ABCE is an isosceles trapezoid and ACDE is a rectangle.AB=10 and EC=20.What is the length of AE.

• ACDE rectangle bisects the right angle BAF and EF=10cm and using the sin45 we get AE=10root2
• 10 years agoHelpfull: Yes(5) No(1)
• Just a possibility
  
  As ABCE is an isosceles trapezoid and EC is twice times AB, we can consider the given trapezoid as half of a regular hexagon(as length of diagonal in regular hexagon is always double the side).
  Thus, sides AB=AE=BC. => AE = 10.
• 10 years agoHelpfull: Yes(5) No(2)
• ab=10,ec=20,be=ec then be=20
  then ae*ae=400-100
  ae=17.5
• 10 years agoHelpfull: Yes(4) No(3)
• draw trapezoid ABCE then draw rectangle ABDE such that common sides/points of rectangle n trapezoid coincide.
  then u ll see that base EC of trapezoid becomes diagonal of rectangle.
  so usin g property:
  diagonals of rectangle bisects each other. u get following
  In Triangle AOE ( O is point of intersection of diagonals)
  AO =OE=10
  Angle AOE =60 degrees
  so Angle A & E = 60
  hence its an equilateral triangle.
  so AE =10
  
  hope it works!!
• 10 years agoHelpfull: Yes(4) No(2)
• in isosceles trapezoid abce ae ..., its also make isosceles tringle with in isosceles trapezoid we know angle aec is 60 degree so perpendicular distance of ato ec line is 5root3. so that ae=10, and ec=20, area of rectangle is 10root3
  
• 10 years agoHelpfull: Yes(1) No(2)
• in isosceles trapezoid abce ae ..., its also make isosceles tringle with in isosceles trapezoid we know angle aec is 60 degree so perpendicular distance of ato ec line is 5root3. so that ae=10, and ec=20, area of rectangle is 10root3
  
• 10 years agoHelpfull: Yes(1) No(1)
• Draw the isosceles trapezoid ABCE, where the length of the sides AB and EC are respectively 10 and 20.
  
  It is given that ACDE is a rectangle (draw the diagonal AC, from C draw a line parallel to the side AE of the trapezoid, from E draw another line parallel to AC, and label the point of intersection with D).
  
  Look at the right triangle EAC. Let the length of EA be x, and the length of AC be y. So we have,
  
  x^2 + y^2 = 20^2 or
  y^2 = 20^2 - x^2.
  
  From A, draw the perpendicular AF to EC. Let the length of AF be z.
  Look at the right triangle EFA, where the length of EF is 5. Here we have,
  
  x^2 = z^2 + 5^2 or
  z^2 = x^2 - 5^2
  
  Look at the right triangle CFA, where the length of CF is 15. Here we have,
  
  y^2 = z^2 + 15^2 or
  z^2 = y^2 - 15^2
  
  So we have:
  
  z^2 = y^2 - 15^2
  y^2 = 20^2 - x^2
  
  z^2 = y^2- 15^2 or
  z^2 = 20^2 - x^2 - 15^2
  
  Also we have:
  
  z^2 = x^2 - 5^2
  z^2 = 20^2 - x^2 - 15^2
  
  Then,
  
  x^2 - 5^2 = 20^2 - x^2 - 15^2
  2x^2 = 20^2 - 15^2 + 5^2
  2x^2 = 400 - 225 + 25
  2x^2 = 200
  x^2 = 100
  x = √100 (ignore the negative sign, because the length is positive)
  x = 10
  
  Thus, the length of AE is 10.
• 8 years agoHelpfull: Yes(1) No(0)