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(B) Given the semi-vertical angle $\alpha = 45^{\circ}$ and the radius of the base $r = 14 	ext{ cm}$.The slant height $l$ is given by $l = \frac{r}{

Vedclass · Accepted Answer

$8$

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(B) Given the semi-vertical angle $\alpha = 45^{\circ}$ and the radius of the base $r = 14 	ext{ cm}$.The slant height $l$ is given by $l = \frac{r}{\sin \alpha} = \frac{r}{\sin 45^{\circ}} = r\sqrt{2}$.The total surface area $A$ of the cone is given by $A = \pi r(r + l)$.Substituting $l = r\sqrt{2}$,we get $A = \pi r(r + r\sqrt{2}) = \pi r^2(1 + \sqrt{2})$.To find the approximate error in $A$,we differentiate $A$ with respect to $r$:$\frac{dA}{dr} = 2\pi r(1 + \sqrt{2})$.The approximate error $dA$ is given by $dA = \frac{dA}{dr} \cdot dr$,where $dr = \frac{\sqrt{2}-1}{11} 	ext{ cm}$.$dA = 2\pi r(1 + \sqrt{2}) \cdot \left(\frac{\sqrt{2}-1}{11}ight)$.Substituting $r = 14$:$dA = 2\pi(14)(1 + \sqrt{2}) \cdot \left(\frac{\sqrt{2}-1}{11}ight)$.Since $(1 + \sqrt{2})(\sqrt{2} - 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1$,we have:$dA = 2\pi(14) \cdot \frac{1}{11} = \frac{28\pi}{11}$.Wait,re-evaluating the expression: $2\pi(14)(1) / 11 = 28\pi/11 \approx 8$. Given the options,the calculation assumes $\pi$ is approximated or cancelled in the context of the problem's specific numerical structure. Re-checking: $2 	imes 14 	imes (1) / 11$ is not $8$. Let's re-examine the area formula: $A = \pi r^2 + \pi r l = \pi r^2 + \pi r (r\sqrt{2}) = \pi r^2(1+\sqrt{2})$. $dA = 2\pi r(1+\sqrt{2})dr$. With $r=14$ and $dr = (\sqrt{2}-1)/11$,$dA = 2\pi(14)(1)(1/11) = 28\pi/11$. If we take $\pi \approx 22/7$,then $dA = 28 	imes (22/7) / 11 = 4 	imes 2 = 8$. Thus,the error is $8 	ext{ sq. cm}$.

The semi-vertical angle of a right circular cone is $45^{\circ}$. If the radius of the base of the cone is measured as $14 \text{ cm}$ with an error of $\left(\frac{\sqrt{2}-1}{11}\right) \text{ cm}$,then the approximate error in measuring its total surface area is (in $\text{sq. cm}$):

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