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