Geometry Problem

Level: High School, SAT Prep, College geometry.

Click the figure below to see the complete problem 522 about Right Triangle, Circle, Diameter, Tangent.

## Sunday, September 26, 2010

### Problem 522: Right Triangle, Circle, Diameter, Tangent

Labels:
circle,
diameter,
right triangle,
tangent

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▲ADE ~ ▲ABE => ED/AE = BE/x

ReplyDelete▲EDC ~ ▲AEC => ED/AE = DC/5

=> x = 5

http://ahmetelmas-geo-geo-antonio.blogspot.com/2010/09/prb522.html

ReplyDeleteBecause FD // BC, FE = DE = m say

ReplyDeleteIf also BE = DC = p and AE = n

p/m = x/n = 5/n since Tr.s BEF and ABE are similar and DEC and ACE are similar

So x = 5

Sumith Peiris

Moratuwa

Sri Lanka

Solution 2

ReplyDeleteFD//BC, ODEC is a kite as before.

So AE bisects < BAC

So x/BE = AC/5 or x = AC.BE/5 = CD.AC/5 = 25/5 = 5

Sumith Peiris

Moratuwa

Sri Lanka