In trapezoid ABCD with base BC and AD, the diagonals meet at point K, BK: KD = 2: 5. Find the area of a triangle AKD if the area of a triangle is BKC = 8
Let us prove that triangles ВКС and AKD are similar. In triangles, the angles ВKС and AKD are equal, as are the vertical angles of intersecting straight lines BD and AC. The angles KDA and KBC are also equal, since they are crosswise at the intersection of parallel lines D and BC of the secant BD, therefore the triangles AKD and ВKС are similar in the first sign of similarity.
Since, by condition, BK / KD = 2/5, the coefficient of similarity of triangles is K = 2/5.
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient, then:
Sbkc / Sakd = K ^ 2.
8 / Sackd = (2/5) ^ 2 = 4/25.
Sacd = 8 * 25/4 = 50 cm2.
Answer: The area of the ACD triangle is 50 cm2.
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