Geometry Practice Test 5

∠CAB + ∠DBA = 180°
[Angles on the same side of a traversal are supplementary]
$\frac{1}{2}$ ∠CAB + $\frac{1}{2}$ ∠DBA = $\frac{180°}{2}$
∠OAB + ∠OBA = 90°
[∵ OA and OB bisect angle ∠CAB and ∠DBA respectively.]
In ΔAOB,
∠AOB = 180° ∠OAB - ∠OBA
∠AOB = 90°
By using Pythagoras theorem,
AB = $\sqrt{\mathit{O}{\mathit{A}}^2+\mathit{O}{\mathit{B}}^2}$
AB = $\sqrt{6^2+{9.1}^2}$
AB = $\sqrt{36+82.81}$
AB = $\sqrt{118.81}$
AB = 10.9 cm