Construct an equilateral triangle whose altitude is 4.5cm.
Draw a line PQ.
Take any point L on this line.
Construct perpendicular AL on PQ.
Cut a line segment AD from D equal to 4.5 cm.
Make angles equal to 30° at A on both sides of AD, say ∠CAD and ∠BAD where B and C lie on XY.
Then, ABC is the required triangle.
Since ∠A = 30° + 30° = 60° and AL ⊥ BC,
Therefore, ∆ABC is an equilateral triangle with altitude AL = 4.5 cm.
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