528π = 2 . π . r (r + 13)
528π = 2 π (r2 + 13r)
528π/2π = r2 + 13r
264 π = r2 + 13r
r2 + 13r – 264 = 0
a = 1
b = 13
c = -264
r1.2 = -b ±√b2 -4ac/2a
r1.2 = -13 ±√132 -4 . 1 . -264/2.1
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528π = 2 . π . r (r + 13)
528π = 2 π (r2 + 13r)
528π/2π = r2 + 13r
264 π = r2 + 13r
r2 + 13r – 264 = 0
a = 1
b = 13
c = -264
r1.2 = -b ±√b2 -4ac/2a
r1.2 = -13 ±√132 -4 . 1 . -264/2.1
Editor: Suci Annisa Caroline
Sumber: buku.kemdikbud.go.id
