Question

pliss bantuu, yang tau jawaban soal ini yaa​

Answer 1

Jawaban:

Jarak CD=¾×8=6m

Jarak AD=⅔×6=4m

Misal seberang sungai adalah titik E

CD/AB=DE/EA

6/8=x/x+4

6(x+4)=8x

6x+24=8x

24=2x

x=12m

Maka lebar sungai adalah 12meter.

Semoga bermanfaat

Answer 2

Materi : Kesebangunan dan Kekonggruenan

∆PCD { a¹ = CD , t¹ = PD }

∆PBA { a² = BA , t² = PA = AD + PD }

__________________________/

⅔AD = CD [ 2 . AD = 3 . CD ]

¾CD = AB [ 3 . CD = 4 . AB ]

---

2 . AD = 3 . CD = 4 . AB

Maka AD : CD : AB = 2 : 3 : 4

---

AB = 8 m

CD = 6 m

AD = 4 m

____________________________/

∆PCD { a¹ = 6 m , t¹ = PD }

∆PBA { a² = 8 m , t² = 4 m + PD }

----

a¹/a² = t¹/t²

6/8 = PD/( 4 + PD )

8( PD ) = 6( PD + 4 )

8( PD ) - 6( PD ) = 24

2( PD ) = 24

PD = 12 m

Semoga bisa membantu

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