1.Luas selimut kerucut adalah 550 cm2. Jika panjang jari-jari alas 7 cm, berapa tinggi kerucut tersebut? 2. Keliling alas sebuah kerucut adalah 31,4 cm. Jika pa

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Mapel : Matematika
Tentang : Bangun Ruang
Pertanyaan yg diajukan : 2
Pembahasan :

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Diketahui :
- LS = 550 cm²
- r = 7 cm

Ditanya :
- t

LS = πrs
550 = 22/7 x 7 x s
550 = 22 x s
s = 550/22
s = 25 cm

t² = S² - r²
t² = 25² - 7²
t² = 625 - 49
t = √576
t = 24 cm

Jawaban : 24 cm

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Diketahui :
- KA = 31,4 cm
- S = 13 cm

Ditanya :
- V

Ka = πd
31,4 = 3,14 x d
d = 31,4/3,14
d = 10 cm
r = 10/2 = 5 cm

t² = S² - r²
t² = 13² - 5²
t² = 169 - 25
t = √144
t = 12 cm

V = ⅓ x La x t
V = ⅓ x 3,14 x 5 x 5 x 12
V = 3,14 x 5 x 5 x 4
V = 3,14 x 25 x 4
V = 3,14 x 100
V = 314 cm³

Jawaban : 314 cm³

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Heil Manstein!

luas selimut kerucut = 550 cm²
π × r × s = 550
[tex] \frac{22}{7} [/tex] × 7 × s = 550
22 × s = 550
s = 550 ÷ 22 = 25 cm
tinggi kerucut = [tex] \sqrt{ s^{2} - r^{2} } [/tex] = [tex] \sqrt{ 25^{2} - 7^{2}} = \sqrt{625 -49} = \sqrt{576} = 24[/tex] cm 

2) keliling alas = 31,4
    2 × π × r = 31,4
    2 × 3,14 × r = 31,4
    r = 5 cm
    s = 13 cm
    t = [tex] \sqrt{ s^{2} - r^{2}} = \sqrt{ 13^{2} - 5^{2}} = \sqrt{169-25} = \sqrt{144} = 12 [/tex] cm
    volume kerucut = [tex] \frac{1}{3} [/tex] × π × [tex] r^{2} [/tex] × t
                              = [tex] \frac{1}{3} [/tex] × 3,14 × [tex] 5^{2} [/tex] × 12
                              = 314 cm³