Find the altitude of an equilateral triangle | vedclass

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Question

(B) Let $ABC$ be an equilateral triangle with side length $a = 8 \, cm$.In an equilateral triangle,the altitude $AD$ drawn from vertex $A$ to the base

Vedclass · Accepted Answer

$4 \sqrt{3}$

Vedclass · Answer

(B) Let $ABC$ be an equilateral triangle with side length $a = 8 \, cm$.In an equilateral triangle,the altitude $AD$ drawn from vertex $A$ to the base $BC$ bisects the base.Therefore,$BD = CD = \frac{BC}{2} = \frac{8}{2} = 4 \, cm$.In the right-angled triangle $ABD$,by Pythagoras theorem:$AB^2 = AD^2 + BD^2$$8^2 = AD^2 + 4^2$$64 = AD^2 + 16$$AD^2 = 64 - 16 = 48$$AD = \sqrt{48} = \sqrt{16 	imes 3} = 4 \sqrt{3} \, cm$.Alternatively,the altitude of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} 	imes 	ext{side}$.$h = \frac{\sqrt{3}}{2} 	imes 8 = 4 \sqrt{3} \, cm$.

Find the altitude of an equilateral triangle of side $8 \, cm$. (in $cm$)

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