Two circles centre's O1 and O2 are of the same radius, 8.75 cm. The two circles intersect one another at P and at Q. The common chord PQ is of length 8.75 cm. Calculate the area common to the two circles.

The two circles have the same radius, $r = 8.75$ cm.
The common chord PQ has a length of 8.75 cm.
The area common to the two circles consists of two identical circular segments.

Consider one circle with center $O_1$. The radii $O_1P$ and $O_1Q$ are both equal to the radius $r = 8.75$ cm.
The common chord PQ is also 8.75 cm.
Therefore, the triangle $\triangle O_1PQ$ has sides $O_1P = O_1Q = PQ = 8.75$ cm. This means $\triangle O_1PQ$ is an equilateral triangle.
The angle subtended by the chord PQ at the center $O_1$ is $\theta = \angle PO_1Q = 60^\circ$.
In radians, $\theta = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3}$ radians.

The area of a sector is given by the formula $A_{sector} = \frac{1}{2}r^2\theta$.

Since $\triangle O_1PQ$ is an equilateral triangle with side length $s = 8.75$ cm, its area is given by $A_{triangle} = \frac{\sqrt{3}}{4}s^2$.

The area of one circular segment is the area of the sector minus the area of the triangle.

The common area to the two circles is composed of two such identical segments.

Now, substitute the numerical values:

Rounding to two decimal places:

The area common to the two circles is $\text{13.87 cm}^2$.