complete the activity.
To find the length of the side and perimeter of an equilateral traingle whose height √ 3cm .
solution: Let ∆ABc be the given equilateral traingle Therefore L B = 60°
AD! BC , B-D-C
In ∆ ABC LB = 60° , L ADB = 90°
L BAD =
∆ ABD is a 30°- 60° - 90°
√3/2 AB _
AB =
side of equilateral traingle =
perimeter of ∆ ABC = 3×AB = cm.
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3 sqrt 2
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