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A solid cubical block has dimension as shown in the figure and the blo
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02 Sep 2015, 05:45
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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. \(29+9 \sqrt 2\) Attachment:
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Re: A solid cubical block has dimension as shown in the figure and the blo
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02 Sep 2015, 06:54
anik19890 wrote: 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 root 2 b. 27 c. 27+ root 9 d. 28 e. 29+9 root 2
i can not understand this problem. please help me someone What question # is this from the official guide. Which official guide is this from? Official guide or the quant review? Please mention the complete thing.You need to find the total area of the 2 halves created by the black plane added to the figure. Doing this, you will get 6 distinct regions that will be the 1. rectangle shaded in black with dimensions 3 \(\sqrt(2)\) * 3 2. 2 slanted rectangles with dimensions 3*3 3. 2 Triangles, 1 at the top and 1 at the bottom with area : 0.5*3*3 Thus the total surface area of the 1 of the 2 halves created: 3 \(\sqrt(2)\) * 3 + 2*0.5*3*3 + 2*3*3 = 9+18+9 \(\sqrt(2)\) = 27+ 9 \(\sqrt(2)\) . A is the correct answer.



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Re: A solid cubical block has dimension as shown in the figure and the blo
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02 Sep 2015, 20:02
Engr2012 wrote: anik19890 wrote: 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 root 2 b. 27 c. 27+ root 9 d. 28 e. 29+9 root 2
i can not understand this problem. please help me someone What question # is this from the official guide. Which official guide is this from? Official guide or the quant review? Please mention the complete thing.You need to find the total area of the 2 halves created by the black plane added to the figure. Doing this, you will get 6 distinct regions that will be the 1. rectangle shaded in black with dimensions 3 \(\sqrt(2)\) * 3 2. 2 slanted rectangles with dimensions 3*3 3. 2 Triangles, 1 at the top and 1 at the bottom with area : 0.5*3*3 Thus the total surface area of the 1 of the 2 halves created: 3 \(\sqrt(2)\) * 3 + 2*0.5*3*3 + 2*3*3 = 9+18+9 \(\sqrt(2)\) = 27+ 9 \(\sqrt(2)\) . A is the correct answer. thank you very much. but i can not understand ``one of the resulting halves```. what does it mean to the structure?



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Re: A solid cubical block has dimension as shown in the figure and the blo
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02 Sep 2015, 20:13
anik19890 wrote: Engr2012 wrote: anik19890 wrote: 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 root 2 b. 27 c. 27+ root 9 d. 28 e. 29+9 root 2
i can not understand this problem. please help me someone What question # is this from the official guide. Which official guide is this from? Official guide or the quant review? Please mention the complete thing.You need to find the total area of the 2 halves created by the black plane added to the figure. Doing this, you will get 6 distinct regions that will be the 1. rectangle shaded in black with dimensions 3 \(\sqrt(2)\) * 3 2. 2 slanted rectangles with dimensions 3*3 3. 2 Triangles, 1 at the top and 1 at the bottom with area : 0.5*3*3 Thus the total surface area of the 1 of the 2 halves created: 3 \(\sqrt(2)\) * 3 + 2*0.5*3*3 + 2*3*3 = 9+18+9 \(\sqrt(2)\) = 27+ 9 \(\sqrt(2)\) . A is the correct answer. thank you very much. but i can not understand ``one of the resulting halves```. what does it mean to the structure? You still have not provided the question number of this question from the official guide. Dividing into halves means that the shaded plane divides the cube into 2 equal shapes/halves having identical properties.



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Re: A solid cubical block has dimension as shown in the figure and the blo
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03 Apr 2016, 23:55
Engr2012 wrote: anik19890 wrote: 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 root 2 b. 27 c. 27+ root 9 d. 28 e. 29+9 root 2
You need to find the total area of the 2 halves created by the black plane added to the figure. Doing this, you will get 6 distinct regions that will be the
1. rectangle shaded in black with dimensions 3 \(\sqrt(2)\) * 3 2. 2 slanted rectangles with dimensions 3*3 3. 2 Triangles, 1 at the top and 1 at the bottom with area : 0.5*3*3
Thus the total surface area of the 1 of the 2 halves created: 3 \(\sqrt(2)\) * 3 + 2*0.5*3*3 + 2*3*3 = 9+18+9 \(\sqrt(2)\) = 27+ 9 \(\sqrt(2)\) . A is the correct answer. May be Typo mistake.The distinct regions will be 5.
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Re: A solid cubical block has dimension as shown in the figure and the blo
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06 Apr 2018, 06:57
Bunuel VeritasPrepKarishma please share your approach. Thanks.



