10th grade Geometry question, inscribed triangle.?

Question

Each side of an inscribed equilateral triangle has a length of 18 meters. Find the length of one of the minor arcs to the nearest 100th. Please help!

Answer 1

If the side of the triangle is 18, the radius of the circle is 6√3 (this is found with a 30-60-90 relationship, where the radius of the circle is the hypotenuse of a little 30-60-90 triangle in which the longer leg, opposite 60°, is half the length of the triangle =9). The arc has measure 120° using the inscribed angle theorem. So the length of the arc is:

120/360 * 2πr
= 1/3 * 2π * 6√3
= 4π√3 meters
= 21.77 meters

Answer 2

Each angle of the inscribed triangle is 60 degrees, if we draw a line from one the vertexes to the center of the circle we will bisect the angle. Now drop a perpendicular from the center to one of the sides of the triangle and we will bisect that side.

We have a right triangle with an angle of 30 degrees and a side of 9 meters

The hypotenuse of the right triangle is the radius

cos 30 deg = 9/radius

0.86603 = 9/radius
radius = 10.3923 m

Now we know that the circumference = 2πr

C = 2π*10.3923 = 65.2968

The triangle vertexes are the end points of 3 equal arcs of 21.77 meters.
.

Answer 3

IF it is inscribed in a circle:
let the radius of said circle be r.
Consider the triangle formed by two of the triangle's vertices and the centre of the circle.
By the cosine rule,
18*18 = rr + rr - 2 rr cos 120d
so 324 = 3rr; r = sqrt 108 = 6rt3.

By the formula arc length = r * angle in radians,
arc = 6rt3 * 120*pi/180
=4pi*rt3, about 21.766m

Answer 4

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