Question

Is there a more efficient (faster) method of solving this geometric question?

Consider a rectangle ABCD .
Let M be a point on the segment AB such that AM= 6 cm and MB= 12 cm .
Let N be a point on the segment BC such that BN= 8 cm and NC= 12 cm .
Let P be a point on the segment CD such that CP= 6 cm and PD= 12 cm .
Let Q be a point on the segment AD such that DQ= 8 cm and QA= 12 cm .
Let O be the point of intersection of MP and NQ .
Find the area of the quadrilateral MONB.

The book answer seems unnecessary complex:
It is easy to see that MNPQ is a parallelogram whose area is equal to the area of the rectangle minus the areas of the four right triangles situated on each corner of the rectangle. So the area of this parallelogram is equal (6+ 12) * ( 8 +12) - (8 * 12/2) - (8 * 12/2) -(6 * 12/2)-(6 * 12/2) =192.

Hence the answer is 192/4 + the area of the triangle MNB
= 192/4 + 8 * 12/2=96

Answer 1

It does seem unnecessary complex, I would have done it like this:

Area of triangle MOB = \((12 \times (8 + 12) \div 2 ) \div 2 = 60\)
Area of triangle BON = \((8 \times (6 + 12) \div 2 )\div 2 = 36\)
Area of the quadrilateral MONB = \(60 + 36 = 96\)

Efficient enough?

Answer 2

It is, I like it thanks mate.

Answer 3

No problem, you are welcome. :-)