Lp kerucut:
s²=(r)²+(36-12)²
s²=(7)²+(24)²
s²=49+576
s=√625
s=25cm²
--------------------
Lp=Πrs
Lp=22/7×7×25
Lp=22×25
Lp=550cm²
Lp tabung:
Lp=Πdt+Πr²
Lp=22/7×14×12+22/7×7×7
Lp=22×2×12+154
Lp=44×12+154
Lp=528+154
Lp=682cm²
Lp total:
682+550
=1.232cm²(c.)
Penjelasan dengan langkah-langkah:
BANGUN RUANG LENGKUNG
d = 14 cm → r = 7 cm
π = 22/7
t.tabung = 12 cm ← t
t.kerucut = 36 – 12 = 24 cm
s.kerucut = √(t.kerucut² + r²)
s.kerucut = √(24² + 7²)
s.kerucut = √(576 + 49)
s.kerucut = √625
s.kerucut = 25 cm ← s
Luas permukaan bangun gabungan kerucut dan tabung dicari menggunakan rumus Luas selimut kerucut dan Luas Permukaan tabung tanpa tutup.
Lp = πrs + πr(r + 2t)
Lp = πr(r + s + 2t)
Lp = 22/7 × 7 × (7 + 25 + 2×12)
Lp = 22 × 7/7 × (32 + 24)
Lp = 22 × 1 × 56
Lp = 11 × 2 × 56
Lp = 11 × 112
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Penyelesaian:
Lp kerucut:
s²=(r)²+(36-12)²
s²=(7)²+(24)²
s²=49+576
s=√625
s=25cm²
--------------------
Lp=Πrs
Lp=22/7×7×25
Lp=22×25
Lp=550cm²
--------------------
Lp tabung:
Lp=Πdt+Πr²
Lp=22/7×14×12+22/7×7×7
Lp=22×2×12+154
Lp=44×12+154
Lp=528+154
Lp=682cm²
--------------------
Lp total:
682+550
=1.232cm²(c.)
Penjelasan dengan langkah-langkah:
BANGUN RUANG LENGKUNG
d = 14 cm → r = 7 cm
π = 22/7
t.tabung = 12 cm ← t
t.kerucut = 36 – 12 = 24 cm
s.kerucut = √(t.kerucut² + r²)
s.kerucut = √(24² + 7²)
s.kerucut = √(576 + 49)
s.kerucut = √625
s.kerucut = 25 cm ← s
Luas permukaan bangun gabungan kerucut dan tabung dicari menggunakan rumus Luas selimut kerucut dan Luas Permukaan tabung tanpa tutup.
Lp = πrs + πr(r + 2t)
Lp = πr(r + s + 2t)
Lp = 22/7 × 7 × (7 + 25 + 2×12)
Lp = 22 × 7/7 × (32 + 24)
Lp = 22 × 1 × 56
Lp = 11 × 2 × 56
Lp = 11 × 112
Lp = 1.232 cm²