Kakak bantu lagi matematika ini Kelas : 8 smp Bab : Teorema pythagoras

Limas T.PQRS

rusuk alas (persegi) = PQ = 8 cm
tinggi limas = TO = 12 cm
OU = 1/2 . PQ = 4 cm

TU = √(TO² + OU²)
TU = √(12² + 4²)
TU = √160
TU = √16 × √10
TU = 4√10 cm

∆TQR
QS = diagonal sisi persegi = 8√2 cm
OQ = 1/2 . QS = 4√2 cm

TQ = √(TO² + OQ²)
TQ = √(12² + (4√2)²)
TQ = √(144 + 32)
TQ = √176
TQ = √16 × √11
TQ = 4√11 cm

TQ = TR

Keliling ∆TQR
= TQ + TR + QR
= 4√11 + 4√11 + 8
= 8 + 8√11
= 8 (1 + √11) cm

Luas ∆TQR
= 1/2 × QR × TU
= 1/2 × 8 × 4√10
= 16√10 cm²

panjang ou = 5 cm
pake triple phytagoras

8 : 12 : 13

jika dibuktikan

12² + 5² = TU²
√144 + 25 = TU
√169 = TU
13 = TU

b. kel segitiga TQR tinggal ditambah
13 + 13 + 8 = 34
luas segitiga TQR
1/2 . alas . tinggi
1/2.8.13 = 52