Question 114

The altitude of an equilateral triangle of side $$\ \frac{2}{\sqrt{3}}\ $$cm is:

Solution

Side of equilateral triangle = AB = BC = CA = $$\frac{2}{\sqrt{3}}$$ cm

Also, BD = $$\frac{BC}{2}=\frac{1}{\sqrt3}$$ cm

$$\therefore$$ $$AD=\sqrt{AB^2-BD^2}$$

= $$\sqrt{\frac{4}{3}-\frac{1}{3}}$$

= $$\sqrt{\frac{3}{3}}=\sqrt1=1$$ cm

=> Ans - (D)

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