An equilateral triangle if its altitude is 3.2 cm.

Explanation

Steps of construction :
1. Draw a line l.
Mark any point D on the line l.
3. At point D, draw `vec(D)X _|_ l` and cut DA = 3.2 cm from `vec(D)X`.
4. At the point A, construct AB and AC which meets the l at points B and C respectively such that
`angle = 30^(@)` and `angleDAC = 30^(@)`
Then, `DeltaABC` is the required equilateral triangle
because `angleABC = 180^(@) - (90^(@) + 30^(@)) = 60^(@)`.
`angleACB = 180^(@) - (90^(@) + 30^(@)) = 60^(@)`.
and `angleBAC = 30^(@) + 30^(@) = 60^(@)`