Question

Suppose given △ABC with a line segment DE connecting sides AB and BC so that DE ∥ AC . The lengths of the sides of △DBE are three times shorter than the lengths of the sides of △ABC. What is the area of trapezoid ADEC if the area of △ABC is 27 cm^2?

Answer 1

The area of trapezoid ADEC is 24 cm^2, determined by subtracting the area of △DBE from the area of △ABC.

Step-by-step explanation:

To find the area of trapezoid ADEC when △ABC has an area of 27 cm^2 and knowing DE ∥ AC and that △DBE has sides three times shorter than △ABC, we can apply the properties of similar triangles and the fact that parallel lines cut by transversals produce proportional segments.

Because DE ∥ AC and △DBE is similar to △ABC, the height of △DBE will also be three times shorter than that of △ABC, leading to the formula for the area of a triangle (1/2 × base × height), allowing us to conclude that the area of △DBE is 1/9 of the area of △ABC, which is 27 cm2. Therefore, the area of △DBE is 27 cm2 / 9 = 3 cm^2.

Since trapezoid ADEC is composed of △ABC minus △DBE, its area would be 27 cm^2 - 3 cm^2 = 24 cm^2.

Answer 2

24 cm^2

Step-by-step explanation:

See the attached diagram

ABC is a triangle, D, E are points on AB and BC. DE is parallel to BC>

Also DBE has sides 1/3 times that of ABC

We have triangles ABC and DBE are similar with proportionality of sides as 1:3

Hence area of triangle DBE = 1/9 times area of ABC = 3 sq cm.

Area of trapezium ADEC = Area of triangle ABC - area of triangle DBE

=27-3 = 24 cm^2

Answer is 24 cm^2