
MATERI KERUCUT
Presentation
•
Mathematics
•
9th Grade
•
Hard
Bambang Handoko
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7 Slides • 0 Questions
1
materi KERUCUT
By Bambang Tri Handoko
2
r = jari-jari kerucut
t = tinggi kerucut
s = garis pelukis
s2 = r2 + t2
Bambang Tri Handoko
UNSUR KERUCUT
3
Ada 2 bagian
1 = Selimut kerucut yang berbentuk juring lingkaran
2 = alas kerucut berbentuk lingkaran
Bambang Tri Handoko
JARING-JARING KERUCUT
4
Luas = Luas Selimut kerucut + Luas alas
= π r s + π r2
= π r ( s + r)
Bambang Tri Handoko
LUAS PERMUKAAN KERUCUT
5
Volume Kerucut = ⅓ Volume Tabung
= ⅓ π r2 t
Bambang Tri Handoko
VOLUME KERUCUT
6
Penyelesaian:
s2 = 72 + 242 = 49 + 576 = 625
s = 25
Luas permukaan kerucut = π r ( s + r)
= 22/7 x 7 x ( 7 + 25) = 22 x 32 = 704
Jadi luas permukaan kerucut 704 cm2
Bambang Tri Handoko
SOAL 1
Jari-jari alas kerucut = 7 cm dan tingginya = 24 cm, Hitunglah luas permukaan kerucut dengan π = 22/7
7
Penyelesaian:
t2 = s2 - r2 = 172 - 82 = 289 - 64 = 225
t = 15
Volume kerucut = ⅓ π r2 t
= ⅓ x 3,14 x 82 x 15 = 1004, 8
Jadi volume kerucut 1004,8 cm3
Bambang Tri Handoko
SOAL 2
Jari-jari kerucut = 8 cm dan panjang garis pelukisnya 17 cm, tentukan volume kerucut tersebut dengan π = 3,14
materi KERUCUT
By Bambang Tri Handoko
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