A regular hexagon is circumscribed about the ring surrounding the clock face. The diameter of the ring is 10in. Find the perimeter if the clock face.
Please included work/steps to solve. Thanks!
I assume that you want the perimeter of the hexagon, since the perimeter of the circle would be real easy.
From the centre sketch one of the six isosceles triangles that make up the hexagon.
Draw in the height, which would be 5 inches.
the central angle is 60°
The height creates a 30-60-90 triangle, and we can find the shortest side of that right-angled triangle using simple ratios of 1:√3:2
x/5 = 1/√3
x = 5/√3
so each side of the hexagon is 10√3
and the perimeter would be 60/√3 or 20√3 after rationalizing.
To find the perimeter of the clock face, we need to find the length of one side of the regular hexagon.
First, let's draw a diagram to visualize the situation:
Now, we can use the information given to solve the problem.
We know that the diameter of the ring surrounding the clock face is 10 inches. The diameter of a circle is twice the length of its radius, so the radius of the circle is 10/2 = 5 inches.
Since the hexagon is circumscribed about the circle, we can draw lines from the center of the circle to each vertex of the hexagon. This will create 6 congruent radii, dividing the hexagon into 6 equilateral triangles.
An equilateral triangle has all three sides equal in length. Therefore, each side of the hexagon is equal to the radius of the circle.
Therefore, the length of one side of the hexagon is 5 inches.
Since a hexagon has 6 sides, we can calculate the perimeter by multiplying the length of one side by the number of sides:
Perimeter of the clock face = Length of one side x Number of sides
= 5 inches x 6
= 30 inches.
Therefore, the perimeter of the clock face is 30 inches.