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The surface area of a sphere is 4πR2, where R is the radius of the sph

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The surface area of a sphere is 4πR2, where R is the radius of the sph  [#permalink]

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New post 08 Feb 2015, 11:53
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The surface area of a sphere is 4πR2, where R is the radius of the sphere. If the area of the base of a hemisphere is 3, what is the surface area of that hemisphere?
(A) 6/π
(B) 9/π
(C) 6
(D) 9
(E) 12
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Re: The surface area of a sphere is 4πR2, where R is the radius of the sph  [#permalink]

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New post 08 Feb 2015, 12:01
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Given Area of the base of a hemisphere is 3 = PI * R^2
Thus R = Sqrt ( 3 / PI ) .

Surface area of whole sphere = 4*PI*R^2 .
= 4 * PI * 3 / PI
= 12 .
Since the hemisphere is half of a sphere the Surface area of the hemisphere = 12 / 2
= 6 ( curved part , not including the flat rounded base ) .

But the total surface area = 6 + Area of the base of a hemisphere .
= 6 + 3
= 9.
Answer is D !!
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Re: The surface area of a sphere is 4πR2, where R is the radius of the sph  [#permalink]

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New post 08 Feb 2015, 12:03
Stanislau wrote:
The surface area of a sphere is 4πR2, where R is the radius of the sphere. If the area of the base of a hemisphere is 3, what is the surface area of that hemisphere?
(A) 6/π
(B) 9/π
(C) 6
(D) 9
(E) 12


Surface area of the hemisphere is = (2πR2 + πR2 ( area of the base) ) =3πR2

Given

πR2 = 3

therefore surface area of the sphere = 3*3 = 9
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Re: The surface area of a sphere is 4πR2, where R is the radius of the sph  [#permalink]

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New post 17 May 2015, 18:11
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This is an interesting question with a nice trap answer built in. The first step is to realize that the area of the base of the hemisphere can be represented by the area of a circle formula: A = pie*r^2. Because this equals 3, we know that pie*r^2 = 3. It is tempting at this point to then try to solve for r, but if you look at the area of the sphere formula you will see that it is better to leave it in pie*r^2 form. Always check before doing math to make sure that it is an efficient approach, and here solving for r will only waste time.

Because the area of a sphere is 4*pie*r^2, we know that the area of the outer portion of the hemisphere will be 2*pie*r^2, or 6. This 6 is the trap answer choice, because you cannot forget the base of the sphere (the 3) on the bottom. You can visualize this by thinking about an orange. The entire surface area of this orange is 12, but if you cut it in half and look at just a half of it, you have to take the bottom part (the center of the orange) into account, not just the peel around that half. Therefore 6+3 = 9, giving us answer choice D.

I hope this helps!
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The surface area of a sphere is 4πR2, where R is the radius of the sph  [#permalink]

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New post 16 Jul 2016, 23:56
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StanS wrote:
The surface area of a sphere is 4πR2, where R is the radius of the sphere. If the area of the base of a hemisphere is 3, what is the surface area of that hemisphere?
(A) 6/π
(B) 9/π
(C) 6
(D) 9
(E) 12


It is slightly tricky question, if there is a concentration lapse on the part of the test taker.

Surface area of sphere = \(4πr^2\)
Therefore curved surface area of Half of the sphere (Hemisphere)=\(\frac{1}{2}*4πr^2 =2πr^2 .................[eq 1]\)

Now the area of the base of the hemisphere is 3
The base of the hemisphere is a circle
therefore
\(πr^2=3\)
\(r^2=\frac{3}{π}\)

Put this value of \(r^2=\frac{3}{π}\)in [eq 1] to find the surface of hemisphere
\(2π*\frac{3}{π} = 6\)

Here comes the tricky part. Now since the hemisphere has a curves surface area and ALSO A FLAT surface area at the base; the total surface area will be the sum of these two
Total surface area of the hemisphere will be 6+3=9
Answer is D
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The surface area of a sphere is 4πR2, where R is the radius of the sph  [#permalink]

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New post 20 Jan 2018, 09:38
classic GMAT trap to get into answer 6 ( i got trapped :))

Yes answer should be 9, we have to add base area 3 of the hemisphere to outer surface area of 6 to get 9
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Re: The surface area of a sphere is 4πR2, where R is the radius of the sph  [#permalink]

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Re: The surface area of a sphere is 4πR2, where R is the radius of the sph   [#permalink] 17 Sep 2019, 09:14
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