A reservoir in the form of the frustum of a right circular cone contai

1.	A reservoir in the form of the frustum of a right circular cone contains 44 × 107 litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir.(Take π = \(\frac{22}7\))
Answer» Let height and slant height of the frustum be h and l. Radius, r’ = 100 m, r’’ = 50 m Volume of frustum = 44 × 10⁷ litres = 44 × 10⁴ m³ \(\frac{1}3\times\frac{22}7(100^2 + 50^2 + 100\times50)h\) = 44 x 10⁷ ^{h = \(\frac{{44}\times{10}^4 \times{7}\times{3}}{{22} \times{17500}}\) = 24 m} Slant height, l = \(\sqrt{(r' - r")^2 + h^2}\) ⇒ l = \(\sqrt{(100- 50)^2 + 24^2}\) ⇒ l = 55.46 m Lateral surface area of the frustum = π l (r’ + r’’) = \(\frac{22}7\times55.46 \times150\) = 26145.4 m²

A reservoir in the form of the frustum of a right circular cone contains 44 × 107 litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir.(Take π = \(\frac{22}7\))

Answer»

Let height and slant height of the frustum be h and l.

Radius, r’ = 100 m, r’’ = 50 m

Volume of frustum = 44 × 10⁷ litres = 44 × 10⁴ m³

\(\frac{1}3\times\frac{22}7(100^2 + 50^2 + 100\times50)h\) = 44 x 10⁷

^{h = \(\frac{{44}\times{10}^4 \times{7}\times{3}}{{22} \times{17500}}\) = 24 m}

Slant height, l = \(\sqrt{(r' - r")^2 + h^2}\)

⇒ l = \(\sqrt{(100- 50)^2 + 24^2}\)

⇒ l = 55.46 m

Lateral surface area of the frustum = π l (r’ + r’’)

= \(\frac{22}7\times55.46 \times150\)

= 26145.4 m²