# Surface area of a cone

1. Aug 22, 2015

### DGK

1. The problem statement, all variables and given/known data

Determine a simplified, factorised expression, in terms of the radius (r), for the surface area of a cone where diameter (D) = perpendicular height (h)

2. Relevant equations

A = πr (r + √(h^2 + r^2))

3. The attempt at a solution

h=D=2r

A = πr (r + √(2r^2 + r^2))

A/π = r (r + √(2r^2 + r^2))

Last edited: Aug 22, 2015
2. Aug 22, 2015

### andrewkirk

You can factorise the inside of your square root, which will set the scene for further simplification. Ideally you want to get r outside of the square root.

3. Aug 22, 2015

### DGK

A/pi = r (r + sqrt( r (2r + r))

Is this right? I really don't know what I'm doing!!

4. Aug 22, 2015

### DGK

See above for my reply....sorry I'm new to this!!

Last edited: Aug 22, 2015
5. Aug 22, 2015

### Nathanael

You could've factored out another r from the red part. Then use $\sqrt{a^2b}=\sqrt{a^2}\sqrt{b}=a\sqrt{b}$.

Also no need to divide both sides by pi, you want to find A not A/pi.

And what is h2? You seem to say its 2r2.

6. Aug 22, 2015

### andrewkirk

DGK there's a further problem that the formula you give for A in the OP under '2. Relevant Equations' is wrong. Show the working by which you got to that formula, and somebody will show you where you went wrong.

7. Aug 23, 2015

### ehild

You miss the parentheses around 2r.

8. Aug 23, 2015

### ehild

The relevant equation A = πr (r + √(h^2 + r^2)) is correct.

9. Aug 23, 2015

### andrewkirk

It is if you include the base. I assumed he wasn't.

10. Aug 23, 2015

### ehild

Why? It would be the lateral surface area.

11. Aug 23, 2015

### HallsofIvy

Staff Emeritus
Here's another way of looking at it. A cone with base radius r and height h has "slant height" $\sqrt{r^2+ h^2}$ by the Pythagorean theorem. Imagine cutting a slit from the base to the tip of the cone, then flattening it. (A cone is a "developable surface" and can be flattened without warping.)

It will flatten to part of a disk with radius $\sqrt{r^2+ h^2}$. To see what part, look at the two circumferences. Since the cone had base radius r, the circumference of that base is $2\pi r$. A circle with radius $\sqrt{r^2+ h^2}$ has circumference $2\pi\sqrt{r^2+ h^2}$.

[ mod edit ]

Last edited by a moderator: Aug 23, 2015
12. Aug 25, 2015

### John Verghese

Isn't the relevant equation
πr^2 + πrL
L = √(r^2+h^2)
= √(r^2+(2r)^2)
L^2 = r^2+(2r)^2
= (3r)^2
L = √3r

13. Aug 25, 2015

### ehild

It is wrong. Expand (2r)^2. It is not 2r^2!

14. Aug 25, 2015

### John Verghese

Then:
TSA = πr^2 + πr√3r
= πr^2 + √3πr^2
= πr^2(1+√3)

15. Aug 25, 2015

### John Verghese

Ahhh ok thanks heaps (this exact question is in my assignment)
L^2 = r^2 + (2r)^2
= r^2 + 4r^2
= 5r^2
L = √5r
or
L = 5√r
?

16. Aug 25, 2015

### ehild

Which one do you think? How do you apply the square root to a product? What is √(ab)?

17. Aug 25, 2015

### John Verghese

so is it:
L = √5r
or
L = 5√r

18. Aug 25, 2015

### ehild

You have to know. Answer my question: √(ab)=???

19. Aug 25, 2015

### John Verghese

The next part of the question is:
A rectangular prism has:
- Height equivalent to the slant height (L) of the cone in part a
- Length twice the diameter of the cone in part a
- width 5 times the radius of the cone in part a
Determine a simplified factorised expression, in terms of the radius (r), for the volume of the rectangular prism.

All i have so far is:
V = LxWxH
H = √5r or 5√r
L = 2D = 2(2r) = 4r
W = 5r
V = 4r x 5r x ...
and I don't know where to go from there.