Question

The semivertical 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

• A. $401 \sqrt{2} \, \pi$
• B. $400 \sqrt{2} \, \pi$
• C. $402 \sqrt{2} \, \pi$
• D. $405 \sqrt{2} \, \pi$

Answer 1

Answer: (A)

Given, semi vertical angle $\theta=45^{\circ}$

In $\Delta A O B$,
$ \tan 45^{\circ}=\frac{r}{h} $
$1=\frac{r}{h}$
$ \Rightarrow r=h $
and $l^{2}=r^{2}+h^{2}$
$l=\sqrt{h^{2}+h^{2}}\,\,\,[\because r=h]$
$l=\sqrt{2 h^{2}}=\sqrt{2} \,h$
Lateral surface area
$(S)=\pi r l $
$=\pi h \cdot \sqrt{2}\, h $
$ S =\sqrt{2}\, \pi h^{2} \,\,\,[\because h=20.025\, cm ]$
$=\sqrt{2}\, \pi(20.025)^{2} $
$=\sqrt{2} \,\pi \times 401$
$S =40 1 \sqrt{2}\, \pi$