1)
Obw. 2πr
2 x π x 6cm=12π cm²
r=⅓h
6=⅓h
h=6:⅓=18cm h=(a√3)/2
18=(a√3)/2
a√3=18·2=36
a=36/√3=12√3cm---dł.boku trojkata
O=3·12√3=36√3cm
odp:obwod tego trojkata rowny 36√3cm
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1)
Obw. 2πr
2 x π x 6cm=12π cm²
r=⅓h
6=⅓h
h=6:⅓=18cm h=(a√3)/2
18=(a√3)/2
a√3=18·2=36
a=36/√3=12√3cm---dł.boku trojkata
O=3·12√3=36√3cm
odp:obwod tego trojkata rowny 36√3cm