In trapezoid ABCD, the bases of AD and BC are equal to 15 cm and 5 cm, respectively, o – the point

In trapezoid ABCD, the bases of AD and BC are equal to 15 cm and 5 cm, respectively, o – the point of intersection of the diagonals of the trapezoid. find the ratio of the areas of the triangles BOC and DOA.

Since ABCD is a trapezoid, AD is parallel to BC.

The BOC triangle is similar to the DOA triangle in the first similarity feature:

The BOC is equal to the DOA since they are vertical.
The angle BCO is equal to the angle OAD, since they are crosswise with parallel AD and BC and secant CA.
Let’s describe the relationship of similar sides:

BC / AD = OC / OA = BO / OD = K;

The area ratio of similar triangles is K2;

SBOC / SDOA = K ^ 2;

BC / AD = 5/15;

SBOC / SDOA = (5/15) 2 = 52/152 = 25/225 = 1/9;

Answer: 1/9