# Math Help - Trigonometric Functions - finding the surface Area(found already) and volume of cone

1. ## Trigonometric Functions - finding the surface Area(found already) and volume of cone

This question is way over my head, its the last question in my txt book for this chapter and the only one I dont fully understand and get.

Q. 10. A sector of a circle with radius 5cm and an angle of PI/3(as in u know the mathematical pi) subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find its exact surface area and volume.

25 PI / 6

(ive figured surface area out already)
my working

ok i know the exact value of the surface area, but how do i get the volume?

the answer in the back for volume is

V = 125 35 PI
----------------
648 cm cubed

the decimal form of this answer is = 3.585245522

(not 35 is only under the square root just to clarify)

so just figuring out the volume of this stupid cut out sector is my problem now : (

thx all help appreciated

2. hi there,

i am not sure what is the answer you wrote there but here is how i will do it:

Given a circle of radius 5 cm and a sector of $\frac{\pi}{3}$ is taken out,

Circumference of the circle without the cut off = $2\pi r$
Circumference of the circle with the cut off = $2\pi r - S$

where,
$S$ = arc length = $r\theta$
$\theta$ in radians

therefore,

Circumference of the circle with the cut off
$=2\pi r - S$

$=2\pi r - r\theta$

$=r(2\pi - \theta)$

$=r(2\pi - \frac{\pi}{3})$

$=r\pi (2 - \frac{1}{3})$

$=r\pi \frac{5}{3}$

this circumference of the circle with the cut off is actually the circumference of the circle of the cone itself:

therefore,

circumference of circle of cone
$=r\pi \frac{5}{3}$

$=2\pi R$

$R = \frac{25}{6}$

where R = radius of the circle of the cone

the height of the cone can be found using Pythagoras theorem: (view attachment)

$r^2 = h^2 + R^2$

$h^2 =r^2- R^2$

$h^2 =5^2- (\frac{25}{6})^2$

$h^2 =\frac{275}{36}$

$h = 2.76385centimeters$

now that you have the radius, R and the height of the cone you can simply substitute them into the volume of a cone formula:

$
Volume, V = \frac{1}{3}\pi R^2 h
$

$
Volume, V = \frac{1}{3}\pi (\frac{25}{6})^2(2.76385)
$

$
Volume, V = 50.2482 cm^3
$

3. thx for the reply but, your answer is incorrect even if i punch into the calculator the answer they gave in a fraction it doesnt equal to what the answer you have given, i also double checked that i wrote the question correctly which it is.

so yea im sorry but your working out wasn't very helpful at all ( i hope im not sounding mean)

hi there,

i am not sure what is the answer you wrote there but here is how i will do it:
it was my working out for the 1st part (which was finding the exact surface area) which didnt come out nicely so i removed it and uploaded a pic of my paper working out.

4. I see you already got the surface area, so I won't bother there. It seems that the answers does not include the base area of the cone for the surface area though.

Surface Area = $\pi rs + \pi r^2$ where r is the radius of the base of the cone, s is the length of the side of the cone.

However, since the base area is not included:

S.A. = $\pi rs = \frac{25\pi}{6}$ , s = 5 in this case, as the radius of the original circle is the side of the cone.
$r = \frac{25\pi}{6(5)\pi}$

= $\frac{5}{6}$

Using the pythagorean theorem, we find h, the height of the cone (refer to Danshader's diagram. Note: he labeled s as r, r as R)

h = $s^2 - r^2$= $5^2 - \left(\frac{5}{6}\right)^2$

= $\sqrt { \frac{875}{36}}$= $\frac{5\sqrt{35}}{6}$

Volume of cone is $\frac{1}{3} \pi r^2 h$

= $\frac{1}{3} \pi \left(\frac{5}{6}\right)^2 \left(\frac{5\sqrt{35}}{6}\right)$

= $\frac{125\sqrt{35}}{648} cm^3$

5. thx , u explained it quite well to me.