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KesebangunanΔACB ~~ ΔCBD
a) sudut yang sama besar
krn ΔACB siku siku di C dan ΔCBD siku siku di D, maka:
<C = < D = (siku - siku)
<B = <B (beritmpit)
<A = <DCB
b. sisi yg bersesuai
CB ~ DB
AC ~ CD
AB ~ CB
c.
AB = 25 , AD = 5 --> AD +DB = AB
DB = 25-5 = 20
CD² = AD x DB
CD² = 5(20)= 100
CD = √100
CD + 10 cm
3.
ΔRST ~~ ΔRPQ
<R = <R
<T = < Q
<S = <Q
RS = 6
RT = 4
PQ = 14
RQ = RT + TQ = 4 +8 =12
..
TS : RS = PQ : RQ
TS : 6 = 14 : 12
TS = 6 (14)/(12)
TS = 7
RP : RQ = RT : RS
RP : 12 = 4 : 6
RP = 4(12)/(6)
RP = 8
RP = RS + SP
SP = RP - RS
SP = 8 - 6
SP = 2