Question

In triangle ABC, D is a point on line AB and E is a point on line AC such that DE is parallel to BC. If BC = 20 centimeters, and the area of the trapezoid (trapezium) DBCE is one-fourth the area of triangle ABC, find DE.

Answer 1

Step-by-step Answer:

If trapezium DBCE is one-fourth of the area of the triangle ABC, that means that the area of triange ADE is (1 - 1/4) = three-fourth of ABC.

Since DE is parallel to BC, we can prove that triangles ADE and ABC are similar.

Similar triangles have corresponding sides proportional, and area is proportional to the square of the linear proportions.

From this we can conclude that

(DE/BC)^2 = 3/4

DE/BC = sqrt(3/4) = sqrt(3)/2

DE = sqrt(3)/2 * 20 = 10 sqrt(3) = 17.320508 cm (to 6 decimal places).

On the subject of similar figures/volumes, if we know the ratio of linear dimensions (such as the side of a cube) as k, then the ratio of AREA of similar squares would be k^2.

Example: A square would have a side of 8 (area=8^2=64), and a (similar) square has a side of 10 (area = 10^2= 100).  The

ratio of areas = 64/100 = (8^2/10^2) = (8/10)^2

Here 8/10 is the ratio of linear dimensions = k, and the ratio of areas is k^2 = (8/10)^2 = 64/100.

The same works for cubes (or similar volumes) where the volumes would be proportional to k^3.