Approximate the volume of the regular triangular pyramid if the height is approximately 11.5. (Remember that the base of a regular triangular pyramid must be an equilateral triangle, not necessarily congruent to the sides of the pyramid.) 828 239 717

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Approximate the volume of the regular triangular pyramid if the height is approximately 11.5. (Remember that the base of a regular triangular pyramid must be an equilateral triangle, not necessarily congruent to the sides of the pyramid.) 828 239 717

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V = 1/3 (base area)(height) V = 1/3 (base area)(12) I need help finding the base area
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Since the base is an equilateral triangle you can solve for the base area by using the formula A=(bxh)/2
So, A = (12*12)/2 = 72

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V = 1/3 (72)(12) = 288 <------------ But that isn't the correct answer. I got that answer last time I did this problem and it wasn't correct. I was told that I needed to use Hero's formula but I don't know how or why?!
SA: 1/2bh(number of sides) SA: 1/2(12)(12)(4) SA: 1/2(144)(4) SA: 72*4 SA: 288
That solves for the Surface area not the volume. @NathalyN
Hero's formula is needed to solve this problem but I don't know how.
@jim_thompson5910 Can you please help solve this problem?
Here is a link to how to use that formula: https://www.mathsisfun.com/geometry/herons-formula.html I've never heard of this before but I hope it helps
`Remember that the base of a regular triangular pyramid must be an equilateral triangle` the length of the bottom side is 12, so the other 2 sides must also be 12 too plug s = 12 into the area of an equilateral triangle formula \[\Large A = \frac{\sqrt{3}}{4}*s^2\]
Okay so, \[\frac{ \sqrt{3} }{ 4 }*12^2\] = 62.4
62.3538290724796 which turns into 62.4 if you round to 1 decimal place looks good
that's the area of the base
Volume of pyramid = (1/3)*(area of base)*(height of pyramid)
So, (1/3) * 62.4 * 12 = 249.6 <--------- that isn't one of the answer options. It is closest to 239. Also, how did you find that formula for an equalateral triangle. Also, can I do this problem using Hero's formula because that is what I was told to use to get the correct answer. @jim_thompson5910
You can use hero's formula, but you'd get the same result if you used the formula I posted above this page shows how the formula is derived http://mathcentral.uregina.ca/QQ/database/QQ.09.03/jared1.html
also why did you use 12 for the height, when the height is 11.5?
Okay, thank you. Sorry, I didn't mean to use 12. I recalculated it and it and got 239. Thank you so much for your help! :)
Do you mind helping me with one more problem using the same image? Calculate the surface area of a regular triangular pyramid with the base edges of length 12 and a slant height of 12. (Remember that the base of a regular triangular pyramid must be an equilateral triangle, not necessarily congruent to the sides of the pyramid.)
SA = Base area + 1/2 bln
SA = 62.4 + 1/2 (12)(12)(4)
you have one base face and 3 lateral faces
we already have the area of the base face from last time the area of each face is (1/2)*b*h b = base of lateral face triangle = 12 h = height of lateral face triangle = 12
the area of each lateral face is (1/2)*b*h b = base of lateral face triangle = 12 h = height of lateral face triangle = 12
Wait, I thought it had 4 sides not including the base. It is really hard to tell from the picture... So, it is SA = 62.4 + (1/2)(12)(12)(3)
So the answer is 278.4?
Rule for a pyramid (any pyramid) there is always one base face (usually on the ground) the lateral faces are triangles and there are n of them where the base is an n-gon (example: n = 4, square pyramid has 4 lateral faces)
yeah approximately
Okay, so if it doesn't say that it is a square pyramid then it has 3 lateral faces?
the key word "triangular" in "triangular pyramid" means the base is a triangle
if they said "rectangular pyramid", then the base is a rectangle (4 lateral sides)
if they said "pentagonal pyramid" then the base is a pentagon with 5 lateral sides
Oh! I see. I was really confused because by the picture it looks like there are four lateral sides. Thank you so much for your help! You clear everything up brilliantly! :D
yeah it's hard to convey 3D shapes on a flat 2D page
if you had these (maybe from a kit or something) http://bit-player.org/wp-content/uploads/2012/10/tetrahedron-geomags-06601.jpg then it might be easier to visualize
I do have something similar to those. Thanks!
no problem

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