If one of the sides of the triangle is 3 cm less than the other, the height divides the third side into

If one of the sides of the triangle is 3 cm less than the other, the height divides the third side into segments of 5 cm and 10 cm, then the perimeter of the triangle is

Consider a triangle ABC.

Let us drop the height BH from the vertex B of triangle ABC to the side AC.

By the condition of the problem, we have that

BC = AB + 3, AH = 5 cm, CH = 10 cm.

We denote the length of AB by X. Then BC = X + 3.

Triangles ABH and BCH are rectangular because BH is the height of triangle ABC.

Therefore, by the Pythagorean theorem, we write down two equations for triangles ABH and BCH:

AB ^ 2 = BH ^ 2 + AH ^ 2, X ^ 2 = BH ^ 2 + 5 ^ 2 = BH ^ 2 + 25,

BC ^ 2 = BH ^ 2 + CH ^ 2, (X + 3) ^ 2 = BH ^ 2 + 10 ^ 2 = BH ^ 2 + 100.

From the first equation we get: BH ^ 2 = X ^ 2 – 25. Then from the second equation we have:

(X + 3) ^ 2 = BH ^ 2 + 100 = X ^ 2 – 25 + 100 = X ^ 2 + 75,

X ^ 2 + 6 * X + 9 = X ^ 2 + 75,

6 * X = 66,

X = 11.

Perimeter P of triangle ABC:

P = AB + BC + AC = X + X + 3 + AH + CH = 2 * X + 3 + 5 + 10 = 2 * X + 18 =

= 2 * 11 + 18 = 40.

Answer: P = 40 cm.