If a twelve sided regular polygon is inscribed in a circle of radius 3 centimeters, then the length of each side of the polygon is
18 - 9√3
This question asks for the length of a side of a regular twelve-sided polygon (a dodecagon) when it is inscribed in a circle with a radius of 3 centimeters.
We can visualize the regular dodecagon inscribed in the circle. The vertices of the dodecagon lie on the circle. If we connect the center of the circle to each vertex of the dodecagon, we divide the dodecagon into 12 identical isosceles triangles.
The angle at the center of the circle for each of these 12 triangles is the total angle around the center divided by the number of triangles (which is the number of sides of the polygon).
Central Angle \( \theta = \frac{360^\circ}{12} = 30^\circ \)
Now, consider one of the isosceles triangles. We have two sides equal to the radius \( r=3 \) cm, and the angle between them is \( \theta = 30^\circ \). We need to find the length of the base, \( s \).
The Law of Cosines states that for a triangle with sides \( a, b, c \) and the angle \( C \) opposite side \( c \): \( c^2 = a^2 + b^2 - 2ab \cos(C) \).
Applying this to our triangle:
Substituting these values into the formula:
\( s^2 = 3^2 + 3^2 - 2(3)(3) \cos(30^\circ) \)
Let's calculate the value:
\( s^2 = 9 + 9 - 2(9) \cos(30^\circ) \)
\( s^2 = 18 - 18 \cos(30^\circ) \)
We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \).
Substituting the value of \( \cos(30^\circ) \):
\( s^2 = 18 - 18 \left( \frac{\sqrt{3}}{2} \right) \)
\( s^2 = 18 - 9\sqrt{3} \)
So, the square of the side length (\( s^2 \)) is \( 18 - 9\sqrt{3} \) square centimeters.
The actual length of the side \( s \) would be \( \sqrt{18 - 9\sqrt{3}} \) cm.
Looking at the options provided:
The calculated value for \( s^2 \) is \( 18 - 9\sqrt{3} \). This matches Option 2. Although the question asks for the length \( s \), the provided option matches the value of \( s^2 \).
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