Find the lateral area the regular pyramid.
L. A. =

- Falling_In_Katt

Find the lateral area the regular pyramid.
L. A. =

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- Falling_In_Katt

##### 1 Attachment

- misssunshinexxoxo

This is a very awesome tutorial http://www.virtualnerd.com/tutorials/?id=PreAlg_11_01_0031

- misssunshinexxoxo

The lateral area of a regular pyramid is 1/2p*l where p is the perimeter of the base and l is the slant height.
The lateral formula for a cone is also 1/2p*l where p is the perimeter (or in this case the circumference) of the base and l is the slant height. It can also be shown as pi*r*l, where r is the radius of the base, or 1/2pi*d*l, where d is the diameter of the radius of the base. However, note that this is only the lateral area formula for a right cone. There can be many formulas for the lateral area of an oblique cone.
And what they mean when they say they want the answer in terms of pi is that they do not want you to do any process to it. They want the pi to be in the answer. For example, instead of multiplying 3*pi, the answer would instead be 3pi. Or instead of adding 3 + pi, the answer would actually be 3 + pi.

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## More answers

- misssunshinexxoxo

If you'd like more help; please let me know. Truly feel that tutorial should shine some light

- Falling_In_Katt

okay

- perl

the base of that pyramid is a hexagon, and a hexagon is made up of 6 equilateral triangles

- perl

If we use this formula 1/2p*L
We know p = 6*6
We have to find the slant height L

- Falling_In_Katt

Wouldn't the slant height be 8?

- perl

8 is the altitude or height of the pyramid

- Falling_In_Katt

Okay so how do I find the slant height?

- perl

First let's identify the slant height L . I drew it in purple
http://prntscr.com/7qqtha

- Falling_In_Katt

okay

- perl

to find L we can construct a right triangle

- perl

http://prntscr.com/7qqvbz

- Falling_In_Katt

okay

- perl

we know the leg 8, that is given. the other leg we don't know

- Falling_In_Katt

correct

- perl

but that leg is the altitude for an equilateral triangle as shown here
http://prntscr.com/7qqwob

- Falling_In_Katt

okay

- perl

and what do we know about equilateral triangles

- perl

an equilateral triangle can be divided into two triangles of 30 60 90 degrees

- Falling_In_Katt

yes

- perl

the ratio of the sides of a 30 60 90 triangle is \( 1 : \sqrt 3 : 2 \)

- Falling_In_Katt

okay

- perl

|dw:1436461601747:dw|

- perl

the height of this triangle is the leg we need

- Falling_In_Katt

okay

- perl

so we need
3 : h : 6
to have same ratio as
1 : sqrt(3) : 2

- perl

3 : 3 √3 : 6 works

- perl

if you divide through by 3, you get
1 : √3 : 2

- Falling_In_Katt

Is 3 the height?

- perl

3√3 is the height

- perl

|dw:1436462137479:dw|

- Falling_In_Katt

So I would do
\[\frac{ 1 }{ 2 }(6*6)(3\sqrt{3})\]

- perl

almost, now we have to L

- Falling_In_Katt

oh okay

- perl

http://prntscr.com/7qr6c4

- perl

We can use pythagorean theorem to find L

- perl

8^2 + ( 3 √ 3)^2 = L^2

- Falling_In_Katt

91?

- perl

correct, that would be L^2
take the square root of that

- perl

L^2 = 91
L = sqrt(91)

- perl

So the lateral area will be
$$\Large \frac{ 1 }{ 2 }(6*6)(\sqrt{91})$$

- perl

I hope that was clear.

- Falling_In_Katt

Thank you!

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