An isosceles triangle with its vertices on a circle is shown in this picture.

What is the perimeter of the circle?
The figure is shown below:

Let OA = radius of circle.
AD = 1 cm
OA = r
∴ OD = 1-r
OA = OB = r
In triangle OBD,
∠ODB = 90°
AB = AC and AD is perpendicular on BC from point A.
∴ BD = DC = ![]()
∴ By pythagoras theorem,
(Hypotenuse)2 = (One side)2 + (Other side)2
∴ (OB)2 = (OD)2 + (BD)2
∴ r2 = (1-r)2 + (
)2
∴ r2- (1-r)2 = ![]()
∴ (r + {1-r})(r-{1-r}) =
{a2-b2 = (a + b) × (a-b)}
∴ 1 × (2r-1) = ![]()
∴ 2r = ![]()
∴ r = ![]()
If r = radius of circle
Circumference = ![]()
∴ Perimeter = ![]()
Couldn't generate an explanation.
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