The semi-vertical angle of a cone is 45^{circ | vedclass

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(A) In $	riangle AOB$,the semi-vertical angle $	heta = 45^{\circ}$.$	an 45^{\circ} = \frac{r}{h} \implies 1 = \frac{r}{h} \implies r = h$.Using the

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$401 \sqrt{2} \pi$

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(A) In $	riangle AOB$,the semi-vertical angle $	heta = 45^{\circ}$.$	an 45^{\circ} = \frac{r}{h} \implies 1 = \frac{r}{h} \implies r = h$.Using the slant height formula $l = \sqrt{r^2 + h^2}$,we get $l = \sqrt{h^2 + h^2} = \sqrt{2h^2} = h\sqrt{2}$.The lateral surface area $S = \pi r l$.Substituting $r = h$ and $l = h\sqrt{2}$,we get $S = \pi (h)(h\sqrt{2}) = \sqrt{2} \pi h^2$.Given $h = 20.025 \ cm$,we have $h^2 = (20.025)^2 = 401.000625 \approx 401$.Thus,$S \approx \sqrt{2} \pi (401) = 401 \sqrt{2} \pi \ cm^2$.Therefore,option $A$ is correct.

The semi-vertical angle of a cone is $45^{\circ}$. If the height of the cone is $20.025 \ cm$,then the approximate value of its lateral surface area (in sq. $cm$) is

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