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by rupsk » Thu Dec 22, 2011 3:55 pm
In the figure to the right, if point C is the center of the circle and DB = 7, what is the length of DE in triangle EDB?

(1) x = 60°

(2) DE || CA[/img]
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by neelgandham » Thu Dec 22, 2011 4:20 pm
In the figure to the right, if point C is the center of the circle and DB = 7, what is the length of DE in triangle EDB?
(1) x = 60°
If x =60, then the triangle ACB is an equilateral triangle. Since we don't know the values of AE or any of the angles related to it Insufficient!
(2) DE || CA
DE||CA Implies triangles BAC and BED are similiar because
Angle ABC = Angle EDB
Angle BAC = Angle BED (Corresponding angles are equal)
Angle BCA = Angle BDE (Corresponding angles are equal)

So, BC/BD = AC/DE
Radius/Diameter = Radius /DE
DE = Diameter = 7 units
Sufficient!

IMO B
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by Anurag@Gurome » Thu Dec 22, 2011 6:08 pm
rupsk wrote:In the figure to the right, if point C is the center of the circle and DB = 7, what is the length of DE in triangle EDB?

(1) x = 60°
(2) DE || CA[/img]
C is the center of the circle implies that CD, CB, CA are the radii of the circle. AC = BC implies that ABC is an isosceles triangle.

(1) x = 60º implies that ABC is an equilateral triangle but this does not give any info for the triangle DEB; NOT sufficient.

(2) DE || CA implies that BED and BAC are similar triangles, as angle B is common and the base angles are equal.
So, AC : ED = BC : BD
3.5 : ED = 3.5 : 7 or DE = 7; SUFFICIENT.

The correct answer is B.
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by ArunangsuSahu » Fri Dec 30, 2011 10:00 am
Statement 1:
ABC is Equilateral. No Info about DEB.. INSUFFICIENT

2) DE||CA and C is midpoint of DB. Therefore by midpoint Th CA=1/2 DB and CA=CB SO DB=DE from SIMILAR properties of triangles
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