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Mathematics 21 Online
OpenStudy (anonymous):

Can Someone Please Help Me with some Geometry??

OpenStudy (anonymous):

For these 3 pictures.. i need to find the surface area

OpenStudy (anonymous):

the first one is composed of 2 triangles: 4x5 + 4x5 = 40m 3 rectangles: 4x14 + 5x14 + \[14\times \sqrt{4^{2} + 5^{2}}\] and you add all of them up

OpenStudy (anonymous):

so the first one would be 89.64?

OpenStudy (radar):

Area of triangle is BH/2, in this figure that equals (4 x 5)/2=20/2=10 sq M, but their are two so the surface area of the triangular portion would be 20 sq meter.

OpenStudy (anonymous):

for the first one i got 2(20)+Ph.. i need to find out what the perimeter is

OpenStudy (radar):

The suface of the sides are as indicated by tacolover. 5x14= 70 sq M for one side. 4x14=56 sqM for another side and 14 x sq rt(16+25) \[\sqrt{41}\times14\]

OpenStudy (anonymous):

im totally lost... the formula i have to use is 2B+Ph

OpenStudy (radar):

Final answer is: 10+10+70+56+89.6=235sq M

OpenStudy (anonymous):

so that would be for the first one?

OpenStudy (radar):

That will work too

OpenStudy (radar):

yes as long as you remember that the B is a triangle and their are two of them.

OpenStudy (radar):

P is perimeter and you must user all three sides.

OpenStudy (radar):

You want to look at the next figure?

OpenStudy (anonymous):

yeah.. so far for the next one i got 2(28)+Ph

OpenStudy (radar):

It is a cylinder or like a can. Too find the total surface area you calculate the area of the ends which are circles (don't forget their are two of them. Then the body is actually a rectangle which has one side equal to the circumference of the circle times the height.

OpenStudy (radar):

Do you agree?

OpenStudy (anonymous):

yeah.. sorry for some reason i put up the wrong picture.. but i need help with that one to

OpenStudy (anonymous):

so id use 2*pi*r^2+2*pi*r*h

OpenStudy (radar):

Do you know how calculate the area of circle whose diameter is 8cm or radius would be 1/2 of that or 4 cm.??

OpenStudy (radar):

The area of the ends would be as you have but it would be like 2(2*pi*r^2). Then the can (not the ends) would be h*C 8*piD=8*pi*8

OpenStudy (radar):

I am sorry I got the area wrong for the ends. The formula is pi*r^2 so the ends would be 2(pi*r^2) that is for both ends

OpenStudy (anonymous):

so for r would it be half the diameter (8).. so 2(3.14*4^2)

OpenStudy (radar):

Yes that gives you the area for the ends of the cylinder. both

OpenStudy (anonymous):

so 100.48

OpenStudy (radar):

Let me check yes for the ends. Now this must be added to the surface area of the body of the can pi*D*h

OpenStudy (anonymous):

so 3.14*8*12=301.44

OpenStudy (radar):

Right now add the end area to your 301.44

OpenStudy (anonymous):

than i would add 100.48+301.44=401.92

OpenStudy (radar):

Yes that would give you the total surface area of the can, I forgot that the heigth was 12 and put 8 but you caught it!

OpenStudy (radar):

and that is in sq cm

OpenStudy (anonymous):

okay

OpenStudy (radar):

Do you need any help on the 3rd figure.

OpenStudy (anonymous):

no but will you check this one for me

OpenStudy (anonymous):

i got 326.25m^2

OpenStudy (radar):

Does it have a bottom and a top? It is hard to tell from the drawing!

OpenStudy (anonymous):

yea

OpenStudy (radar):

I used calculator and got 326.73, so I would say you used the correct procedure and don't forget what the units are.......sq M

OpenStudy (anonymous):

ok thank you

OpenStudy (anonymous):

can you also check this one.. i got 254.34 ft^2

OpenStudy (radar):

I agree with your answer of 254.34 sq ft.

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