Question : The base of a right pyramid is an equilateral triangle, each side of which is 20 cm. Each slant edge is 30 cm. The vertical height (in cm) of the pyramid is:
Option 1: $10 \sqrt{\frac{23}{3}}$
Option 2: $5 \sqrt{3}$
Option 3: $10 \sqrt{3}$
Option 4: $5 \sqrt{\frac{23}{3}}$
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Correct Answer: $10 \sqrt{\frac{23}{3}}$
Solution : The base of the right pyramid is an equilateral triangle, each side, (a) = 20 cm Each slant edge, (E) = 30 cm According to the question Each side of an equilateral triangle = 20 cm We know that, circumradius of the equilateral triangle = $\frac{a}{\sqrt{3}}$ ⇒ AD = $\frac{20}{\sqrt{3}}$ h = $\sqrt{(AE^2 – AD^2)}$ h = $\sqrt{30^2 – (\frac{20}{\sqrt{3}})^2}$ ⇒ h = $\sqrt{900 – (\frac{400}{3})}$ ⇒ h = $\sqrt{\frac{2700-400}{3}}$ ⇒ h = $\sqrt{\frac{2300}{3}}$ ⇒ h = $10\sqrt{\frac{23}{3}}$ $\therefore$ The vertical height of the pyramid is $10\sqrt{\frac{23}{3}}$cm. Hence, the correct answer is $10\sqrt{\frac{23}{3}}$cm.
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