Jawab:
TC = AB = BC = CD = AD = 9 cm
TP = 2/3 × 9 = 6 cm
PC = 1/3 × 9 = 3 cm
ΔBDT ⊥ ΔTOC ⇒ maka jarak P ke ΔBDT = panjang PP'
OC = 1/2 AC
= 1/2 × 9√2
= [tex]\frac{9}{2} \sqrt{2} \ cm[/tex]
PP' : OC = TP : TC
[tex]PP':\frac{9}{2} \sqrt{2} =6:9[/tex]
[tex]PP'=\frac{6}{9} \times\frac{9}{2} \sqrt{2}[/tex]
[tex]=3\sqrt{2} \ cm[/tex]
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Jawab:
TC = AB = BC = CD = AD = 9 cm
TP = 2/3 × 9 = 6 cm
PC = 1/3 × 9 = 3 cm
ΔBDT ⊥ ΔTOC ⇒ maka jarak P ke ΔBDT = panjang PP'
OC = 1/2 AC
= 1/2 × 9√2
= [tex]\frac{9}{2} \sqrt{2} \ cm[/tex]
PP' : OC = TP : TC
[tex]PP':\frac{9}{2} \sqrt{2} =6:9[/tex]
[tex]PP'=\frac{6}{9} \times\frac{9}{2} \sqrt{2}[/tex]
[tex]=3\sqrt{2} \ cm[/tex]