A regular pentagon is a polygon with 5 equal sides and 5 equal interior angles. An inscribed circle fits perfectly inside, touching the midpoint of each side. The center of this circle is the same as the center of the pentagon.
To find the angle subtended by one side at the center, we consider the lines connecting the center to the two vertices forming that side. These lines, along with the side itself, form an isosceles triangle.
The sum of angles around any point (like the center of the pentagon) is always $360^\circ$. A regular pentagon can be divided into 5 identical isosceles triangles, with their vertices meeting at the center.
The angle subtended by each side at the center is equal because the pentagon is regular.
Therefore, the angle subtended by one side of the regular pentagon at the center of the inscribed circle is $72^\circ$.
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.