In an acute-angled triangle ABC, the heights AK and CE = 12cm, BE = 9cm, AK = 10cm are drawn.

In an acute-angled triangle ABC, the heights AK and CE = 12cm, BE = 9cm, AK = 10cm are drawn. Find the area of triangle ABC.

Since CE is the height of triangle ABC, triangle CEB is rectangular, in which, by condition, CE = 12 cm, BE = 9 cm.Then, by the Pythagorean theorem, BC ^ 2 = CE ^ 2 + BE ^ 2 = 144 + 81 = 225 cm.

AC = 15 cm.

Then the area of the triangle ABC will be equal to:

Savs = AK * VS / 2 = 10 * 15/2 = 75 cm2.

Answer: The area of triangle ABC is 75 cm2.