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= phi × r × s
3,14 × 3 × 5 = 47,1
mencari s
s² = t² + r²
= 4² + 3²
= 16 + 9 = 25
s = akar dri 25 adalah 5
l.permukaan kerucut
= phi × r (r+s)
= 3,14 × 3 (3+5)
= 9,42 (8)
= 75,36
diket r=3 cm π=3,14 t=4 cm s=?
s=√3²+4²
=√9+16 =√25= 5, jadi s= 5
L. selimut kerucut= πrs =3,14x3x5 = 3,14x15= 47,1 (luas selimut kerucut)
rumus L. permukaan kerucut = πr²+πrs
(3,14x3²)+(47,1)
28,26+47,1=75,36 (L.permukaan kerucut)