On the BC side of the rectangle ABCD, point K is chosen so that the area of the triangle ABK is 1/3 of the area

On the BC side of the rectangle ABCD, point K is chosen so that the area of the triangle ABK is 1/3 of the area of the entire rectangle. What part of the total area. of the rectangle is the area of the triangle CDK.

We denote the area of ​​the triangle ABK through S1, the area of ​​the triangle DCК through S2, and the area of ​​the triangle AKD through S3.

Then:

S1 = (ВK * AB) / 2.

S2 = (СK * AB) / 2.

S3 = (AD * AB) / 2.

Let’s find the sum S1 + S2.

S1 + S2 = (BK * AB) / 2 + (CK * AB) / 2 = AB / 2 * (BK + CK).

Since ВK + SK = AD, then S1 + S2 = (AD * AB) / 2 = S3.

The sum of the areas S1 + S2 is equal to half the area of ​​the rectangle, since the heights of the triangles AВK and KСD are equal, AB = CD, and the sum of the legs ВK + KС = AD.

S1 + S2 = S / 2.

Since, by hypothesis, S1 = S / 3, then

S / 3 + S2 = S / 2.

S2 = S / 2 – S / 3 = (3 * S – 2 * S) / 6 = S / 6.

Answer: The area of ​​the KСD triangle is one-sixth of the AВСD rectangle.