In trapezoid ABCD (AD and BC base), the diagonals intersect at points O, AD = 12 cm BC = 4 cm.Find the area of the triangle BOC if the area of the triangle AOD is 45 cm2.
Let us prove that the triangles BOC and AOD are similar.
The angles BOC and AOD are equal as the vertical angles of intersecting straight lines AC and BD.
The angles OAD and OCB are equal as criss-crossing angles at the intersection of parallel straight lines AD and BC of the secant AC.
Therefore, the triangles BОС and AOD are similar in the first attribute, two angles, then the coefficient of similarity of triangles is: AD / BC = 12/4 = 3.
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
Saod / Svos = K ^ 2.
45 / Swax = 9.
Swax = 45/9 = 5 cm2.
Answer: Swax = 5 cm2.
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