Please kak bantu jawab, carilah luas permukaan bangun ruang tersebut,, yang b sm c
butuuh bgt,, please bantu
Galladeaviero
B) t kerucut = 24 cm r = 7 cm S kerucut = √(t²+r²)= √24²+7²= 25 Luas Permukaan = luas selimut kerucut + luas 1/2 bola LP= π r s + 2 π r² LP = π r (s + 2r) LP = 22/7. 7 ( 25 + 14)= 22(39)= 858 cm² . c) S kerucut = 5 r = 1/2 (8) = 4 t tabung = 16 Luas permukaan = Luas 1/2 bola + luas selimut tabung + luas selimut kerucut LP = 2 π r² + 2 π r t + π r s LP = π r ( 2r + 2t + s) LP = (3,14 x 4)(8 + 32+5) = 12,56 (45)= 565,2 cm²
t kerucut = 24 cm
r = 7 cm
S kerucut = √(t²+r²)= √24²+7²= 25
Luas Permukaan = luas selimut kerucut + luas 1/2 bola
LP= π r s + 2 π r²
LP = π r (s + 2r)
LP = 22/7. 7 ( 25 + 14)= 22(39)= 858 cm²
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c)
S kerucut = 5
r = 1/2 (8) = 4
t tabung = 16
Luas permukaan = Luas 1/2 bola + luas selimut tabung + luas selimut kerucut
LP = 2 π r² + 2 π r t + π r s
LP = π r ( 2r + 2t + s)
LP = (3,14 x 4)(8 + 32+5) = 12,56 (45)= 565,2 cm²