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Re: A solid cubical block has dimension as shown in the figure and the blo
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10 Jul 2018, 10:29
anik19890 wrote: a solid cubial 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 root 2 b. 27 c. 27+ root 9 d. 28 e. 29+9 root 2
i can not understand this problem. please help me someone Hello pushpitkc what is the tempereture under sun today ? i know that Surface Area of Cube is \(6a^2\) so plug in 3 into this formula and get 54. if cube is divided into halves then i get 54/2 = 27 so what am i missing ? thanks



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Re: A solid cubical block has dimension as shown in the figure and the blo
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10 Jul 2018, 21:33
dave13 wrote: anik19890 wrote: a solid cubial 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 root 2 b. 27 c. 27+ root 9 d. 28 e. 29+9 root 2
i can not understand this problem. please help me someone Hello pushpitkc what is the tempereture under sun today ? i know that Surface Area of Cube is \(6a^2\) so plug in 3 into this formula and get 54. if cube is divided into halves then i get 54/2 = 27 so what am i missing ? thanks Hey dave13It has been hot and humid here 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. Attachment:
Diagram.png [ 18.18 KiB  Viewed 1004 times ]
When you add the individual areas you will get the total area, which is \(18 + 9 + 9\sqrt{2} = 27 + 9\sqrt{2}\) Hope this helps you!
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A solid cubical block has dimension as shown in the figure and the blo
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11 Jul 2018, 09:16
pushpitkc wrote: dave13 wrote: anik19890 wrote: a solid cubial 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 root 2 b. 27 c. 27+ root 9 d. 28 e. 29+9 root 2
i can not understand this problem. please help me someone Hello pushpitkc what is the tempereture under sun today ? i know that Surface Area of Cube is \(6a^2\) so plug in 3 into this formula and get 54. if cube is divided into halves then i get 54/2 = 27 so what am i missing ? thanks Hey dave13It has been hot and humid here 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. Attachment: The attachment Diagram.png is no longer available When you add the individual areas you will get the total area, which is \(18 + 9 + 9\sqrt{2} = 27 + 9\sqrt{2}\) Hope this helps you! pushpitkc thanks for explanation but i just dont get why you get area of rectanle \(9\sqrt{2}\) and area of two face 18 what faces ? why you marked you marked some parts in red ? and how do you get right triangle This is how I see when cube is cut into halves. So when it is cut I see two equilateral triangles, two rectangles and bottom surface of square shape which occurred as a result of samurai cutting it into halves
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A solid cubical block has dimension as shown in the figure and the blo
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11 Jul 2018, 18:26
Still not following this. How does one get the dimensions of 3 * 3\sqrt{2} for the area of the shaded black region? @bunel



