tolong dijawab yah bagi yg bisa :)

TC= 4,8(phytagoras 1,4,4,8,5,0)
volume bandul= V.limas+V.1/2bola
=1/3.phi.r^2.t+1/2.4/3.phi.r^3
=1/3.phi.r^2(t+2r)(disederhanakan biar mudah)
=1/3.22/7.1,4.1,4(4,8+2(1,4)
=1/3.6,16(7,6)
=15605,3

r=2,8/2=1,4 cm
s=0,5 dm = 5 cm

a).
t²=s²-r²=5²-1,4²=25-1,96=23,04
t=√23,04=4,8 cm

tc=t=4,8 cm

b).
Volume bandul
=Volume½bola + Volume kerucut
=⅔πr³ + ⅓πr²t
=πr² (2r/3 + t/3)
=22/7. 1,4. 1,4 ( 2. 1,4/3 + 4,8/3)
=6,16 (2,8/3 + 4,8/3)
=6,16 ( 7,6/3)
=46,816/3
=15,6 cm³

Luas permukaan bandul
=Luas½bola + Luas selimut kerucut
=2πr² + πrs
=πr (2r+s)
=22/7. 1,4 ( 2. 1,4 + 5)
=4,4(2,8+5)
=4,4(7,8)
=34,32 cm²