If abce is an isosceles trapezoid and acde is a rectangle what is the length of ae when ab equals 10 and ec equals 20?

Answer 1

Draw the isosceles trapezoid ABCE, where the length of the bases AB (on the top) and EC are respectively 10 and 20.

From A and B draw the perpendiculars to the base EC of the trapezoid, and label the point of intersections with F and G). The rectangle ABGF is formed, where the length of FG is 10 (since the two opposite sides of a rectangle are congruent). Then the lengths of EF and GC are 5 (since the trapezoid is isosceles).

Draw the diagonal AC. From C and E draw the parallel lines respectively to AE and AC, and label the intersection point with D. So the rectangle ACDE is formed. Thus, the triangle EAC is a right triangle, where the angle A is 90 degrees (as the angle of a rectangle), and AF is the altitude drawn at the hypotenuse EC. We have a theorem that states: "If an altitude is drawn to the hypotenuse of a right triangle, then each leg is the geometric mean between the hypotenuse and its touching segment on the hypotenuse". So we have,

EC/AE = AE/EF, which yields AE2 = EC*EF, and so AE = √(20*5) = 10.

or using the same diagram,

Since the diagonals of a rectangle bisect each other (let's say they bisect at H), then AH is the median of the right triangle EAC, drawn to the hypotenuse EC, so its length is the half of the hypotenuse. So the length of AH is 10. Since the trapezoid is isosceles, then the diagonal BE also will form a right triangle EBC, where the median BH also it is 10. So the triangle ABH is equilateral, where angle A is 60 degrees, and it is congruent with angle H of the triangle AHE, as two alternate interior angles. Thus, the isosceles triangle AHE is also equilateral, where AE = 10.