PPolygon Problems & Solutions
Explore various problems related to polygons with detailed solutions. Learn more about regular polygons.
Problem 1
A regular hexagon is inscribed in a circle of radius 10 cm. Find the length of one side of the hexagon.
Solution:
- Angle \( \angle AOB \) = \( \frac{360^\circ}{6} = 60^\circ \)
- Since \( OA = OB = 10 \) cm, triangle \( OAB \) is isosceles.
- Since \( \triangle OAB \) has equal angles, it is equilateral, so \( AB = 10 \) cm.
Problem 2
A regular pentagon is circumscribed around a circle of radius 6 cm. Find the length of one side of the pentagon (approximate to two decimal places).
Solution:
- Let t be the size of angle AOB, hence: \( t = \frac{360^\circ}{5} = 72^\circ \)
- Triangle MOB is a right triangle; using trigonometry: \( \tan( \frac{t}{2} ) = \frac{MB}{OM} \)
- \( MB = OM \tan( \frac{t}{2} ) = 6 \tan(36^\circ) \)
- Side of pentagon = \( 2 \times MB = 2 \times 6 \times \tan(36^\circ) \approx 8.7 \) cm.
Problem 3
Find the area of a dodecagon with side length 6 mm (approximate to one decimal place).
Solution:
- A dodecagon is a regular polygon with 12 sides and the central angle t opposite one side of the polygon is given by.
- \( t = \frac{360^\circ}{12} = 30^\circ \)
- Using the formula: \( A = \frac{1}{4} n s^2 \cot( \frac{180^\circ}{n} ) \)
- Substituting \( n = 12, s = 6 \), we get \( A \approx 403.1 \) mm².
Problem 4
Show that as the number of sides \( n \) of a polygon inscribed in a circle of radius \( R \) increases, the area of the polygon approaches the area of the circle.
Solution:
- We now use the formula for the area when the side of the regular polygon is known
- Area of the polygon: \( A_n = \frac{1}{2} n R^2 \sin( \frac{2\pi}{n} ) \)
- For large \( n \), \( \sin( \frac{2\pi}{n} ) \approx \frac{2\pi}{n} \), so \( A_n \approx \pi R^2 \), which is the area of the circle.
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