ACD is a triangle(A in lower left corner and C, D going : PS Archive
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# ACD is a triangle(A in lower left corner and C, D going

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ACD is a triangle(A in lower left corner and C, D going [#permalink]

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02 Apr 2004, 07:38
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ACD is a triangle(A in lower left corner and C, D going clockwise). Each side is length 3. B is point on AC such that AB is length 1. BE is a line perpendicular to AC drawn from point B to E. E is on side AD of the triangle. What is the area of the region BCDE?

A. 9/4
B. (7/4)sqrt(3)
C. (9/4)sqrt(3)
D. (9/4)sqrt(3)
E. 6 + sqrt(3)
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02 Apr 2004, 07:54
B)
AE = 2*1 = 2
BE = (AE*sqrt3) / 2 = sqrt3
area of ABE = 1/2 sqrt3
height of triangle ACD = (3sqrt3) / 2 = 3/2 sqrt3
area of ACD = [3*(3/2)*sqrt3] / 2 = 9/4 sqrt3

area of BCDE = 9/4 sqrt3 - 1/2 sqrt3 = 7/4 sqrt3
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02 Apr 2004, 17:07
asagem99 wrote:
ACD is a triangle(A in lower left corner and C, D going clockwise). Each side is length 3. B is point on AC such that AB is length 1. BE is a line perpendicular to AC drawn from point B to E. E is on side AD of the triangle. What is the area of the region BCDE?

ACD is equilateral, area = sqrt(3)9/4
ABE is a right angled traingle(30-60-90), are = sqrt(3)/2
Answer = sqrt(3)/2 ((9/2)-1) = 7/4 sqrt(3)
Re: PS - Triangle   [#permalink] 02 Apr 2004, 17:07
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