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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