In trapezoid ABCD (BC || AD), point N divides AD in a ratio of 5: 1, counting from vertex A.

| March 8, 2021 | education

In trapezoid ABCD (BC || AD), point N divides AD in a ratio of 5: 1, counting from vertex A. Point K is the intersection of BD and NC. Find the ratio of the area of the triangle ABD to the area of the trapezoid ABCD if BK: KD = 3: 2.

Consider triangles ВСK and DНK in which the angle ВКС = DKН as vertical, and the angle CBK = НDE as crosswise. Then BC / DН = ВK / DК = 3/2.

DН = 2 * BC / 3.

Since AH / DH = 5/1, then AH = 10 * BC / 3.

Then AD = AН + DH = 12 * BC / 3 = 4 * BC.

Let’s draw the height of ВM, which is common for the trapezoid and triangle ABD.

Savsd = (ВС + АD) * ВМ / 2 = (ВС + 4 * ВС) * ВМ / 2 = (5 * ВС) * ВМ / 2 cm2.

Savd = АD * ВМ / 2 = 4 * ВС / 2 cm2.

Then Savd / Saavsd = (4 * ВС / 2) / ((5 * ВС) * ВМ / 2) = 4/5.

Answer: The area ratio is 4/5.



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