Geometry - Circles

This topic has expert replies
Source: — Problem Solving |

User avatar
Master | Next Rank: 500 Posts
Posts: 241
Joined: Wed Jan 27, 2010 10:10 pm
Location: Chennai
Thanked: 23 times
Followed by:2 members
GMAT Score:690

by sars72 » Thu Jan 28, 2010 12:30 pm
i'm giving this a shot.. geometry is not my forte..

using property of similar triangles, AB/BC = AD/ DE

--> 4/6 = 6/DE

--> DE = 9

User avatar
Legendary Member
Posts: 2109
Joined: Sun Apr 19, 2009 10:25 pm
Location: New Jersey
Thanked: 109 times
Followed by:79 members
GMAT Score:640

by money9111 » Thu Jan 28, 2010 2:04 pm
how do you know the triangles are similar?
My goal is to make MBA applicants take onus over their process.

My story from Pre-MBA to Cornell MBA - New Post in Pre-MBA blog

Me featured on Poets & Quants

Free Book for MBA Applicants


User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pm
Location: Arabian Sea
Thanked: 125 times
Followed by:2 members

by ajith » Thu Jan 28, 2010 3:52 pm
money9111 wrote:how do you know the triangles are similar?
https://www.andrews.edu/~calkins/math/we ... 13.htm#SIM
Similar Triangles
If three sides of a triangle are proportional to the three sides of another triangle,
then the triangles are similar (SSS Similarity Theorem).

If the angles (two implies three) of two triangles are equal,
then the triangles are similar (AA Similarity Theorem).
Note: this applies not only to ASA, AAS=SAA, but also to AAA situations.

If two sides of a triangle are proportional to two sides of another triangle and the
included angles are congruent, then the triangles are similar (SAS Similarity Theorem).


Now, in this case the triangles in Question are ABC ande ADE

Angle AED = angle BED = angle DCB = Angle ACB (angles inscribed in the same arc and equal)
Angle ADE = angle CDE = angle CBE = Angle CBA (angles inscribed in the same arc and equal)

Now by AA theorem the triangles ABC ande ADE, are similar

I hope, I answered your question
Always borrow money from a pessimist, he doesn't expect to be paid back.

User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

by sanju09 » Fri Jan 29, 2010 5:19 am
skorolkova wrote:1)

Image


[/img]
Angles in the same segment are equal. So, if BD is the chord, then ∠BCD = ∠BED; and if CE is the chord, ∠CBE = ∠CDE.

Also, ∠BAC = ∠DAE, as vertically opposite angles are equal.

With the help of above discussion, we can conclude that ∆ABC ~ ∆ADE, in that strict correspondence.

∴ AB/AD = BC/DE = CA/EA.

Let's take the needful only, i.e.

AB/AD = BC/DE

4/6 = 6/DE or DE = [spoiler]9[/spoiler].
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 am
Thanked: 137 times
Followed by:5 members

by thephoenix » Fri Jan 29, 2010 6:03 am
money9111 wrote:how do you know the triangles are similar?
in tri ADE and ABC

<ade=<ABC ( ANGLE MADE BY TWO ON A CIRCLE ARE ALWAYS EQUAL)

FOR THE SAME REASON
<AED=<ACB

<DAE=<BAC(VERTICAL OPP ANGLES)

SO TRI ADE IS SIMILAR TO TRI ABC

THEREFORE AD/AB=DE/BC(PROP OF SIM TRI)

DE=6*6/4=9
HTH