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AB = BC = 12 cm
TO = 8 cm
ditanya : O->TBC...?
jawab :
OP = AB/2
OP = 12/2
OP = 6 cm
PT = √(TO^2 + OP^2)
PT = √(8^2 + 6^2)
PT = √(64 + 36)
PT = √100
PT = 10 cm
O->TBC = OR = x
*gunakan konsep mencari luas segitiga
• (OP)(TO)(1/2) = (PT)(OR)(1/2)
*OP dan PT = alas
*TO dan OR = tinggi
• (6)(8)(1/2) = (10)(x)(1/2)
• (6)(8) = 10x
• 10x = 48
• x = 48/10
• x = 24/5
• x = 4 4/5 #D
maka
OP = 1/2 · 12 = 6 cm
TO = 8 cm
Sehingga
TP² = TO² + OP²
= 8² + 6²
= 64 + 36
= 100
TP = 10 cm
Jarak O ke bidang TBC = OR
Perhatikan ΔTOP
Berlaku,
LΔ = LΔ
⇒ 1/2 · TO · OP = 1/2 · TP · OR
⇒ 8 · 6 = 10 · OR
⇒ OR = 48/10
⇒ OR = 24/5
⇒ OR = 4 4/5 cm
Jawab : D
Terimakasih semoga membantu