Let me first redraw the figure so that it becomes more of a "to scale" figure and we can at least visualize how the actual quadrilateral will look like.

∆ABM is an isosceles right-angled triangle with angle M = 90°

Hence, BM = AM = AB/(√2) = 8/(√2) = 4√2

Quadrilateral AMCN is a rectangle.

Hence, CN = AM = 4√2

And, AN = MC = BC - BM = (16 - 4√2)

∆CDN is a 30-60-90 triangle angle N = 90°

Hence, DN = (√3)CN = (√3)*(4√2) = 4√6

Now, area of the quadrilateral ABCD

= (area of ∆ABM) + (area of rectangle AMCN) + (area of ∆CDN)

\(= (\frac{1}{2}*AM*BM) + (AM*MC) + (\frac{1}{2}*CN*DN)\)

\(= (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{2})) + ((4\sqrt{2})*(16 - 4\sqrt{2})) + (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{6}))\)

\(= 16 + 64\sqrt{2} - 32 + 16\sqrt{3}\)

\(= 64\sqrt{2} - 16 + 16\sqrt{3}\)

Attachments

qua.jpg [ 11.53 KiB | Viewed 6304 times ]

_________________

Anurag Mairal, Ph.D., MBA

GMAT Expert, Admissions and Career Guidance

Gurome, Inc.

1-800-566-4043 (USA)

+91-99201 32411 (India)

http://www.facebook.com/Gurome