Let OA = 2 a, OB = 6 a + 5 b and OC = 3 b, where O is the origin. If the area of the parallelogram with adjacent sides OA and OC is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to:
- 32
- 38
- 35
- 40
Answer
(3) 35
Solution
[figure]
The parallelogram on OA and OC has area 15, so |2 a 3 b| = 15, i.e. | a b| = 52.
The area of a quadrilateral is half the magnitude of the cross product of its diagonals. Here OB = 6 a + 5 b and AC = OC - OA = 3 b - 2 a (dashed), so
Area = 12 |(6 a + 5 b) (3 b - 2 a) |
= 12 |18 a b - 10 b a | (since a a = b b = 0)
= 12 28| a b| = 14 52 = 35
Answer: (3)