diketahui sebuah kerucut mempunyai s=25cm t=24cm tentukan jari jari diameter luas selimut kerucut luas permukaan kerucut dan volume

jari2 alas kerucut (r)
= √(s²-t²)
= √(25²-24²)
= √(625-576)
= √49
= 7 cm

diameter (d) = 2 x r = 2 x 7 = 14 cm


Ls.kerucut
= phi x r x s
= 22/7 x 7 x 25
= 550 cm²


Lp.kerucut
= phi x r x (r+s)
= 22/7 x 7 x (7+25)
= 22 x 32
= 704 cm²


V.kerucut
= 1/3 x La x t
= 1/3 x (phi x r²) x t
= 1/3 x (22/7 x 7 x 7) x 24
= 8 x 154
= 1.232 cm^3