Geometry Problem

Click the figure below to see the complete problem 598.

## Friday, April 29, 2011

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## Friday, April 29, 2011

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Geometry Problem 598: Trapezoid, Perpendicular, Equal Angles, Metric Relation

## Link List

Online Geometry theorems, problems, solutions, and related topics.

Labels:
angle,
angle bisector,
congruence,
metric relations,
perpendicular,
trapezoid

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Let H in AD that AEH is rectangle. Then AEB is congruent to AEH (angle ABE=90) and EH=3. Now BH is perpendicular to AE and thus is parallel to CD. Then HD=BC=2 and finally Pitagoras to EHD gives x=ED=sqrt(13).

ReplyDeleteMIGUE.

In ∆ABE, angle at A is α, angle at E is 90° - α. So angle at B is 90°

ReplyDeleteSince AD ∥ BC we have ∠ABC is 180° - 2α.

So ∠EBC is 90° - 2α

Two angles in ∆BCE being α and 90° - 2α,

the third angle ∠BCE = 90°+ α

Applying Sine Rule in ∆BCE,

BE : sin (90° + α ) = 2 : sin α

3 sin α = 2 cos α,

So tan α = 2/3 and

AE = 3/sin α = 3 / [2 /√13] = 3√13 / 2

From the right angled ∆ADE, tan α = x / AE

Hence x = AE tan α = (3√13/2).(2/3)

(i.e.)x = √13

Let AB,DC meet at F. Triangles AFE and ADE are congruent ASA so EF = x

ReplyDeleteNow BC = BE = 2 hence applying Pythagoras to triangle BEF,

x = √(2^2 + 3^2) =√13

Sumith Peiris

Moratuwa

Sri Lanka