CPM Homework Help : INT2 Problem 12-83

In a circle, chord $\overline{AB}$ has length $10$ units, while $m\overarc{AB} = 60°$ . What is the area of the circle? Draw a diagram and show all work.

Step 1:

The measure of the intercepted arc is equal to the measure of the central angle.
$m\overarc{AB} = 60° = m∠C$

Step 2:

$AC=BC$ , because both are radii, so $ΔABC$ is isosceles.

Isosceles triangles have congruent base angles, so $m∠A = m∠B$ .

$m∠A + m∠B + m∠C = 180°$
but $m∠C = 60°$ , $m∠A = m∠B$
so $m∠B + m∠B + 60° = 180°$
$2m∠B = 120°$
$m∠B = 60° = m∠A$

Circle, center C, with points, A, &, B, line segments from, C, to, B, from, C, to, A, from, A, to, B. Segment, A, B, labeled, 10, arc, A, B, labeled, 60 degrees.

Answer:

$π(10)^2 = 100π \approx 314.16\text{ units}^2$

Step 3:

Because the angles are all congruent, the triangle is equilateral.

So the length of the radius ( $AC$ ) is $10$ .