7.3


Objectives:


Define and use the formula to get the furface area of a regular pyramid
Define and use the formula for the volume of any pypramid

Definitions:
Pyramid:Is a polyhedron that has a base which is a polygon also with 3 or more lateral faces
Base: Faces that formwhen you traslate one figur through space.
Lateral Face: The images formed when translating a figure through space (in a pyramids case the lateral faces triangles
Vertex of the pyramid: is the point that the lateral faces meet at
Base edge: That edge that the lateral fase has in common with the base
Lateral edge: The intersecont of two of the lateral faces on the pyramid
Altitude: Is a segent that has endpoints on the center of the base that is perpendicular to the base and the other end point on the vertex of the pyramid
Height: THe length of the altitude
Regular Pyramid: A pyramid that has a base that is a regular polygon thats lateral faces are congruent (equal) isosceles triangles
Slant Height: Is the lenght of he altitude of the the lateral faces
Examples of Pyramids:
211448109_cab9cff84e.jpg

Surface area and volume of a Regular Pyramid


The equation for surface area of a regular pyramid is:
S=1/2lp+b
L=The area of the lateral face
As in: say you laid out a net for the above pyramids and the length of each base was 3 and the length of each triangal for the center of the base to the tip was 6! Like this drawing:
To get to what we need from L we must find the area of each of the the lateral faces or triangles.... to do that we must use the area of a Triangle equation which is
for_wiki.JPG
A=1/2BH
So we fill in the equation...
A=1/2(3)(6)
Plug it into a calculator
A=9
now we multipuly that by 4 to get all the areas and get...
9x4=36
So now our first equation looks like this...
S=1/2(36)(p)+B
Now the P
the p= the primeter of the base so...
To get the perimeter of the base we just add all the base edges together (or the base of the triangles)...Using the above picture we can see that the bases edges are all 6..so..
6+6+6+6=24
Equation now....
S=1/2(36)(24)+B
Now to find B..
B is very easy to find it is just the area of the base...
A=BH but sense this is a square the area is B²
so the area would be...
B=6²
B=36
so now our equation will look like this now....
S=1/2(36)(24)+36
plug that into your calculator and get...
S=468 units²

Volume.....


The equation of the volume of a pyramid is...
V=1/3Bh...
With h being the height of the altitude and b being the area of the base
Using the same example as in the surface area problem to start our equation will look like this..
V=1/336h...because....we have already figured out that the area of the base =36
but we only know the slant height and not the height of the pyramid so....we must first start with a picture that look like this
now_too.JPG
Sense we know that by definetion that a altitude is a segment that intersects the base in the center with a perpendicular bisecter we know that it is a right angle or 90* so...now we know what kind of triangle it is....a 30,60,90..which means that we can use the 30,60,90 triangle congruence postulate to find the altitude height..first we must know the postulate draw this figure....
33333.JPG
Sense we know what the hypotonuse is 6(from previous pictures)we know that the shorter leg is
6/2=3 and if we know what the short leg is we can get the long let ...which is ....
3(ratical or sq.root symbol)3
back to the first equation lets fill in what we know
V=36(3(ratical )3
Plug in to your calculator...(the ratical will be 1st) Press 2nd then 2nd) hit the button that says x² and you will have a ratical)
V=187.06units ³


Now you know....
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