2.Diketahui suatu kerucut memiliki jari jari 5 Cm dan tinggi 12 cm tentukan A.luas selimut kerucut B.luas permukaan kerucut C.volume kerucut BANTU JAWAB SAMA CA

a. luas selimut kerucut = π.r.s
cari terlebih dahulu s (garis pelukis)
s2 = r2 + t2
s2 = 5(pangkat 2)+12(pangkat 2)
s2 = 25 + 144
s2 = √169
s. = 13
π .r.s
= 3.14 x 5x 13
= 204,1 cm3

b. luas permukaan kerucut = π.r ( r + s)
= 3,14 x 5 ( 5+ 13)
= 282,6 cm3

c. volume kerucut =1/3.π.r2.t
= 1/3 x 3,14 x5x5x12
= 314cm3

a.
[tex]K_{alas}=2\pi r\\=2\pi\times 5\\=10\pi\\\\R=pelukis=\sqrt{t^2-r^2}\\=\sqrt{12^2-5^2}\\=7\\\\K_{lingkaran}=2\pi R\\=2\pi\times 7\\=14\pi\\\\\theta=2\pi\times \frac{K_{selimut}}{K_{lingkaran}}\\=2\pi\times\frac{10\pi}{14\pi}\\=\frac{10\pi}{7}\\\\L_{selimut}=\frac{1}{2}\theta R^2\\=\frac{1}{2}\times\frac{10\pi}{7}\times 7^2\\=35\pi[/tex]

[tex]L_{alas}=\pi r^2\\=\pi\times 5^2\\=25\pi\\\\L_{kerucut}=L_{alas}+L_{selimut}\\=25\pi+35\pi\\=60\pi\\\\V_{kerucut}=\frac{1}{3}\pi r^2 t\\=\frac{1}{3}\pi\times 5^2\times 13\\=\frac{325\pi}{3}[/tex]