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  • Find the altitude of an equilateral triangle of side 8 cm.

Find the altitude of an equilateral triangle of side 8 cm.

 

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Let ABC be an equilateral triangle of side 8 cm

i.e. AB = BC = AC = 8 cm and AD \perp BC

Then D is the mid-point of BC.

\therefore BD=DC=\frac{1}{2}BC=\frac{8}{2}=4cm

Apply Pythagoras theorem in triangle ABD we get

\left ( AB \right )^{2}=\left ( BD \right )^{2}+\left ( AD \right )^{2}

\left ( 8 \right )^{2}=\left ( 4 \right )^{2}+\left ( AD \right )^{2}

64=16+\left ( AD \right )^{2}

64-16=\left ( AD \right )^{2}

48=\left ( AD \right )^{2}

\sqrt{48}=AD

4\sqrt{3}=AD

\text{Hence the altitude of the equilateral triangle is} 4\sqrt{3}cm

Posted by

infoexpert23

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