Materi : Kesebangunan dan Kekonggruenan
∆PQR { a¹ = PQ , t¹ = PR , s¹ = QR = 16 }
∆PQS { a² = SQ = 8 , t² = ST , s² = PQ }
a¹/a² = s¹/s²
PQ/SQ = QR/PQ
PQ × PQ = QR × SQ
PQ² = 16 × 8
PQ = √[ 2⁴ × 2³ ]
PQ = √[ 2⁶ × 2 ]
PQ = [2³]√2
PQ = 8√2 cm
Semoga bisa membantu
[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
PQ² = 16(8)
PQ² = 128
PQ = √128
PQ = √2⁶(2¹)
PQ = √(2⁶(2)
PQ = 2^(6/2)√2
PQ = 2³√2
PQ = (2(2(2)√2
PQ = (4(2)√2
PQ = 8√2 CM
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Materi : Kesebangunan dan Kekonggruenan
∆PQR { a¹ = PQ , t¹ = PR , s¹ = QR = 16 }
∆PQS { a² = SQ = 8 , t² = ST , s² = PQ }
Panjang PQ
a¹/a² = s¹/s²
PQ/SQ = QR/PQ
PQ × PQ = QR × SQ
PQ² = 16 × 8
PQ = √[ 2⁴ × 2³ ]
PQ = √[ 2⁶ × 2 ]
PQ = [2³]√2
PQ = 8√2 cm
Semoga bisa membantu
[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
Verified answer
PQ² = 16(8)
PQ² = 128
PQ = √128
PQ = √2⁶(2¹)
PQ = √(2⁶(2)
PQ = 2^(6/2)√2
PQ = 2³√2
PQ = (2(2(2)√2
PQ = (4(2)√2
PQ = 8√2 CM
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