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The semi-vertical angle of a cone is 45°. If the height of the cone is 20.025, then its approximate lateral surface area, is

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a

401 \sqrt{2} π

b

400 \sqrt{2} π

c

399 \sqrt{2} π

d

none of these

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detailed solution

Correct option is A

Let $r$ be the radius, $h$ be the height and $I$ be the slant

height of a cone of semi-vertical angle 45°. Then,

$r = h$ and $I = \sqrt{2 h}$

Let $h = 20 and h + Δ h = 20.025 \Rightarrow Δ h = 0.025$

Let $S$ be the lateral surface. Then,

$\begin{array}{l} S = \sqrt{2} π h^{2} [∵ S = π r l \Rightarrow S = π h \times \sqrt{2} h] \\ \Rightarrow \frac{d S}{d h} = 2 \sqrt{2} π h \\ \Rightarrow {(\frac{d S}{d h})}_{h = 20} = 40 \sqrt{2} π \\ ∴ Δ S = \frac{d S}{d h} Δ h \Rightarrow Δ S = 40 \sqrt{2} π \times 0.025 = \sqrt{2} π \end{array}$

Also, at $h = 20,$ we have $S = 400 \sqrt{2 π}$

$∴ S + Δ S = 400 \sqrt{2} π + \sqrt{2} π = 401 \sqrt{2} π$

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