Question

PLEASE HELP FIND THE LATERAL AREA https://media.glynlyon.com/g_geo_2012/8/groupi78.gif

Answer 1

let's take a peek at the bottom of the pyramid, it looks like

Answer 2

using pythagorean theorem, can you get the "green" line there?

Answer 3

b^2 is 27

Answer 4

right and \(\bf \sqrt{27} \implies 3 \sqrt{3} = \text{green line}\)

Answer 5

now lemme stand up the pyramid, keeping in mind that line from the center of the hexagonal bottom, the "apothem" , is \(\bf 3 \sqrt{3}\)

Answer 6

same thing, using pythagorean theorem, what would be the missing line of the yellow triangle there?
namely the so-called "slant height"

Answer 7

so wait, would the \[3\sqrt{3}\] squared be \[9\sqrt{3}\] ??

Answer 8

\(\bf (3 \sqrt{3})^2 \implies 9(\sqrt{3^2}) \implies 9 \times 3\)

Answer 9

gotcha!! haha ok, so C^2 would be 91

Answer 10

yes and that gives us the "nose of the triangular face of the pyramid"
the triangles are the the sides or laterals of the pyramid, their area summed up will be the Lateral Area of the pyramid

\(\bf c^2 = 8^2 + (3 \sqrt{3})^2 \implies c = \sqrt{64 + 9(\sqrt{3^2})} \implies c= \sqrt{91}\\
\text{area of a triangle } = \cfrac{1}{2} \times base \times height\\
\text{base = 6, height = } \sqrt{91}\\
\cfrac{1}{2} \times 6 \times \sqrt{91}\)

Answer 11

you find that Area, that'd be the area of one of the triangular faces
you have 6 faces in a hexagon, you can just multiply that amount by 6 and that's the Lateral Area of the pyramid

Answer 12

THANK YOU SO MUCH!!!

Answer 13

yw

Answer 14

could you also help me find the volume of this??? https://media.glynlyon.com/g_geo_2012/8/groupi75.gif

Answer 15

the procedure is about the same, if we use the bottom of the pyramid and the missing side from

Answer 16

without having to look much, using the 30-60-90 rule, the ratio of the missing side there is just the smaller side times \(\sqrt{2}\)

Answer 17

or \(\bf 2\sqrt{2}\)

then we stand up the pyramid and we use the same yellow triangle as before
going from the apothem upwards and outwards

this time you'll notice the missing side is the height of the pyramid, lemme sketch it quick

Answer 18

acck. one sec.... I meant to say the ratio is the smaller times \(\bf \sqrt{3}\)

Answer 19

so our green line will be \(\bf 2\sqrt{3}\)

Answer 20

so do I do Pythagorean theorem?

Answer 21

yes \(\bf c^2 = a^2 + b^2 \implies c^2 - a^2 = b^2 \implies
\sqrt{c^2 - a^2 }= b^2\)

Answer 22

in this case you have the hypotenuse and the adjacent side, so the height will be the opposite side

Answer 23

you must be wondering, so who cares about the height? heheh
well, volume of a pyramid is \(\bf \cfrac{1}{3} (\text{hexagonal bottom})(\text{height})\)

Answer 24

so would it be 6^2 = \[2\sqrt{3}\] + b^2 ???

Answer 25

\(\bf \sqrt{6^2 - (2\sqrt{3})^2 }= b^2 \implies \sqrt{24} \implies 2\sqrt{6}\)

Answer 26

so that's the height, so now let's find the hexagonal bottom area
well, it's a regular polygon, a hexagon
area of a regular polygon = \(\bf \cfrac{1}{2}(apothem)(perimeter)\\
\cfrac{1}{2}(2\sqrt{3})(\text{6 sides 4 units long each})\\
\cfrac{1}{2}(2\sqrt{3})(4 \times 6)\)

Answer 27

so \[24\sqrt{3}\] cut in half, so \[12\sqrt{3}\] ??
sorry if its wrong, I moved before I ever learned square roots and the place I moved to already learned them so I suck at them

Answer 28

\(\bf \cfrac{1}{\cancel{2}}(\cancel{2}\sqrt{3})(4 \times 6) \implies 24 \sqrt{3}\)

Answer 29

gotcha! thanks so much!!

Answer 30

\(\bf \cfrac{1}{3} (\text{hexagonal bottom})(\text{height})\\
\cfrac{1}{3}24\sqrt{3} \times 2\sqrt{6} \implies
\cfrac{1}{3} 48 \sqrt{18} \implies \cfrac{48}{3} 3\sqrt{2}\\
\cfrac{48}{\cancel{3}} \cancel{3}\sqrt{2}\)

Answer 31

yw