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

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

okay

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

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

Wouldn't the slant height be 8?

8 is the altitude or height of the pyramid

Okay so how do I find the slant height?

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

okay

to find L we can construct a right triangle

http://prntscr.com/7qqvbz

okay

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

correct

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

okay

and what do we know about equilateral triangles

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

yes

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

okay

|dw:1436461601747:dw|

the height of this triangle is the leg we need

okay

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

3 : 3 √3 : 6 works

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

Is 3 the height?

3√3 is the height

|dw:1436462137479:dw|

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

almost, now we have to L

oh okay

http://prntscr.com/7qr6c4

We can use pythagorean theorem to find L

8^2 + ( 3 √ 3)^2 = L^2

91?

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

L^2 = 91
L = sqrt(91)

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

I hope that was clear.

Thank you!