In rectangle ABCD, point M divides side AB in a 2: 1 ratio. And point K is the middle of the BC side.

In rectangle ABCD, point M divides side AB in a 2: 1 ratio. And point K is the middle of the BC side. Find the ratio of the area of the MBK triangle to the area of the ABCD rectangle.

Let the length of the segment BM = X cm, then, by condition, the length of the segment AM = 2 * X cm.

We denote the length of the segment BK = Y cm, and since, by condition, BK = CK, then BC = 2 * Y cm.

The area of the rectangle ABCD will be equal to: Savsd = AB * BC = 3 * X * 2 * Y = 6 * X * Y cm2.

The area of the MBK triangle is equal to: Svmk = BM * BK / 2 = X * Y / 2 cm2.

Then: Svmk / Savsd = (X * Y / 2) / (6 * X * Y) = 1/12.

Answer: The area ratio is 1/12.