In an isosceles trapezoid, the midline is 4, the diagonal is 5, find the area.

Draw a segment СР through the vertex С, parallel to the diagonal ВD. Since ВСРD is a parallelogram, then СР = ВD = 5 cm, DP = ВС.

The diagonals of an isosceles trapezoid are equal, then the ACP triangle is isosceles.

The base AP of an isosceles triangle ACP is equal to: AP = (AD + DP) = (AD + BC).

Since the middle line KM = (BC + AD) / 2, then AR = 2 * KM = 8 cm.

The height of the CH is common for the ACP triangle and the trapezium, then Strap = Sav.

The height of CH is also the median of the ACP triangle, then AH = AP / 2 = 8/2 = 4 cm.

Then CH ^ 2 = AC ^ 2 – AH ^ 2 = 25 – 16 = 9.

CH = 3 cm.

Then Strap = AR * CH / 2 = 8 * 3/2 = 12 cm2.

Answer: The area of ​​the trapezoid is 12 cm2.