It is currently 16 Dec 2017, 05:15

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

ACD is a triangle(A in lower left corner and C, D going

Author Message
Intern
Joined: 26 Jan 2004
Posts: 35

Kudos [?]: [0], given: 0

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

Show Tags

02 Apr 2004, 07:38
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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)

Kudos [?]: [0], given: 0

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4284

Kudos [?]: 545 [0], given: 0

Show Tags

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
_________________

Best Regards,

Paul

Kudos [?]: 545 [0], given: 0

Senior Manager
Joined: 02 Mar 2004
Posts: 327

Kudos [?]: 2 [0], given: 0

Location: There

Show Tags

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)

Kudos [?]: 2 [0], given: 0

Re: PS - Triangle   [#permalink] 02 Apr 2004, 17:07
Display posts from previous: Sort by