Bunuel wrote:

In the figure above, inscribed polygon PQRST is equilateral. If the diameter of the circle is 10, then the length of arc PQR is

(A) 10π

(B) 4π

(C) 2π

(D) 4π/5

(E) 2π/5

Attachment:

2017-11-15_1035.png

rocko911 - in fairness, if I recall correctly (and I may not), many properties of pentagons inscribed in circles involve trig.

Not true here, but you do have to know (or derive): An equilateral \(n\)-gon inscribed in a circle divides the circle's circumference into \(n\) pieces.

The distance between the vertices is the same. The vertices lie on the circle. The chords that intercept the arcs are equal. If two chords are equal in measure, then their corresponding minor arcs (PQ and QR) are equal in measure.

Here, an inscribed equilateral pentagon's vertices divide the circumference into 5 arcs.

We are asked for the arc length of arc PQR. That length is a fraction of the circumference's length.

Count arc segments: from P to Q, then Q to R, there are 2 arc segments out of 5 total. Arc PQR hence is \(\frac{2}{5}\) of the circumference of the circle.

Circumference of circle?

Diameter = 10 = 2r. So r = 5

Circumference =

\(2\pi*r=10\pi\)Length of arc PQR?

\((\frac{2}{5} *

10\pi) = 4\pi\)

Answer B

Hope that helps.

_________________

At the still point, there the dance is. -- T.S. Eliot

Formerly genxer123