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A solid cubical block has dimension as shown in the figure and the blo
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11 Jul 2018, 22:02
dave13 wrote: pushpitkc wrote: dave13 wrote: Hello pushpitkc what is the tempereture under sun today ? i know that Surface Area of Cube is \(6a^2\) so plug in 3 into this formula and get 54. if cube is divided into halves then i get 54/2 = 27 so what am i missing ? thanks Hey dave13It has been hot and humid here 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. Attachment: Diagram.png When you add the individual areas you will get the total area, which is \(18 + 9 + 9\sqrt{2} = 27 + 9\sqrt{2}\) Hope this helps you! pushpitkc thanks for explanation but i just dont get why you get area of rectanle \(9\sqrt{2}\) and area of two face 18 what faces ? why you marked you marked some parts in red ? and how do you get right triangle This is how I see when cube is cut into halves. So when it is cut I see two equilateral triangles, two rectangles and bottom surface of square shape which occurred as a result of samurai cutting it into halves Hey dave13One of the faces(flat surface) of the cube is a square. PropertyIf the side of the square is x, the diagonal of the square is \(x\sqrt{2}\) So, in the square where the length of the side is 3, the diagonal is \(3\sqrt{2}\) The triangle has 3 sides  \(3,3,3\sqrt{2}\). This makes the triangle an isosceles right triangle as the sides are in the ratio \(1:1:\sqrt{2}\)(Not iscoceles) Each of the rightangled triangle has the area of \(\frac{1}{2}*\) Product of the legs = \(\frac{1}{2}*3*3 = 4.5\) Therefore, the sum of the areas of both the triangles will be \(4.5*2 = 9\) Hope this helps you! surfingpirate  Hope this explanation clears your confusion as well
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Re: A solid cubical block has dimension as shown in the figure and the blo
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11 Jul 2018, 22:33
anik19890 wrote: 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. \(29+9 \sqrt 2\) Attachment: The attachment 4587ans A.jpg is no longer available We should find the total surface area of the figure below: Notice that it has five faces: Two 3 by 3 squares. Area = 2*3^2 = 18 Two right triangles which make up one 3 by 3 square (purple face + opposite face). Area = 3^2 = 9 One face which is 3 by \(3\sqrt{2}\) (pink face). Area = \(3*3\sqrt{2}=9\sqrt{2}\) The total surface area \(= 18 + 9 + 9\sqrt{2}=27+9\sqrt{2}\). Answer: A. Hope it's clear. Attachment:
bP5XZ.png [ 38.79 KiB  Viewed 889 times ]
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A solid cubical block has dimension as shown in the figure and the blo
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14 Jul 2018, 02:02
Bunuel wrote: anik19890 wrote: 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. \(29+9 \sqrt 2\) We should find the total surface area of the figure below: Notice that it has five faces: Two 3 by 3 squares. Area = 2*3^2 = 18 Two right triangles which make up one 3 by 3 square (purple face + opposite face). Area = 3^2 = 9 One face which is 3 by \(3\sqrt{2}\) (pink face). Area = \(3*3\sqrt{2}=9\sqrt{2}\) The total surface area \(= 18 + 9 + 9\sqrt{2}=27+9\sqrt{2}\). Answer: A. Hope it's clear. Bunuel, pushpitkc, hello there thanks for great explanation:) i have one question Regarding this part One face which is 3 by \(3\sqrt{2}\) (pink face). Area = \(3*3\sqrt{2}=9\sqrt{2}\) So when diagonal divides square into halves we get TWO ISOLESCES RIGHT triangles with BASE 3 and HEIGHT 3. so area one ISOLESCES RIGHT TRIANGLE IS base * heght /2 > 3*3/2 = 4.5 same rule applies to the second ISOLESCES RIGHT TRIANGLE So total area of pink face is \(4.5+4.5 = 9\), so my question if we have found area of pink face why do we attach \(\sqrt{2}\) to \(9\) can you please explain the logic ? thanks and have a good weekend



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Re: A solid cubical block has dimension as shown in the figure and the blo
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14 Jul 2018, 03:51
dave13 wrote: Bunuel wrote: anik19890 wrote: 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. \(29+9 \sqrt 2\) We should find the total surface area of the figure below: Notice that it has five faces: Two 3 by 3 squares. Area = 2*3^2 = 18 Two right triangles which make up one 3 by 3 square (purple face + opposite face). Area = 3^2 = 9 One face which is 3 by \(3\sqrt{2}\) (pink face). Area = \(3*3\sqrt{2}=9\sqrt{2}\) The total surface area \(= 18 + 9 + 9\sqrt{2}=27+9\sqrt{2}\). Answer: A. Hope it's clear. Bunuel, pushpitkc, hello there thanks for great explanation:) i have one question Regarding this part One face which is 3 by \(3\sqrt{2}\) (pink face). Area = \(3*3\sqrt{2}=9\sqrt{2}\) So when diagonal divides square into halves we get TWO ISOLESCES RIGHT triangles with BASE 3 and HEIGHT 3. so area one ISOLESCES RIGHT TRIANGLE IS base * heght /2 > 3*3/2 = 4.5 same rule applies to the second ISOLESCES RIGHT TRIANGLE So total area of pink face is \(4.5+4.5 = 9\), so my question if we have found area of pink face why do we attach \(\sqrt{2}\) to \(9\) can you please explain the logic ? thanks and have a good weekend Pink face is a RECTANGLE, which is 3 by \(3\sqrt{2}\).
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Re: A solid cubical block has dimension as shown in the figure and the blo
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14 Jul 2018, 05:01
Bunuel, I got this problem in a GMAT Prep test. Someone should add the tag pertaining to the question source.




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