In right-angled triangle OAD, the angle at D is 90^°. By applying the Pythagoras theorem, OA² = OD² + AD², which simplifies to OD² = 25 - 16 = 9, so OD = 3 cm.

In figure 10.3 if oa is equals to 5 cm ab = 8 cm

Answer: 2 cm (Option a)

Step-by-step solution

Step 1: Identify given measurements and radius

In the given circle with centre $O$ , segment $OA$ is a radius, so $OA = 5\text{ cm}$ . Line segment $CD$ lies along the radius or diameter passing through the centre $O$ perpendicular to chord $AB$ . Therefore, the total length from $O$ to the circle along this radius is $OC = 5\text{ cm}$ .

Step 2: Use perpendicular bisector theorem on chord AB

The perpendicular drawn from the centre of a circle to a chord bisects the chord. Since $OD \perp AB$ , point $D$ bisects $AB$ , giving $AD = \frac{1}{2} AB = 4\text{ cm}$ .

Step 3: Apply Pythagoras theorem in triangle OAD

In right-angled triangle $\triangle OAD$ , the angle at $D$ is $90^\circ$ . By applying the Pythagoras theorem, $OA^2 = OD^2 + AD^2$ , which simplifies to $OD^2 = 25 - 16 = 9$ , so $OD = 3\text{ cm}$ .

Step 4: Calculate the length of CD

Since point $D$ lies on the radius $OC$ , the length $CD$ is the difference between the full radius $OC$ and the distance from the centre $OD$ . Thus, $CD = 5 - 3 = 2\text{ cm}$ . This corresponds to option (a).

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