anonymous
  • anonymous
Megan bought an outdoor wall light fixture. The fixture has semicircular ends at the top and bottom as shown. The side of the fixture that faces the wall is open. What is the exterior surface area of the light fixture? Answer 78π in2 84π in2 108π in2 120π in2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
apoorvk
  • apoorvk
Just need to find the surface area of each semicircle part, and add 'em up. Or rather, two same semicircles make a circle! so, find out the are of a circle which has a diameter of '12in'. Then, add the area of the rectangular part --> 12x14 - right? What do you think now?
anonymous
  • anonymous
i did something way off ... but im not sure what i found the formula for a cylinder and i plugged in my numbers and calculated it and got 753.6 ....

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apoorvk
  • apoorvk
it's not a cylinder! its like a lampshade kinda thing - two semicircular shades, and a rectangular background.
anonymous
  • anonymous
i know but if we added the other half... aha ... ok so what equation do i use?
apoorvk
  • apoorvk
find the area of the rectangular part, then are of each semi-circular shade and add em up!
anonymous
  • anonymous
ok the area of the rectangular part is 168 now i need to find the area of each semi-circular shade ... how do i do that?
anonymous
  • anonymous
sorry ive been working since 6 this morning non-stop... i dont think im trying as hard ... cuz this probably something i should get
apoorvk
  • apoorvk
its okay - i am woozy too :P. area of semi-circle = (1/2) area of circle plus you could think this way - two same semicircles make a circle, so just find the area of the the equivalent circle..
anonymous
  • anonymous
ok well i know the formula for the area of a circle is pi*r2 so since the diamiter is 12 the radius would be 6 plug that in and i got 113.04
anonymous
  • anonymous
uugh but adding those up gives me to big of a number hmmm
apoorvk
  • apoorvk
113 + 168 isn't right?
anonymous
  • anonymous
well it adds up to 281 which isnt one of my options
apoorvk
  • apoorvk
what are your options?
anonymous
  • anonymous
78π in2 84π in2 108π in2 120π in2
apoorvk
  • apoorvk
i am sorry, it's too late here, and i am too woozy - i misunderstood the whole thing - it must be a cylinder itself - i am really sure this time though. @Diyadiya will help you out. Thanks :) and apologies for misguiding - didn't mean it.
Diyadiya
  • Diyadiya
I got one answer :D
anonymous
  • anonymous
it's ok thanks for the help =) and ok =)
apoorvk
  • apoorvk
yeah would be -the last one - just help her understand.
Diyadiya
  • Diyadiya
First find the area of two semicircles area of semicircle = (pi * r^2)/2 =pi*(6)^2/2 =18pi 2 semicircles =18pi+18pi=36pi
Diyadiya
  • Diyadiya
did you understand til this ?
anonymous
  • anonymous
yeahh when we go to the semi circle stuff i think is when i started to get confused
Diyadiya
  • Diyadiya
Now its half of a cylinder
Diyadiya
  • Diyadiya
and i guess the rectangular part is open
Diyadiya
  • Diyadiya
You just have to find the lateral surface area of half of cylinder ( 2*pi*r*h) half cylinder so lateral surface area = 2*pi*r*h/2 =Pi*r*h can you do it?
Diyadiya
  • Diyadiya
@cbrusoe ??
anonymous
  • anonymous
yeahh i can do it thank-you =)
Diyadiya
  • Diyadiya
can you tell me the lateral surface area of it ?
Diyadiya
  • Diyadiya
take your time np
anonymous
  • anonymous
ok when i calculate 2*pi*r*h i got 263.76
Diyadiya
  • Diyadiya
don't take the value of pi .. just keep pi like that since your answer contains pi
Diyadiya
  • Diyadiya
so what will be the lateral surface area ?
anonymous
  • anonymous
ok so 2*r*h and my diamiter is 12 so radius would be 6 and height is 14 so 2*6*14 = 168pi?
Diyadiya
  • Diyadiya
Yes this is the lateral surface area of a cylinder but we have only half of a cylinder so divide it by 2
Diyadiya
  • Diyadiya
and add the area of 2 semicircles
anonymous
  • anonymous
ok so 84 plus the are of 2 semicircles
Diyadiya
  • Diyadiya
Yes
Diyadiya
  • Diyadiya
so what is it ?
anonymous
  • anonymous
120
anonymous
  • anonymous
i may have calcuulated that last part wrong

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