Triangle ABC and triangle NBK have a common angle AB = 26 cm, BC = 21 cm, BN = 13 cm, BK = 14 cm.

Triangle ABC and triangle NBK have a common angle AB = 26 cm, BC = 21 cm, BN = 13 cm, BK = 14 cm. Find the area ABC / area NBK by applying the area ratio theorem for triangles with equal angles.

As you know, the area of a triangle ABC = 1/2 * AB * BC * sin (<B), this implies a theorem about the ratio of the areas of triangles with an equal angle: the areas of triangles are related as the product of the sides that make up this angle, respectively.

For our case, the sides of angle B are in the triangle ABC, AB and BC, in the triangle NBK these are the sides BN and BK.

Then the area ABC / area NBK = (1/2) * (AB * BC) / (1/2) * (NB * BK) = (26 * 21) / (13 * 14) = (26/13) * (21/14) = 2 * 3/2 = 3.

Answer: the area ratio is 3.