Diagonals AC and BD of trapezoid ABCD intersect at point O. The areas of triangles AOD

univerkov | September 23, 2021 | education

Diagonals AC and BD of trapezoid ABCD intersect at point O. The areas of triangles AOD and BOC are 25 cm ^ 2 and 16 cm ^ 2, respectively. Find the area of the trapezoid.

Since the bases of the trapezoid are parallel, the angle ACB = CAD as criss-crossing angles at the intersection of parallel lines AD and BC secant AC.

Angle BOC = AOD as vertical angles at the intersection of diagonals AC and BD.

Then the triangles BOC and AOD are similar in two angles.

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

Saod / Svos = 16/25 = K2.

K = 4/5.

Then BC / AD = 4/5.

The heights of such triangles are: НO / KO = 4/5. Let the height of НK = 9 * X, then HO = 4 * X, KO = 5 * X.

The area of the triangle AOD is equal to: Saod = AD * 5 * X / 2 = 25.

AD = 10 / X.

The area of the triangle BОС is equal to: Sвос = ВС * 4 * X / 2 = 16.

BC = 8 / H.

Then the area of the trapezoid is:

Savsd = (BC + AD) * НK / 2 = (8 / X + 10 / X) * 9 * X / 2 = 81 cm2.

Answer: The area of the trapezoid is 81 cm2.

One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.