In an isosceles triangle, the side is 17 cm, the base is 16 cm. Find the height.

Consider an isosceles triangle ABC with base AC = 16 and sides AB = BC = 17.

Drop the height BH from the vertex of triangle B.

Since triangle ABC is isosceles, the height BH is also the median.

Hence,

AH = CH = AC / 2 = 16/2 = 8.

Consider a right-angled triangle ABH. By the Pythagorean theorem we get:

BH ^ 2 = AB ^ 2 – AH ^ 2 = 17 ^ 2 – 8 ^ 2 = 9 * 25,

BH = 3 * 5 = 15.

Area S of triangle ABC:

S = 1/2 * AC * BH = 1/2 * 16 * 15 = 120.

Therefore, we can find the heights AH1 and BH2 on the sides:

S = 1/2 * AH1 * BC,

AH1 = 2 * S / BC = 2 * 120/17 = 240/17.

BH1 = AH1 = 240/17.