Odpowiedź :
a=6cm
jezeli dany trojkat rownoboczny o boku a
to r=1/6a√3---->P=πr²=1/12a²π=3πcm² (wpisane)
to R=1/3a√3---->P=πR²=1/3a²π=12πcm² (opisane)
jezeli dany trojkat rownoboczny o boku a
to r=1/6a√3---->P=πr²=1/12a²π=3πcm² (wpisane)
to R=1/3a√3---->P=πR²=1/3a²π=12πcm² (opisane)
a)r=⅓h=(a√3)/6
r=6√3/6=√3 cm
P=πr²=3,14*(√3)²=3,14*3=9,42cm²
b)=R=⅔H=(a√3/)3=6√3/3=2√3 cm
P=πR²=3,14*(2√3)²=3,14*4*3=37,68cm²
r=6√3/6=√3 cm
P=πr²=3,14*(√3)²=3,14*3=9,42cm²
b)=R=⅔H=(a√3/)3=6√3/3=2√3 cm
P=πR²=3,14*(2√3)²=3,14*4*3=37,68cm²