\sqrt{3}}{2}\)

E. \(50\sqrt{3}\)

Attachment:
image023.jpg

Since the adjacent sides are equal, opposite angles are parallel and internal angles are not 90, it is a rhombus.
In rhobus, diagonals bisect each other at rt angles.
Suppose diagonals bisect at O.
In triangle OAB, OA=\(5\sqrt{3}\)
OB = 5 as it is a 30-60-90 triangle.
BD=10
Area=1/2* Product of diagonals = \(\frac{1}{2}*10*10\sqrt{3}=50\sqrt{3}\)
Answer E.
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how do u know it is a 30-60-90 triangle.
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