# The height of the right triangle is 10cm. drawn to the hypotenuse is 4 cm, and divides it into segments

**The height of the right triangle is 10cm. drawn to the hypotenuse is 4 cm, and divides it into segments, the difference of which is 6 cm. Find the sides of the triangle.**

Let us prove the similarity of the triangles AСН and BCH.

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

Angle АСВ = 900, then angle ВСН = (90 – (90 – X) = X0.

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

Then AH / CH = CH / BН.

AH * BH = CH * CH = 16 cm.

Let the length of the segment BH = X cm, then AH = (X + 6) cm.

Then (X + 6) * X = 16.

X2 + 6 * X – 16 = 0.

Let’s solve the quadratic equation.

X = BH = 2 cm, then AH = 2 + 6 = 8 cm.

AB = 8 + 2 = 10 cm.

In a right-angled triangle ACН, AC ^ 2 = AH ^ 2 + CH ^ 2 = 64 + 16 = 80.

AC = √80 = 4 * √5 cm.

In a right-angled triangle BCH, BC ^ 2 = BH ^ 2 + CH ^ 2 = 4 + 16 = 20.

AC = √20 = 2 * √5 cm.

Answer: The sides of the triangle are 4 * √5 cm, 2 * √5 cm, 10 cm.