Jawabin nomor 7 kalau bisa beserta caranya

Ef=fg=4:3.Ef=4/3 FG
EG=15CM.EC =25 CM.

EG * 2=Ef*2+fg*2
15*2=(4/3 fg) *2+Fg*2
225=(4/3 fg×4/3 fg) +fg*2
fg*2=225-(4/3 fg :3/4 fg)
fg*2=225-(4/3:3/4)
Fg*2=225-16/9=2.025/9=akar 223,222
Fg=akar 2.009/9=akar 223,222
Fg=14,94cm
Ef=4/3 ×14,94=19,92cm
tinggi balok =Cg=akar (25*2-15*2)
=akar (625-225)
=akar (400)=20 cm


luas permukaan balok
=2(14,94×19,92)+2(14,94×20)+2(19,94×20)
=595,21+597,6+797,6
=1.990,41cm*2

maaf kl salah