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Rumus cepat
Cara cepat !
AO = ⅔ Diagonal Ruang
AO = ⅔ 9√3
AO = 6√3 cm.
AB = 9 cm
maka
AC = s√2 = 9√2 cm
OC = (9/2)√2 cm
maka
TC² = TO² + OC²
= 9² + ((9/2)√2)²
= 81 + 81/2
= 81(1 + 1/2)
= (6/4)(81)
TC = (9/2)√6 cm
Perhatikan TC = TA = (9/2)√6 cm (sama kaki)
Jarak A ke TC = AP
Perhatikan ΔATC, berlaku
TO × AC = AP × TC
⇒ 9 × 9√2 = AP × (9/2)√6
⇒ 9√2 = AP × (1/2)√6
⇒ AP = (9√2)/(1/2)√6
⇒ AP = 18√1/3
⇒ AP = 18/√3
⇒ AP = (18/3)√3
⇒ AP = 6√3 cm
Jawab : C
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