tolong dibantu ya kk matnya terima kasih no 10 ya

LABCD = 2 × LABC
⇔ 125 = 2 × 1 /2 × AC × BQ
⇔ 125 = AC × BQ
⇔ 125 = 25 × BQ
⇔ BQ = 5
Perhatikan ΔAPD, PD = BQ = 5 cm.

Perhatikan ΔAPD, PD = BQ = 5 cm.
AP = √AD^2- PD^2= √13^-5^2
= √169-25 =√ 144
AP = 12 cm
AP = QC = 12 cm.
AC = AP + PQ + QC
⇔ 25 = 12 + PQ + 12
⇔ 25 = 24 + PQ
⇔ PQ = 1
Jadi, panjang PQ = 1 cm.