Diketahui balok ABCD.EFGH, panjang AB=6 cm, BC=8 cm, CG=10 cm. Hitinglah : A. Panjang AF B. Panjang CE C. Volume D. Luas permukaan

A) AF=
[tex] \sqrt{(ab {}^{2} } + cg {}^{2}) [/tex]
[tex] \sqrt{(6 {}^{2} } + 10 {}^{2} )[/tex]
[tex] \sqrt{(36 + 100} )[/tex]
=11,6cm
B) AC=
[tex] \sqrt{(ab {}^{2} } + bc {}^{2} [/tex]
[tex] \sqrt{(6 {}^{2} } + 8 {}^{2} [/tex]
[tex] \sqrt{(36 + 64)} [/tex]
[tex] \sqrt{100} = 10cm[/tex]
CE=
[tex] \sqrt{(ac {}^{2} } + cg{}^{2} )[/tex]
[tex] \sqrt{(10 {}^{2} } + 10 {}^{2} )[/tex]
[tex] \sqrt{200} = 14.1cm[/tex]
C) volume=p×l×t
=6×8×10
=480cm^3
D)Luas permukaan=AB×BC
=6×8
=48cm^2