Question

Find the missing measure for a right circular cone given the following information. Find V if r = 2 and L.A. = 12. (16/3)√2 16√2 4√(38)

Answer 1

Volume of cone is given by formula

$V = \frac{\pi r^{2}h}{3}$-------------------------------------------(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 = $12\pi$ given

Formula for lateral area of cone is $\pi r\sqrt{r^{2}+ h^{2}}$

so we have$12 = \pi r\sqrt{r^{2}+ h^{2}}$

Now plug r as 2 in this equation

$12 \pi = \pi (2)\sqrt{2^{2}+ h^{2}}$

Now solve for h as shown and get h by itself

$\frac{12\pi}{2\pi} = \frac{2\pi\sqrt{4+h^{2}}}{2 \pi}$

$6 = \sqrt{4+h^{2}}$

$6^{2} = \sqrt{4+h^{2}} ^{2}$

$36} = 4+h^{2}$$36 - 4 = 4+h^{2} -4$

$32 = h^{2}$$\sqrt{32} = \sqrt{h^{2}}$

$\sqrt{16 \times 2} = h$$\sqrt{16} \times \sqrt{2} = h$

$4\sqrt{2} = h$----------------------------------------------------(2)

Now plug 2 in r place and $4\sqrt{2}$ in h place in volume formula given in (1)

$V = \frac{\pi r^{2}h}{3}$

$V = \frac{\pi (2)^{2}4\sqrt{2}}{3} [/tex][tex] V = \frac{\pi (4)4\sqrt{2}}{3}$

$V = \frac{16 \pi \sqrt{2}}{3}$

so choice (1) is right answer

$(1) \frac{16 \sqrt{2}\pi}{3}$

correct answer --------------------------------------------------------------------------------------------------

$(2) 16 \sqrt{2} \pi$

incorrect answer as correct volume answer we got as $\frac{16 \sqrt{2}\pi}{3}$ --------------------------------------------------------------------------------------------------

$(3) 4 \sqrt{38} \pi$

incorrect answer as correct volume answer we got as $\frac{16 \sqrt{2}\pi}{3}$ --------------------------------------------------------------------------------------------------------

Answer 2

Answer:

The answer is (16/3)√2

Step-by-step explanation:

I found the answer by following Netta00's awesome step by step.