Please help find surface area
|dw:1399929427740:dw| \[area~ of~ one~ slant ~surface=\frac{ 1 }{ 2 }*15*12=90\] \[total~slant~area=6*90=540 sq. units\] total surface area=Base area+Slant area
what is the base area?
@surjithayer
\[area~ of~ base=6*\frac{ 1 }{ 2 }*15*7.5\sqrt{3}=337.5 *\sqrt{3} sq.units.\]
would my answer be 647 m^2? @surjithayer
sorry i suck at math…..
base area \[=337.5*\sqrt{3}=584.57 sq. units.\]
now add the two to get the total surface area.
1,125 m^2? I think thats right since thats what i got :)
would you please be able to help me with a couple more match questions?
math*
i will try to guide you.
Thank you!! :)) The lateral area of a cone is 574 pi cm^2. The radius is 29 cm. What is the slant height to the nearest tenth of a centimeter?
|dw:1399931753434:dw|
thats a^2+b^2=c^2 right?
do i do 574^2+29^2=c^2?
@whpalmer4
yes, in this case, \[r^2+h^2 = l^2\]
but \[l = 574/29\] not \[l = 574\]
19.8 is my answer?
hold on, I don't think we set this up correctly, let me check it
Can you check that you've stated the problem correctly?
i divided 574 by 29 and i got 19.7 which is basically 19.8
I understand that. However, I don't think you've stated the original problem correctly.
oh sorry i didn't know what you meant. But i will check right now
yep i checked and i wrote the problem correct
Well, something is rotten in Denmark. If the radius is 29 cm, and the lateral area is \(574\pi\) cm^2, that means that \[A= \pi r l = 574\pi = 29*l*\pi\]Solving for \(l\) we get \[l = \frac{574\cancel{\pi}}{29\cancel{\pi}} = 19.79\] However, \[r^2+h^2=l^2\]\[(29)^2+h^2 = l^2\]If this is a right triangle, the hypotenuse (\(l\)) must be the longest side. That is not the case here!
would you like me to tell you all the answer choices i have?
however, if you are missing a decimal point in the radius...
sure
9.9 cm 6.3 cm 19.8 cm 12.6 cm
Yes, there's a mistake in the statement of the problem. \(r\) must be smaller than the slant height!
Oh i guess that's alright if you can't figure it out. But would you be able to help me with some more questions please?
"if I can't figure it out?" No one can figure out that problem as you've stated it; it is an impossibility to have a right triangle where one of the legs is longer than the hypotenuse.
lol sorry i didn't mean it like that…
what is the other problem?
What is the volume of the given prism? Round the answer to the nearest tenth of a centimeter.
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