In a right-angled triangle ABC, the angle is C = 90, CD is the height of the triangle, AC = 5, CB = 10 cm.

In a right-angled triangle ABC, the angle is C = 90, CD is the height of the triangle, AC = 5, CB = 10 cm. Find the ratio of the areas of triangles ACD and CDB.

Since CD is the height of triangle ABC, triangles CDB and ACD are rectangular.

Let us prove that triangles CDB and ACD are similar.

Let the angle ABC = X0, then the angle BAC = (90 – X) 0.

In a right-angled triangle ACD, the angle ACD = (90 – (90 – X)) = X0.

Then the angle CBD of the triangle CDB is equal to the angle ACD of the triangle ACD, therefore, the triangles CDB and ACD are similar in acute angle.

Then the coefficient of similarity of the triangle will be: K = AC / BC = 5/10 = 1/2.

Then Sasd / Ssdv = K2 = 1/4.

Answer: The ratio of the areas of triangles ACD and CDB is 1/4.