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....
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....
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.