Volume of cone is given by formula
[tex] V = \frac{\pi r^{2}h}{3} [/tex]-------------------------------------------(1)
We are already given r =2, so we need to find h and then we can find volume
So missing meausre is height here ------------------------------------------------------------------------------
To find h we will use lateral area = [tex] 12\pi [/tex] given
Formula for lateral area of cone is [tex] \pi r\sqrt{r^{2}+ h^{2}} [/tex]
so we have[tex] 12 = \pi r\sqrt{r^{2}+ h^{2}} [/tex]
Now plug r as 2 in this equation
[tex] 12 \pi = \pi (2)\sqrt{2^{2}+ h^{2}} [/tex]
Now solve for h as shown and get h by itself
[tex] \frac{12\pi}{2\pi} = \frac{2\pi\sqrt{4+h^{2}}}{2 \pi} [/tex]
[tex] 6 = \sqrt{4+h^{2}} [/tex]
[tex] 6^{2} = \sqrt{4+h^{2}} ^{2} [/tex]
[tex] 36} = 4+h^{2} [/tex][tex] 36 - 4 = 4+h^{2} -4 [/tex]
[tex] 32 = h^{2} [/tex][tex] \sqrt{32} = \sqrt{h^{2}} [/tex]
[tex] \sqrt{16 \times 2} = h [/tex][tex] \sqrt{16} \times \sqrt{2} = h [/tex]
[tex] 4\sqrt{2} = h [/tex]----------------------------------------------------(2)
Now plug 2 in r place and [tex] 4\sqrt{2} [/tex] in h place in volume formula given in (1)
[tex] V = \frac{\pi r^{2}h}{3} [/tex]
[tex] V = \frac{\pi (2)^{2}4\sqrt{2}}{3} [/tex
][tex] V = \frac{\pi (4)4\sqrt{2}}{3} [/tex]
[tex] V = \frac{16 \pi \sqrt{2}}{3} [/tex]
so choice (1) is right answer
[tex] (1) \frac{16 \sqrt{2}\pi}{3} [/tex]
correct answer --------------------------------------------------------------------------------------------------
[tex] (2) 16 \sqrt{2} \pi [/tex]
incorrect answer as correct volume answer we got as [tex] \frac{16 \sqrt{2}\pi}{3} [/tex] --------------------------------------------------------------------------------------------------
[tex] (3) 4 \sqrt{38} \pi [/tex]
incorrect answer as correct volume answer we got as [tex] \frac{16 \sqrt{2}\pi}{3} [/tex] --------------------------------------------------------------------------------------------------------
