The $DeltasABC$ and $FGH$ shown below are similar. Find $bar(FH)$ ?

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Answer 1 · 2017-03-14T08:32:59

$bar(FH)=20$

As the corresponding angles in $DeltaABC$ are congruent to the corresponding angles in $DeltaFGH$ , we have

$(AB)/(FG)=(BC)/(GH)=(AC)/(FH)$

Now as $AB=6$ , $AC=8$ , $FG=x+10$ and $FH=4x$ , we have

$6/(x+10)=8/(4x)$

i.e. $6xx4x=8(x+10)$

or $24x=8x+80$

or $16x=80$

i.e. $x=80/16=5$

and hence $bar(FH)=4xx5=20$

The DeltasABCDeltasABC and FGHFGH shown below are similar. Find bar(FH)bar(FH)?