Question

Tolong bantu jawab dengan caranya
jawab yang nomor 1,3,dan 4

Answer 1

1) Δ ABC₁ sebangun Δ C₂DE , karena
<B = < D = siku-siku
<A =< C₂
<C₁  = <E

b. panjang AC
dgn (pitagoras) --> AC² = AB² +BC²
AC² = 20²+15² = 625
AC = √625 = 25

c. Panjang CE danDE
CE/ CD = AC/AB
CE/ 15 = 25/20
CE = (15 x 25)/(20)
CE= 18,75

DE/CD = BC/AB
DE/15 = 15/20
DE = 15(15)/20 = 11,25

3)
AE/AD = EG/DC
8 / (8+4) = EG /9
12 EG = 9(8)
EG = 72/12 = 6  ---> FG =  EF - EG = 14 - 6 = 8

EF = DC(AE) + AB(DE)/ (AE+DE)
14 = {9(8) + AB(4)) /(8+4)
14 =  ( 72 + 4 (AB))/ (12)
72 + 4 AB = 12 (14)= 168
4AB = 168 - 72 = 96
AB = 96/4 = 24

CF / DE = FB/AE
CF/4 = 12/8
CF = 4 (12)/8
CF= 48/8 = 6

4) Δ ABC sebangun Δ FBE
<A =<F
<C= <E
<B = <B
EF = √(BE²+BF²) = 25

BF : BA = 15 : 30 = 1 : 2
BE = 20  --> BC = 40
EF = 25 --> AC = 50

ΔBEF dengan ΔDCF
<B = <B
<F = < F
<E = < C
EF  : CF = 25 : 25 = 1 : 1
BE = CD = 20
BF = DF = 15

ED =  EF + FD = 25 + 15  = 40
AD = AC - CD = 50 - 20 = 30