It is currently 22 Feb 2018, 20:41

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

PS: Similar Triangles

Author Message
Intern
Joined: 03 Feb 2009
Posts: 27
Schools: Duke, UNC, Emory, Ga Tech, Vanderbilt, Indiana, Wash U, Texas, Rollins

Show Tags

05 Mar 2009, 17:42
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.

Please see attached. I am stuck on how to approach this problem.
Attachments

PracTest1_Geo_SimTri.doc [70.5 KiB]

Senior Manager
Joined: 30 Nov 2008
Posts: 483
Schools: Fuqua

Show Tags

05 Mar 2009, 20:20
One of the properties of Similar Triangles worth remembering: -

If two triangles ABC and DEF are similar, then ratio of their areas = square of the (ratio of the corresponding sides).

Coming back to the original question, Area of the left side triable be a. Then Area of the right side triable will be 2a.

Based on the above rule, we have $$2a / a = (S/s)^2$$ ==> $$sqrt(2) = S/s. ==> S = s sqrt(2).$$
Manager
Joined: 17 Dec 2008
Posts: 171

Show Tags

06 Mar 2009, 08:27
Area of any polygon is proportional to square of the side...
The above is not any special rule. Let minimize the number of special rules to learn

$$A1 = 1/2 (sh) => 1/2 * k1* s1^^2 (h = k1s1) A2 = 1/2 (s1h1) => 1/2 *k2 * s2^^2 (h= k2s2)$$

Since they are similar triangles k1=k2 since the angles are the same, h will be same function of s.

$$A1/A2 = (s1/s2)^^2$$
Re: PS: Similar Triangles   [#permalink] 06 Mar 2009, 08:27
Display posts from previous: Sort by