Proposed Problem

Click the figure below to see the complete problem 418 about Triangle, Cevian, Incircle, Excircle, Inradius, Exradius, Congruence.

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Complete Problem 419

Level: High School, SAT Prep, College geometry

## Saturday, January 16, 2010

### Problem 419: Triangle, Cevian, Incircle, Excircle, Inradius, Exradius, Congruence

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draw AH altitude on BC.

ReplyDeletename S point D meet BC, P point D meet AC

from P 325 => AH = 4r = 4∙ 17 = 68

AH∙BC = AC∙BP ( the same area of ABC, express two time )

68∙(5/7)AC = AC∙BP =>

BP = 340/7 (1)

▲DSB ~ ▲PBC => x/BD = (1/2)AC/(5/7)AC

x/BD = 7/10 => BP/x = 17/7 => x/BP = 7/17

x = (7/17)∙BP substitute (1)

x = (7/17)∙(340/7)

x = 20

Once we've AH=4*17=68, we can also say that x = Inradius of Tr. ABC= Area(Tr. ABC)/semiperimeter =(68/2)*(5b/7)/((5b/7+5b/7+b)/2) which directly gives x=20

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bjhvash44@sbcglobal.net

ReplyDeleteIwas solving number 420.Why no comment or solution.Too easy?