Since there are [tex]360^{\circ}[/tex] in a circle, and central [tex]\angle AOB = 90^{\circ}[/tex], you can see that the part "sliced out" by the central angle is [tex]\frac{90^{\circ}}{360^{\circ}} = \frac{1}{4}[/tex] of the whole circle.
This means that the length of the intercepted arc [tex]\stackrel{\frown}{AB}[/tex] is [tex]\frac{1}{4}[/tex] of the circumference of the circle. So,
[tex]\stackrel{\frown}{AB} = \frac{1}{4}(2 \pi r) = \frac{1}{4}(2 \pi (7.5 cm)) = \bf \frac{15 \pi}{4} cm[/tex]
