Kak bantu matematika Kelas : 8 smp Bab : Teorema pythagoras

Balok ABCD.EFGH

AB = 8 cm
BC = 6 cm
CG = 4 cm

Bidang ACGE
AC = √(AB² + BC²)
AC = √(8² + 6²)
AC = √(65 + 36)
AC = √100
AC = 10 cm

Keliling ACGE
= 2(AC + GE)
= 2(10 + 4)
= 2 × 14
= 28 cm

Luas ACGE
= AC × GE
= 10 × 4
= 40 cm²

Diagonal ruang, AG
= √(AB² + BC² + CG²)
= √(8² + 6² + 4²)
= √(64 + 36 + 16)
= √116
= √4 . √29
= 2√29 cm