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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$401 \sqrt{2} \pi$

Vedclass · Answer

(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

Similar Questions

Using differentials,find the approximate value of $\sqrt{0.082}$. Let $f(x) = \sqrt{x}$.

If the error in measuring the side $l$ of an equilateral triangle is $0.01$, then the percentage error in the area of the triangle, in terms of its side $l$ is:

If there is an error of $\pm 0.04 \text{ cm}$ in the measurement of the diameter of a sphere,then the approximate percentage error in its volume,when the radius is $10 \text{ cm}$,is

The approximate value of $\frac{1}{(2.002)^2}$ is

If $y = (1 + \alpha + \alpha^2 + \ldots) e^{nx}$,where $\alpha$ and $n$ are constants,then the relative error in $y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module