The length of an altitude in an equilateral triangle of side ‘a’ cm is______.
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AB = BC = AC = a cm
BD= DC= a /2 cm
AD l_BC
In ∆ ABD , angle d = 90°
by Pythagoras rule
AB2 = BD2 + AD2
AD2 =a2 + AD2
AD =a2 -(a/2)2
AD2 = a2 - a2/4
AD =4a2 -a2 / 4
i.e.AD2= 3a2/4
AD= √3a/2 cm
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