please help Find the total surface area of the figure shown here when D = 50 cm, d = 40 cm, and h = 55 cm.
the given choices : A) 4592 π square centimeters B) 5213 π square centimeters C) 5328 π square centimeters D) 5400 π square centimeters
I used this rule the surface area of cylinder is = \[2\Pi r ^{2} + 2 \Pi r h\]
but the answer was not included in the choices , of course I did something wrong. any help please?
Jody, You can't use that rule because there is a hole in the cylinder, but try to find the circumference on the top outer cylinder circle
ok I'll find it now
I'll tell you why later
so 2 pi r = 2 *3.14 * 25 = 157 cm so this is the circum. of the outer top circle
ok
Ok so let's say that you cut the cylinder clean into a rectangular prism. the length would be 157 cm
ok so L=157 cm
the width [or thickness] is 5 cm, and also I was wrong on the length. you also need the inner circumference, then add 157 to the inner circumference, then divide by 2
ok so for now the W=5 cm
yes
now the length is: [inner cir+outer cir]/2
inner cirum = 2 pi r = 2*3.14*20 = 125.6 after adding it to the 157 cm so the whole circum. will be 282.6 cm
good. now divide by 2 to get 141.3
so 282.6 /2 = 141.3 cm
that is the length
the height is already known
so L= 141.3 cm w = 5 cm
Yep!
yes h = 55 cm L= 141.3 cm W= 5 cm
now calculate [l*w+l*h+h*w]*2
ok
17506
then ?
Hmm Im stumped here. @mathmale Can u help mathmale?
yes this is where I got so confused , I couldn't find the answer in any of the given choices
@mathmale can you please help us here?
ill msg him
I msged him
I see you've already spent a lot of time on this problem, and have explored various means of solving it. Please describe where ... and how ... you'd like for me to begin. Before you do anything else, explain in your own words what the goal of this problem is.
The goal is to find the surface area of a cylinder, shown at the top.
we have the dimensions, but we aren't very sure what went wrong
the way I read this problem, you have a figure, which is not a cylinder, but rather is a solid with a vertical hole through its center. Do y ou agree or disagree?
I agree with you
OK. So let's now list all the surfaces we have: 1) two washers of outside diameter D and inside diameter of d, and ... 2) an outer surface, which, if "unwound" from the solid, would be a rectangle, and ... 3) an inner surface which, if "unwould" ... would also be a rectangle.
Question #1 for you both: Given that the outside diameter is D, what is the area of the outside cylindrical surface?
Well, using 3.14 as Pi, it is 1963.495408
That is with counting the hole's space
Let's please just concentrate on the area of the outside cylindrical surface. What formula are you using to calculate that?
A=Pi*r^2 [area of circle is Pi times the radius squared]
We're not talking about the same thing. By "outside cylindrical surface" I mean ONLY the outside of this cylindrical solid (no area of a circle here, at least, not yet). Hint: What is the circumference of a circle of diameter D?
Ohh... the circumference is 157.0796327
Now we're in agreement. How tall is the figure?
It is 55 cm tall
Right. So, now, what is the area of the outside cylindrical surface (just THAT one surface)? By the way, where is @jody999?
Ok 55*157.1 [rounding] is 8640.5 is the outer area of the figure. P.S. I msg @jody999 so shell come soon
I'm really glad you're interested and involved here...but this was originally Jody's problem. Your result, 8640 square units, is correct for the area of the outside surface of this cylindrical object. Please guide Jody through the process of finding the surface area of the cylindrical hole. (Help her do it herself, referring to your and my discussion.)
Ok, and Thanks!
Message me when you're ready to resume. I may not be available immediately, but if you need help I will provide it ... today. And thank YOU.
Wait.. that's too high... it's not one of her ABC choices
A) 4592 π square centimeters B) 5213 π square centimeters C) 5328 π square centimeters D) 5400 π square centimeters
@Dbzfan836 sorry my laptop battery died
I'm reading the posts now
Jody, even mathmale couldn't find it out.
yes I can see this :( what do you think about this I will email my instructor and ask him for an explanation, right ? so I'll close the question for now and I will reply back when I got something agree?
Yeah, I agree
Gtg
ok thanks a lot for your help with this one , I really appreciate it :) @Dbzfan836
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