Diket balok abcd.efgh , ab=12 cm, bc=5 cm, ae=7 cm. Ditanya jarak a ke bh???

Perhatikan segitiga AHB
AB = 12
AH = √(AD^2 + DH^2) = √(5^2 + 7^2) = √(25 + 49) = √74
HB = √(AH^2 + AB^2) = √(√74^2 + 12^2) = √(74 + 144) = √218 =
Jarak A ke HB = (AB . AH)/HB
= (12 . √74)/√218
= (12 . √37) / √109
= (12 √37) / √109 . √109/√109
= (12 √4033) / 109