Geometry Problem

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Level: High School, SAT Prep, College geometry

## Monday, December 27, 2010

### Problem 558: Quadrilateral, Trisection, Sides, Wittenbauer Parallelogram, Area

Labels:
area,
parallelogram,
quadrilateral,
side,
trisection,
Wittenbauer

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Denote area common to ABCD and KLMN by S’

ReplyDeleteWe denote area XYZ.. by XYZ.. itself.

Using KE’ // AC // NH, it is easy to see

AEH’ = KEE’ + HNH’. Similarly,

BFE’ = LFF’ + KEE’,

CGF’ = MGG’ + LFF’,

DHG’ = NHH’ + MGG’.

Add: S – S’ = 2(KLMN – S’)

KLMN = (S + S’)/2 ….…. (i)

Next,

AEH’ = ABD/9 by similar triangles and AE = AB/3

CGF’ = CBD/9,

BFE’ = ABC/9,

DHG’ = ADC/9.

Add: S – S’ = 2S/9

So S’ = 7S/9 ……… (ii)

Hence by (i) and (ii):

KLMN = [S+(7S/9)]/2 = 16S/18 = 8S/9

1) E'F//EF'//AC & HG'//H'G//AC => E'F//HG' (Thales T )

ReplyDelete=> KLMN parallelogram

2) EF' = 2/3 AC, hp = 2/3 hq (altitudes of par, quad)

=> 1/2 Sp = EF' ∙hp = 2/3∙2/3 = 4/9 S

=> Sp = 4/9S + 4/9S = 8/9 S