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An equilateral triangle, if its altitude is 3.2cm

Answer»

Solution :We know that , in an equilateral triangle all SIDES are equal and all angles are equal i.e.,each angle is of `60^(@)` .
Given , altitudeof an equilateral triangle say ABC is `3.2cm` . To construct the `DeltaABC` use the following steps.
(i) Draw a line PQ.
(ii) Take a point D on PQ and draw a ray `DE bot PQ`.
(iii) Cut the line segment AD of length `3.2cm` from DE.
(iv) Make angles equal to `30^(@)` at A on both sides of AD say `angleCAD and angleBAD` , where B and C LIE on PQ.
(v) Cut the line segment DC from PQ such that DC = AD
(vi) Join AC
Thus , `DeltaABC` is the required triangle.
Justification
Here, `angleA = angleBAD + angleCAD`
= `30^(@) + 30^(@) = 60^(@)`
Also, `AD bot BC`
`:. angle ADB = 90^(@)`
In `DeltaABDangleBAD + angleDBA = 180^(@)` [angle sum property]
`30^(@) + 90^(@) + angleDBA = 180^(@)` [`angleBAD= 30^(@)`,by construction]
`angleDBA = 60^(@)`
Similary,`angleDCA = 60^(@)`
Thus , `angleA = angleB = angleC = 60^(@)`
Hence, `DeltaABC` is an equilateral triangle.


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