Chords AB and CD are drawn in a circle; they intersect at point K, DK = 8 cm, BK = 12 cm.

Chords AB and CD are drawn in a circle; they intersect at point K, DK = 8 cm, BK = 12 cm. The area of triangle AKD is 24 cm. Find the area of triangle CBK

In triangles AKD and СBK, the angle AKD = BKC as vertical angles at the intersection of straight lines AB and CD. The ADK angle is equal to the СBK angle as the angles based on the same arc AC.

Then triangles AKD and СBK are similar in two angles.

Let’s determine the coefficient of similarity of triangles. K = DK / BK = 8/12 = 2/3.

The areas of similar triangles are referred to as the square of the similarity coefficient.

Sacd / Ssvk = (2/3) 2 = 4/9.

Ssvk = 9 * Sacd / 4 = 9 * 24/4 = 54 cm2.

Answer: The area of the triangle AKD is 54 cm2.