Area of Pentagon (Posted on 2008-01-07)

   |AB| = |BC|, |CD| = |DE|,

   /ABC = 150°, /CDE = 30°,

   and |BD| = 2.

	Submitted by Bractals
Rating: 2.0000 (1 votes)
Solution:	(Hide)
Construct line segments AC and CE. Let b = |AB| = |BC|, d = |CD| = |DE|, f = |AC|, and h = |CE|. From isosceles triangles ABC and CDE, we get     /BCA = 15°    and    /DCE = 75° From the given information, we get f = 2*b*cos(15°) h = 2*d*cos(75°) = 2*d*sin(15°) Thus, f*h = 2*b*d*2*cos(15°)*sin(15°) = 2*b*d*sin(30°) = b*d Letting x = /ACE and applying the Cosine Law to triangle BCD, we get 4 = 22 = b2 + d2 - 2*b*d*cos(x + 90°) = b2 + d2 + 2*b*d*sin(x) Hence, Area(ABCDE) = Area(ABC) + Area(CDE) + Area(ACE) = ½*b2*sin(150°) + ½*d2*sin(30°) + ½*f*h*sin(x) = ¼*[b2 + d2 + 2*b*d*sin(x)] = ¼[4] = 1

	Subject	Author	Date
	answer	K Sengupta	2008-02-06 06:18:48
	different approach	Daniel	2008-01-08 00:33:41
	re: two ways (spoiler)	Charlie	2008-01-07 13:53:58
	two ways (spoiler)	Charlie	2008-01-07 13:45:11