Can some one help me out please...

In a quadrilateral ABCD, the diagonals AC and BD intersect at O. Let OA = 2, OB = 2, OC = 3, OD = 4 and AB = 3. The area of the quadrilateral is ?

Im sorry if this is too simple....

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- Oct 17th 2009, 11:52 PMitsmaximhereArea of Quadrilateral given diagonals and 1 side.
**Can some one help me out please...**

**In a quadrilateral ABCD, the diagonals AC and BD intersect at O. Let OA = 2, OB = 2, OC = 3, OD = 4 and AB = 3. The area of the quadrilateral is ?**

**Im sorry if this is too simple....** - Oct 18th 2009, 02:12 AMaidan
The area (K) of a quadrilateral can be found by

$\displaystyle K = \dfrac{1}{2} pq \sin \theta $

the diagonals are p & q

In this case, from the data give:

p = 2+3 = 5 & q = 4+2 = 6

You have given the three sides of a triangle: 2,2,3

Use the cosine law to compute $\displaystyle \angle AOB = \theta $

Probably easier (since $\displaystyle \triangle AOB $ is isosceles)

$\displaystyle \theta = 2 \, sin^{-1} \left ( \dfrac{1.5}{2} \right) $

Compute $\displaystyle \theta $ and $\displaystyle \sin \theta $, then plug them into the equation above.