Polygons and Quadrilaterals

Sample Problems:

Problem: ABCD is a parallelogram and FBED is a square. What is the perimeter of ABCD if BD (not shown) equals 8√2 and AB = 10?

Solution: Since FBED is a square, BF = FD and Triangle BFD is a 45-45-90 triangle.

      

      Since Triangle BFD is a 45-45-90 triangle, s = 8. Then AF can be found with the Pythagorean Theorem.

           a² + b² = c² --> a² + 82 = 102 --> a² = 36 --> a = 6 --> AF = 6

      

Perimeter = AB + BC + CD + DA = 10 + 14 +10 + 14 = 48

Problem: A 6 sided regular polygon (hexagon) is inscribed in a circle of radius 10 cm, find the length of one side of the hexagon. 

Solution:

• Angle AOB is given by
  
  angle (AOB) = 360^o / 6 = 60^o
• Since OA = OB = 10 cm, triangle OAB is isosceles which gives
  
  angle (OAB) = angle (OBA)
• So all three angles of the triangle are equal and therefore it is an equilateral triangle. Hence
  
  AB = OA = OB = 10 cm.