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Let O be the origin and let OA = 2a, OB = 6a + 5b, OC = 3b. If OA and OC are adjacent sides of a parallelogram with area 15 sq. units, find the area (in sq. units) of quadrilateral OABC.

1. 38
2. 40
3. 32
4. 35

Correct answer: 35

Solution

|OA x OC| = |2a x 3b| = 6|a x b| = 15, so |a x b| = 5/2. Now find area of quadrilateral OABC with vertices O, A=2a, B=6a+5b, C=3b. Split along diagonal OB: triangles OAB and OCB. Area(OAB) = (1/2)|OA x OB| = (1/2)|(2a) x (6a+5b)| = (1/2)|10(a x b)| = 5|a x b| = 5*(5/2) = 25/2. Area(OCB) = (1/2)|OC x OB| = (1/2)|(3b) x (6a+5b)| = (1/2)|(-18)(b x a)| = 9|a x b| = 9*(5/2) = 45/2. Total area = 25/2 + 45/2 = 70/2 = 35.

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