Step 1: Understand what is being asked.
A chord $PQ$ of a circle of diameter 28 cm subtends $90^\circ$ at the centre $O$, and we need the sector area $OPCQ$ and the segment area $PCQ$. Instead of treating $\Delta OPQ$ as a simple right triangle with base and height equal to the radius, let us use the general formula for the area of a triangle with two known sides and the included angle.
Step 2: Find the radius.
\[ r = \frac{\text{diameter}}{2} = \frac{28}{2} = 14 \text{ cm} \]
Step 3: Find the sector area.
\[ A_{\text{sector}} = \frac{\theta}{360^\circ}\times\pi r^2 = \frac{90^\circ}{360^\circ}\times\frac{22}{7}\times 14^2 \]
\[ = \frac{1}{4}\times\frac{22}{7}\times 196 = \frac{1}{4}\times 616 = 154 \text{ cm}^2 \]
Step 4: Find the triangle area using the two-sides-and-included-angle formula.
In $\Delta OPQ$, both $OP$ and $OQ$ equal the radius 14 cm, and the angle between them is $\angle POQ=90^\circ$. The area of any triangle with two sides $a,b$ and included angle $C$ is $\frac{1}{2}ab\sin C$, which works for any angle, not just $90^\circ$:
\[ A_{\Delta OPQ} = \frac{1}{2}\times OP\times OQ\times\sin(\angle POQ) = \frac{1}{2}\times 14\times 14\times\sin 90^\circ \]
Since $\sin 90^\circ=1$:
\[ A_{\Delta OPQ} = \frac{1}{2}\times 196\times 1 = 98 \text{ cm}^2 \]
Step 5: Subtract to find the segment area.
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\Delta OPQ} = 154-98 = 56 \text{ cm}^2 \]
Final Answer:
The sector area is $154\ \text{cm}^2$ and the segment area is $56\ \text{cm}^2$, matching option (A).
\[ \boxed{154\ \text{cm}^2,\ 56\ \text{cm}^2} \]