Step 1: Understanding the Concept:
This question is based on the Inscribed Angle Theorem in circle geometry. The theorem states that the measure of an angle inscribed in a circle is half the measure of its intercepted arc.
Step 2: Key Formula or Approach:
The formula derived from the Inscribed Angle Theorem is:
\[ \text{Measure of Inscribed Angle} = \frac{1}{2} \times \text{Measure of Intercepted Arc} \]
Or, rearranging for the arc:
\[ \text{Measure of Intercepted Arc} = 2 \times \text{Measure of Inscribed Angle} \]
Step 3: Detailed Explanation:
We are given the measure of the inscribed angle, \(\angle\)ACB.
\[ \angle\text{ACB} = 65^{\circ} \]
The intercepted arc is AXB.
Using the formula from Step 2:
\[ \text{Measure of arc AXB} = 2 \times \angle\text{ACB} \]
\[ \text{Measure of arc AXB} = 2 \times 65^{\circ} \]
\[ \text{Measure of arc AXB} = 130^{\circ} \]
Step 4: Final Answer:
The measure of the intercepted arc AXB is 130\(^{\circ}\).
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.