In a rectangular trapezoid ABCD, the angle is D = 90. Point K lies on the base of AD so that AK = KD and BK

| February 23, 2021 | education

In a rectangular trapezoid ABCD, the angle is D = 90. Point K lies on the base of AD so that AK = KD and BK is perpendicular to BC, O is the middle of the diagonal BD. Prove: AB: AD = BO: BC. Find: the area of the triangle. ABD, if the area of the pentagon is 30 cm2.

Since, by condition, AK = DK, and ВK is perpendicular to BC, and therefore perpendicular to AD, then ВK is the height and median of triangle ABD, and triangle ABD is isosceles.

Then Svcd = Svka.

In the rectangle DСВК ВD is the diagonal and divides the rectangle into two equal triangles Sdw = Sdkv. Then Strapezium = Savd + Ssvd = 3 * Ssvd.

Svd = 30/3 = 10 cm2, then Svd = 2 * Svd = 2 * 10 = 20 cm2.

In the triangle BOC, CB = DА / 2, ОВ = ВD / 2, the angle CBO = ВDА as criss-crossing, then the triangles BOC and ABC are similar in two proportional sides and the angle between them. Then:

BD / AD = BO / BC, and since BD = AB, then AB / AD = BO / BC, which was required to be proved.

Answer: The area of ​​the triangle ABC is 20 cm2.



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