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Mathematics 12 Online
OpenStudy (anonymous):

Someone please help with surface area of pyramids and cones

OpenStudy (anonymous):

are these answers correct? the first one i got 282.6m^2 the second one i got 175.84in^2

OpenStudy (amistre64):

if you can cut the cone up its lengthand make into part of a circle, the surface area is easy to get :) maybe even easy if I think about it some

OpenStudy (amistre64):

is the 10 inches the length up the side of the cone or the height?

OpenStudy (anonymous):

length

OpenStudy (anonymous):

its actually slant height

OpenStudy (amistre64):

then that 10 inch length is the radius of a larger circle, find its circumference" 2pi10 20pi is its curcumference the small base circle is the length of the circumference that you want to find an area for 2pi4 = 8pi make sense?

OpenStudy (anonymous):

kinda

OpenStudy (anonymous):

\[s =\pi \times r ^{2}+\pi \times r \times l\] is the formula i need to use

OpenStudy (amistre64):

length of an arc = theta(r) 8pi = theta(10) theta = 8pi/10 = 4pi/5 area of sector = theta r^2/2 (4pi/5)(10)(10)/2 200/5 = 40pi fpr the surface area of the cone, do we include that base in this or not?

OpenStudy (amistre64):

40pi + 16pi = 56pi if I did it right :)

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

would = 175.84in^2

OpenStudy (amistre64):

175.92 is what I get... is that close enough?

OpenStudy (amistre64):

175.84 if you use (3.14)

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

how would i find the surface area of this one

OpenStudy (amistre64):

if im doing it right, I get 85pi

OpenStudy (anonymous):

would i use half the diameter?

OpenStudy (amistre64):

if you dont have answers to check against, I would suggest gettting a second opinion :) but yes, radius is half the diameter

OpenStudy (anonymous):

\[s =\pi \times 5^{2}+\pi \times 5 \times 12\]

OpenStudy (amistre64):

i got 25pi + 60pi too :)

OpenStudy (amistre64):

im glad the formula works :)

OpenStudy (anonymous):

witch would = 266.9ft^2

OpenStudy (anonymous):

would you please help me with three more

OpenStudy (amistre64):

i could try.... getting kinda deer in the headlights feeling....must.....carry.........on. . . . . . . ok

OpenStudy (anonymous):

find the surface area of the regular pyramid

OpenStudy (amistre64):

#31 3cm is pointing to what?

OpenStudy (anonymous):

3cm is height i believe

OpenStudy (amistre64):

is it face height or pyramid height?

OpenStudy (anonymous):

the formula for this on is S=B+1/2Pl

OpenStudy (anonymous):

looks like pyramid hight

OpenStudy (amistre64):

pyramid height would have a dotted line stricking down the middle of it..... i think its face height.

OpenStudy (anonymous):

so do i

OpenStudy (amistre64):

4cm^2 + 12 cm^2 = 16 cm^2 for #31

OpenStudy (amistre64):

#32...all I see is a 7 and not much else for details...

OpenStudy (anonymous):

thats all there is

OpenStudy (amistre64):

is that all sides are 7? like an equalateral triangle?

OpenStudy (anonymous):

yea

OpenStudy (amistre64):

then I would do: 4 times (1/2)(49)sin(60) 84.87 m^2

OpenStudy (amistre64):

if you have a side-angle-side, the area can be determined by: (side)(side)(sin(angle))/2 and there are 4 sides so I timesed it by 4 :)

OpenStudy (amistre64):

(side)sin(a) = height, and the other side = base

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

and for the last one its just like the first except with decimals..

OpenStudy (amistre64):

(9.2)^2 + 4(9.2)(13)/2

OpenStudy (amistre64):

84.64 + 239.2 = 323.84

OpenStudy (anonymous):

okay thank you so much

OpenStudy (amistre64):

youre welcome :)

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