A circle with center O has radius 10 cm. Chord AB is drawn such that its length is also 10 cm. C is a point on the major arc AB. Find: a) The special name of triangle AOB. b) The value of AOB. c) The

OA and OB are radii of the circle.
The problem states that the chord AB is equal in length to the radius.
Therefore, $OA = OB = AB$.

A triangle with all three sides equal in length is an equilateral triangle.
The special name of triangle AOB is an equilateral triangle.

Since triangle AOB is an equilateral triangle (from part a), all its interior angles are equal to $60^\circ$.
Therefore, $\angle AOB = 60^\circ$.

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circumference.
Here, $\angle AOB$ is the angle at the center subtended by arc AB, and $\angle ACB$ is the angle at the circumference subtended by the same arc AB.
So, $\angle AOB = 2 \times \angle ACB$.

We found $\angle AOB = 60^\circ$ in part b.

Radius, $r = 7$ cm.
Angle subtended by arc AB at the center, $\theta = \angle AOB = 60^\circ$ (from part b).
$\pi = \frac{22}{7}$.
The formula for the length of an arc is $L = \frac{\theta}{360^\circ} \times 2\pi r$.