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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
A solid cubical block has dimension as shown in the figure and the blo
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17 Dec 2022, 10:42
17 Dec 2022, 10:42
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Question Stats:
75% (02:31) correct
25% (00:28) wrong based on 4 sessions
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A solid cubical block has dimension as shown in the figure and the block is to be cut in half as indicated be the shared region. What is the total surface area of one of the resulting halves of the block?
A. \(27+9 \sqrt 2\)
B. 27
C. \(27+ \sqrt 9\)
D. 28
E. \(29+9 \sqrt 2\)
Show: ::
Attachment:
4587ans A.jpg [ 2.93 KiB | Viewed 886 times ]
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: A solid cubical block has dimension as shown in the figure and the blo
[#permalink]
19 Dec 2022, 09:35
19 Dec 2022, 09:35
Expert Reply
So, the total surface area of a cube is got by adding the individual areas of the 6 faces of the cube. Now, when the cube is cut in half, there are 5 individual areas that are added in order to give us the surface area of the figure formed.
When you add the individual areas you will get the total area, which is \(18 + 9 + 9\sqrt{2} = 27 + 9\sqrt{2}\)
Answer: A
Diagram.png [ 5.14 KiB | Viewed 868 times ]
When you add the individual areas you will get the total area, which is \(18 + 9 + 9\sqrt{2} = 27 + 9\sqrt{2}\)
Answer: A
Show: ::
Attachment:
Diagram.png [ 5.14 KiB | Viewed 868 times ]
gmatclubot
Re: A solid cubical block has dimension as shown in the figure and the blo [#permalink]
19 Dec 2022, 09:35
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