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

Question

2701 views around the world

You can reuse this answer
Creative Commons License

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#