# Area of a sector - Inscribed Angle and Central Angle

I know the formula for the area of a sector of an arc made by central angle is $$\text{Area}_\text{Sector}= \frac{\text{Arc Angle} \times \text{Area of Circle} }{360}$$ Now my question is , Is this formula also applicable for Arcs formed by inscribed angles rather than Central Angles ? (I know that angle of an intercepted arc of an inscribed angle is twice the measure of the inscribed angle.)

Here are some equal angle inscribed angles (they all subtend half the angle of the red arc). It is clear that the area of the last sector is properly contained in the areas of the others.

$\hspace{2.5cm}$ The area of the sector is the sum of the area between the arc and its chord (a constant area) and the area of the triangle, which is $\frac12$ the product of the length of the chord (a constant length) and the altitude of the triangle, which varies as the vertex moves around the circle.

• So in short the formula is only applicable for arcs made by central angles right ? just to be sure – Rajeshwar Jul 11 '12 at 2:54
• @Rajeshwar: yes. – robjohn Jul 11 '12 at 2:54

No. Consider the case of a right angle. For the sector we get a quarter circle. For an inscribed angle, consider the base of the angle as a diameter and now move the other side of the angle so that it contains the quarter circle. The area can be made almost twice the size of the quarter sector.

• So you are saying this formula is only valid if the arc is formed by a central angle ? – Rajeshwar Jul 11 '12 at 0:07
• Yes, that is right. – i. m. soloveichik Jul 14 '12 at 20:43

I'm working on something like this at the moment. In fact, this problem is a special case of what I'm trying to figure out. This is the general formula for these areas, and it's not pretty.

I don't know how to format these posts, but I'll give it a try.

ϕ = angle of vector from the center pointing towards the vertex.
θ_1 = angle of vector from vertex to first chord
θ_2 = angle of vector from vertex to second chord
$$L(\theta_{2}) - L(\theta_{1})$$
$$L( \theta ) =\ \frac{r^{2}}{2} \big(\theta+ (\frac{sin(2\theta- 2\phi )}{2}) - sin(\theta- \phi) \sqrt{1-sin(\theta- \phi)^{2}} -(\theta- \phi) \big)$$