Panjang jari jari alas sebuah kerucut 28 cm,dan luas seluruh kerucut itu 5.544cm.dengan pi:22/7.hitunglah: a.panjang garis pelukis b.tinggi kerucut c.volume kerucut
DB45
A) panjang garis pelukis = s r = 28 cm Luas kerucut = 5.544 πr ( r + s)= 5.544 ²²/₇ (28) ( 28 +s) = 5.544 88 (28+s) = 5.544 28+ s = 63 s = 35 cm
b. tinggi = t t= √(s²-r²) = √(35²-28²)= √441 = 21 cm
c. V = ¹/₃ π r² t V = ¹/₃ (²²/₇) (28²)(21) = 17.248 cm³
16 votes Thanks 27
davidyonggy
Diket: r = 28cm, luas kerucut = 5544cm L.k= phi.r^2+pi.r.s 5544=22/7.28.28+22/7.28.s 5544=2464+88s 5544-2464=88s 3080=88s s=35
a. panjang garis pelukis = 35cm b. tinggi kerucut
c. volume kerucut V=1/3.phi.r^2.t =1/3.22/7.28.28.21=17248cm3
r = 28 cm
Luas kerucut = 5.544
πr ( r + s)= 5.544
²²/₇ (28) ( 28 +s) = 5.544
88 (28+s) = 5.544
28+ s = 63
s = 35 cm
b. tinggi = t
t= √(s²-r²) = √(35²-28²)= √441 = 21 cm
c. V = ¹/₃ π r² t
V = ¹/₃ (²²/₇) (28²)(21) = 17.248 cm³
L.k= phi.r^2+pi.r.s
5544=22/7.28.28+22/7.28.s
5544=2464+88s
5544-2464=88s
3080=88s
s=35
a. panjang garis pelukis = 35cm
b. tinggi kerucut
c. volume kerucut
V=1/3.phi.r^2.t
=1/3.22/7.28.28.21=17248cm3