In an isosceles triangle, the side is 5 mm and the base is 8 mm. Find the height dropped to the base.

Construct an isosceles triangle ABC, in which AB = BC, AC is the base.

Drop the height BH from the vertex B.

In an isosceles triangle, the height drawn to the base is the median of the base, that is, it divides it in half. This means that AH = HC = AC: 2. Since, according to the condition of the problem, the base is AC = 8 mm, then AH = HC = 8: 2 = 4 mm.

Consider a triangle ABH. It is rectangular since the angle is AHB = 90 °. Then, by the Pythagorean theorem, BH2 = AB ^ 2 – AH ^ 2, BH = √ (AB ^ 2 – AH ^ 2).

BH = √ (5 ^ 2 – 4 ^ 2) = √ (25 – 16) = √9 = 3 mm.

Answer: 3 mm.