Question

how do you find the height and slant height of a cone when you are only given the radius ?

Answer 1

Are you given the volume of the cone, or is what you wrote everything you were given?

Answer 2

Because in order to find the slant height, you need the height. But in order to do that, we need to first find the height of the cone. The problem with that is that you don't have the volume or surface area of a cone, to determine the height.
\[Slant.Height=\sqrt{r^2+h^2}\]

Answer 3

I was given the radius of 5ft and lateral area of 65π ft(squared) , and i have to find the slant height, surface area and volume.

Answer 4

lateral area?

Answer 5

the lateral area is given which is 65π

Answer 6

but how do i find the slant height and height with only having the radius and lateral area ?

Answer 7

And you know
\[s=\sqrt{r^2+h^2}\]

Answer 8

So you can find the height now.

Answer 9

Give it a try for me by substituting s with what I wrote down above and see what you get for the height.

Answer 10

what does the "s" stand for?

Answer 11

slant height.

Answer 12

I already gave you the formula for the slant height, so you subsitute the slant height witht he formula.

Answer 13

substitute* the*

Answer 14

i can only substitute the radius because that's all i have. \[s = \sqrt{5^{^{2 }}} + h ^{2}\]

Answer 15

i don't know what to do next.

Answer 16

How would you find the lateral area of a cone?

Answer 17

@wisejay_ Do you know the lateral formula of a cone? If you do, isn't it this:
\[Lateral.Area=\pi\times r \times s\]

Answer 18

@Azteck but that's the thing, i don't need to find the lateral are nor the radius because they are already given. I'm supposed to find the volume, height, surface area, and slant height of the cone. But i don't know how to do that i don't get it.

Answer 19

That's where you're stumped. You need to work backwars. When you're given the lateral area and the radius, you have to ask yourself that if you got the lateral area, you would obviously need to know the formula for the lateral area. And when they ask to find the height and slant height, most of the time, you would start by finding the height. And when you know that the slant height formula is sqrt(r^2+h^2) Then you can substitute the "s" in the lateral area formula with sqrt(r^2+h^2).

Answer 20

backwards*

Answer 21

Okay, here's the formula for the lateral area of a cone. I will let L stand for the lateral area of the cone.
\[L=\pi \times r \times s\]
Since:
\[s=\sqrt{r^2+h^2}\]
YOu can substitute the "s" in the lateral area formula with the above expression:
\[L=\pi\times r\times \sqrt{r^2+h^2}\]

Answer 22

Now you only have one unknown variable which is h. And h is the height of the cone which is one of the things you need to find.

Answer 23

@wisejay_ You there?

Answer 24

yeah i'm here

Answer 25

@Azteck can you help me substitute

Answer 26

I just did it. Didn't you read?

Answer 27

You know what L is which is 65 and you know what r is which is 5. Now rearrangle the equation to make h^2 the subject.

Answer 28

rearrange*

Answer 29

ohh okay

Answer 30

So after you've done that, could you tell me what the height is.

Answer 31

i got 2404.5

Answer 32

That's way off the mark. Could you show me your working please?

Answer 33

Ugh, i give up. I substituted everything in and i still don't get it

Answer 34

Don't give up. Never give up. Maths is s tep by step process. You don't just go all out in the beginning and have nothing left at the end. You have to take it one step at a time. Not everything at once.

Answer 35

Okay, first:
You have this equation:
\[L=\pi \times r \times \sqrt{r^2+h^2}\]
Substitute your given values and you get:
\[65=5\pi\sqrt{25+h^2}\]

Answer 36

Now you need to make h^2 the subject in order to find h. But h^2 is inside the square root sign, so what should you do to get rid of the square root sign?

Answer 37

do i have to multiply 5pie first ?

Answer 38

Nope, what do you normally do to get rid of square root signs or radicals?

Answer 39

i simplify, by getting two number that multiply to them ?

Answer 40

First, be as simple as you can. This question is just using your basics in primary school. What would you do in this example to get rid of radicals.
\[\sqrt{x}=2\]

Answer 41

sorry for getting distracted

Answer 42

well um find the square root ?

Answer 43

Read what I wrote again. I asked you what you should do to get RID of the square root...

Answer 44

It seems that you don't normally focus on what you need to do, except procrastinating. That's what I used to do and I was academically ranked near last in Year 7-8. Then I realised I was a terrible student who would often procrastinate. Don't follow that route to procrastination. It would lead you to nowhere but downhill...

Answer 45

What grade are you in ? @Azteck

Answer 46

I think we're diverting to another topic, can we continue on this question and actually get though it please?

Answer 47

Okay, sorry.

Answer 48

Um square root both sides ?

Answer 49

so you're saying we need to square root, a square root?

Answer 50

Wouldn't that complicate things even more?
\[\sqrt{\sqrt{x}}=\sqrt{2}\]

Answer 51

LMAO no, i mean square both sides ?

Answer 52

Yes, that's what we should do with your question.
\[65=5\pi\sqrt{r^2+h^2}\]
\[65^2=25\pi^2(r^2+h^2)\]

Answer 53

Now expand/distribute the RHS (Right Hand Side for me please?

Answer 54

(Right Hand Side)*

Answer 55

\[4225=246.14(25+h ^{2}) \]

Answer 56

is that wrong ?

Answer 57

i did it on a calculator

Answer 58

Do you know what Expanding/Distributing is?

Answer 59

Never mind that, you're meant to move the term outside of those brackets to the LHS.
\[\frac{65^2}{25\pi r^2}=25+h^2\]

Answer 60

Now make h^2 the subject. Do you know what that means when I say make h^2 the SUBJECT?

Answer 61

sadly i don't @Azteck

Answer 62

You need to place h^2 on it's own on one side.
If you make x the subject in this example:
\[x+y=2\]
You would move the y to the other side:
\[x=2-y\]
That's making x the subject.

Answer 63

Now make h^2 the subject for me please.