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The base of a right pyramid is an equilateral triangle each side of which is 4 m long. Every slant edge is 5 m long. Find the lateral surface area and volume of the pyramid? |
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Answer» Let ‘h’ be the height of the pyramid and ‘a’ the length of each side of the base of an equilateral triangle. Then, slant edge = \(\sqrt{h^2+\frac{a^2}{3}}\) ⇒ 5 = \(\sqrt{h^2+\frac{16}{3}}\) ⇒ 25 - \(\frac{16}{3}\) =h2 ⇒ h2 = \(\frac{59}{3}\) ⇒ h = \(\sqrt{\frac{59}{3}}\) m Slant height = \(\sqrt{h^2+\frac{a^2}{12}}\) = \(\sqrt{\frac{59}{3}+\frac{16}{12}}\) = \(\sqrt{21}\) m ∴ Lateral surface area = \(\frac{1}{2}\)(Perimeter of base × Slant height) = \(\frac{1}{2}\)(4+ 4+ 4)× \( \sqrt{21}\) m2 = \(6 \sqrt{21} \) m2 Volume of the pyramid = \(\frac{1}{3}\)(Area of base × height) = \(\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 4^2 \times \sqrt{\frac{59}{3}}\) m3 = \(\frac{4\sqrt{59}}{3}\) m3 |
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