In the trapezoid ABCD BC and AD are 8 and 12 cm. The diagonal AC = 40 and intersects BD

| September 2, 2021 | education

In the trapezoid ABCD BC and AD are 8 and 12 cm. The diagonal AC = 40 and intersects BD at point O. find AO, CO, the ratio of the areas AOD and BOC.

Let us prove the similarity of the triangles BOC and AOD.

The angle BOC is equal to the angle AOD as the vertical angles of intersecting straight lines AC and BD.

The angles BCO and AOD are equal as criss-crossing angles at the intersection of parallel straight lines BC and AD of the secant AC, therefore, the triangles BOC and AOD are similar, according to the first sign of similarity of triangles.

Then AD / BC = AO / CO.

Let AO = X cm, then CO = (40 – X) cm.

A / BC = X / (40 – X).

12 * (40 – X) = 8 * X.

480 = 20 * H.

X = 480/20 = 24 cm.

AO = 24 cm.

CO = 40 – 24 = 16 cm.

The coefficient of similarity of triangles is: K = 1AD / BC = 12/8 = 3/2.

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

Saod / Svos = K2 = (3/2) 2 = 9/4.

Answer: AO = 24 cm, CO = 16 cm, Saod / Svos = 9/4.



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