1.

An equilateral triangle if its altitude is 3.2 cm.

Answer»

Solution :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^(@)`


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