Find the total area for the regular pyramid. T. A. = \[([][][]+[][]\sqrt{[]})\]

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Find the total area for the regular pyramid. T. A. = \[([][][]+[][]\sqrt{[]})\]

Geometry
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you want the surface area correct?
yes

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|dw:1436465833356:dw|
Do you know how to find the area of the|dw:1436465959427:dw|??
no...
|dw:1436466095899:dw|
x²+6²=10² x=?
4^2
I guess we should start over from when I split the triangle
|dw:1436466297750:dw|
now, you can almost find the area of this small triangle (and then multiply times 2 to get the area of the full triangle), BUT you don't know the hieght.
U can see that the small triangle (formed by splitting the original triangle in halves) is a right triangle, and therefore we can apply a pythegorean theorem (a²+b²=c²) to find the missing height. in this case (as regards to this smaller triangle) we know a leg of 6 units a hypotenuse of 10. Therefore we will set the equation: a² + 6² = 10² (to solve for this missing height (or this missing leg)
can you find this missing side>?
64?
a² + 6² = 10² a² + 36 = 100 a² = 64 good, but that is a² (you found that if you raise this missing side to the second power, then you get 64, BUT, you haven't found the actual side) Take the square root of both sides to find what "a" is going to be.
8
?
yes
the missing height is 8.
|dw:1436467003677:dw|
can you find the area of this triangle? (shaded) |dw:1436467035260:dw|
24?
yes
and the area of the whole triangle would be?
(if a triangle with an area of 24 is only a half of the entire triangle, then the entire triangle is?)
48
yes
What if I told you to find the area of this:|dw:1436467254378:dw|(our 4th face in a pyramid)
I will split this triangle in halves as well.
|dw:1436467297253:dw|
can you find the height using the pythegorian theorem a²+b²=c² ?
ok, you are missing the leg or the hypotenuse here?
Leg?
yes
So, you know one of the legs is 6, and the hypotenuse is 10, but you are missing another leg. Please plug this into a²+b²=c² (note: it doesn't mater which leg is a and which leg is b - the missing or the known leg)
please take a shot to plug in numbers into a²+b²=c² (if you make a mistake it is ok, i will correct)
\[a ^{2}+6^{2}=12^{2} ?\] \[a ^{2}=108\] \[a=6\sqrt{3}\]
yes. a=√108 a=√(36•3) a=6√3 correct
|dw:1436467977322:dw|
This again shows the rule of a 30º-60º-90º, btw .... and in any case, you now have to find the area of 1/2 of this triangle
|dw:1436468071869:dw|
this is a smaller triangle, but a bigger triangle (that is twice greater) would have the area of 6•6√3, without dividing by 2.
So your 4th face is 36√3.
Now, your faces of the pyramid: you have three faces (side faces), of 48 each. AND a base (bottom face) of 36√3. So the total surface area of the pyramid (the sum of all of its face), is?
\[144+36\sqrt{3}\]?
yes
wolfram gives an estimate of 206.35 (or we can estimate √3 using a Taylor polynomial of function f(x)=√x, at a=4.... no I am just kidding.. would be to much....)
you are done, that is your answer.... (for units though, I would add ":square units" or "units²" to denote that we found the area just now - (not volume pressure or other). )
yw
Thank you!!

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