Kesebangunan
Nomor 4
TQ = (SP·RT) / RSsehingga:TQ = (4·9) / 6 = 36/6TQ = 6 cm.
Jika memerlukan penjelasan mengenai perbandingan tersebut, berikut ini penelusurannya.
Dari ΔRPQ dan ΔRST:RP/RS = RQ/RT(RS+SP) / RS = (RT+TQ)/RT1 + SP/RS = 1 + TQ/RTSP/RS = TQ/RTTQ = (SP·RT) / RS_____________
Nomor 5
Dari ΔPAB dan PDC:PA/PD = AB/DC(PD+DA) / PD = AB/DC1 + DA/PD = AB/DCDA/PD = AB/DC – 1PD = DA / (AB/DC – 1)PD = 4 / (8/6 – 1)PD = 4 / (4/3 – 1)PD = 4 / (1/3) = 4 × 3PD = 12 cm.
Atau dengan cara lain.PD/PA = DC/(AB–DC)PD/4 = 6/(8–6)PD/4 = 6/2 = 3PD = 4 × 3 = 12 cm._____________
[tex]\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\\\end{array}[/tex]
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Pembahasan
Kesebangunan
Nomor 4
TQ = (SP·RT) / RS
sehingga:
TQ = (4·9) / 6 = 36/6
TQ = 6 cm.
Jika memerlukan penjelasan mengenai perbandingan tersebut, berikut ini penelusurannya.
Dari ΔRPQ dan ΔRST:
RP/RS = RQ/RT
(RS+SP) / RS = (RT+TQ)/RT
1 + SP/RS = 1 + TQ/RT
SP/RS = TQ/RT
TQ = (SP·RT) / RS
_____________
Nomor 5
Dari ΔPAB dan PDC:
PA/PD = AB/DC
(PD+DA) / PD = AB/DC
1 + DA/PD = AB/DC
DA/PD = AB/DC – 1
PD = DA / (AB/DC – 1)
PD = 4 / (8/6 – 1)
PD = 4 / (4/3 – 1)
PD = 4 / (1/3) = 4 × 3
PD = 12 cm.
Atau dengan cara lain.
PD/PA = DC/(AB–DC)
PD/4 = 6/(8–6)
PD/4 = 6/2 = 3
PD = 4 × 3 = 12 cm.
_____________
[tex]\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\\\end{array}[/tex]