In a circle, we have a 10 unit long chord AB, which intercepts 60^(∘) arc AB. We want to find the area of this circle. First, let's draw a diagram of this situation.
A central angle intercepting arc AB has the same measure as this arc — 60^(∘). The vertex of this central angle is the center of our circle. Let's call the vertex O and add the angle to our diagram.
Since O is the center of our circle, both OA and OB are the circle's radii. This means OA = OB and △ OAB is isosceles. By the Base Angles Theorem, in an isosceles triangle the base angles are congruent. In consequence, m ∠ OAB = m ∠ OBA. Let's call this measure x. In a triangle angles interior angles sum to 180^(∘), so we can write the following equation for x.
x + x + 60^(∘) = 180^(∘)
Let's solve it!
We have that x = 60^(∘), so both m ∠ OAB and m ∠ OBA equal 60^(∘). As a result, both of these angles are congruent to ∠ AOB, so all interior angles in △ OAB are congruent to each other. We conclude that △ OAB is an equilateral triangle and all its side lengths equal 10 — the length of AB. Let's add this information to our diagram.
Now, let's recall the formula for the area A of a circle.
A = π r^2
Here, r is a radius of the circle. In our case, OA is the radius, so r = OA = 10. Let's substitute this value into our formula and find the area.
