A point K is marked in the rhombus ABCD on the side BC such that KC: BK = 3: 1. Find the area ABK

A point K is marked in the rhombus ABCD on the side BC such that KC: BK = 3: 1. Find the area ABK if the area of the rhombus is 48cm ^ 2.

Let the length of the segment VK = X cm, then the length of the segment KS, by condition, is equal to 3 * X cm.

Then the length of the side of the rhombus will be equal to AB = BC = CD = AD = BK + KC = X + 3 * X = 4 * X.

Let the angle ABC = α0.

Then Savsd = AB * BC * Sin α = 4 * X * 4 * X * Sinα = 16 * X2 * Sinα = 48 cm2.

X2 * Sinα = 48/16 = 3.

Then the area of the triangle ABK will be equal to:

Savk = AB * BK / 2 * Sinα = (4 * X * X / 2) * Sinα = 2 * X2 * Sinα.

Then Savk = 2 * 3 = 6 cm2.

Answer: The area of the ABK triangle is 6 cm2.



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