The height of the right-angled triangle ABC, drawn from the vertex of the right angle, divides the hypotenuse

The height of the right-angled triangle ABC, drawn from the vertex of the right angle, divides the hypotenuse into segments equal to 4 cm and 9 cm. Find the area of this triangle.

Let the value of the angle DАС of the triangle ABC be equal to X0, then the angle АСН = (90 – X) 0.

Angle АСВ = 90, then angle ВСН = (90 – (90 – X) = X0.

The acute angles of the right-angled triangles ACD and BCD are equal, then the triangles are similar in acute angle.

Then in similar triangles AD / CD = CD / BD.

CD ^ 2 = AD * BD = 9 * 4 = 36.

CD = 6 cm.

The length of the hypotenuse AB = AD + BD = 9 + 4 = 13 cm.

Determine the area of the triangle ABC.

Savs = AB * CD / 2 = 13 * 6/2 = 39 cm2.

Answer: The area of the triangle is 39 cm2.