Difference between revisions of "Specimen Cyprus Seniors Provincial/2nd grade/Problem 1"

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a) the distance of <math>\Gamma</math> from <math>(\epsilon)</math> is equal to the sum of the distances <math>B, \Delta</math> from <math>(\epsilon)</math>.
 
a) the distance of <math>\Gamma</math> from <math>(\epsilon)</math> is equal to the sum of the distances <math>B, \Delta</math> from <math>(\epsilon)</math>.
  
b)Area(<math>B\Gamma\Delta</math>)=Area(<math>B'\Gamma '\Delta '</math>)
+
b)Area(<math>B\Gamma\Delta</math>)=Area(<math>B'\Gamma '\Delta '</math>).
  
 
== Solution ==
 
== Solution ==

Revision as of 06:06, 12 November 2006

Problem

Let $AB\Gamma\Delta$ be a parallelogram. Let $(\epsilon)$ be a straight line passing through $A$ without cutting $AB\Gamma\Delta$. If $B', \Gamma ', \Delta '$ are the projections of $B, \Gamma, \Delta$ on $(\epsilon)$ respectively, show that

a) the distance of $\Gamma$ from $(\epsilon)$ is equal to the sum of the distances $B, \Delta$ from $(\epsilon)$.

b)Area($B\Gamma\Delta$)=Area($B'\Gamma '\Delta '$).

Solution


See also