In the trapezoid ABCD AD is a larger base. A straight line is drawn through the vertex B, parallel to CD
In the trapezoid ABCD AD is a larger base. A straight line is drawn through the vertex B, parallel to CD at point E, BC = 7 cm., AE-4, see find: a) the length of the middle line of the trapezoid; b) the perimeter of the trapezoid, if the perimeter of the ABE triangle is 17 cm
The quadrangle BCDE is a parallelogram, since BC is parallel to DE, as the base of a trapezoid, CD is parallel to BE by condition. Then BE = CD, BC = DE = 7 cm.
Determine the length of the larger base AD = AE + DE = 4 + 7 = 11 cm.
By condition, the perimeter of the ABE triangle is 17 cm.
Rave = AB + BE + AE = 17 cm.
(AB + BE) = 17 – AE = 17 – 4 = 13 cm.
Since BE = CD, then (AB + BE) = (AB + CD) = 13 cm.
Then the perimeter of the trapezoid is: Ravsd = (AB + CD) + BC + AD = 13 + 7 + 11 = 31 cm.
Determine the length of the midline of the trapezoid.
KM = (BC + AD) / 2 = (7 + 11) / 2 = 9 cm.
Answer: The perimeter of the trapezoid is 31 cm, the middle line is 9 cm.