The area of the triangle ABC is 28.DE the midline. Find the area of the trapezoid ABDE

Since DE is the middle line, DE is parallel to AC.

Let us prove that triangle ABC is similar to triangle DBE.

Both triangles have a common angle B, the angle BAC = BDE as the corresponding angles at the intersection of parallel lines DE and AC secant AB. Then the triangle ABC is similar to the triangle BDE in two angles.

The middle line of the triangle is equal to half of the parallel base, then the coefficient of similarity is: k = AC / DE = 2.

The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.

Saavs / Svde = 2 ^ 2 = 4.

Svde = Savs / 4 = 28/4 = 7 cm2.

Then the area of ​​the trapezoid is equal to: Sades = Saavs – Stwo = 28 – 7 = 21 cm2.

Answer: The area of ​​the trapezoid is 21 cm2.