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There is a question which states

The solid brick figure shown is made of small bricks of side 1. When the large brick is disassembled into its component small bricks , the total surface area of all small bricks is how much greater than surface area of larger brick

enter image description here

Here is how I am attempting it:

Surface Area of Large Brick = $6x^2=6(2)^2=24$

Surface Area of 1 small Brick = $6x^2=6(1)^2=6$

Total Small Bricks = 12 = So Total Surface Area = $72$

Difference = $72-24=48$

But the difference is suppose to be $40$. How ?

(EDIT : Image Edited)

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    Just changed the image. Thanks for the suggestions. Looks like the image was faulty2012-07-27

2 Answers 2

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For the brick you've drawn, the large brick has two $3\times3$ faces and four $2\times 3$ faces, so its surface area is $2(9)+4(6)=42$.

There are $3\times3\times2 = 18$ smaller bricks, each of which has surface area $6$ so the total surface area of the $18$ small bricks is $18(6)=108$.

The difference of the areas, then, is $108-42=66$.


On the other hand, if the brick has dimension $2\times2\times3$, then the large brick has two $2\times2$ faces and four $2\times 3$ faces, so its surface area is $2(4)+4(6)=32$.

There are $2\times2\times3 = 12$ smaller bricks, each of which has surface area $6$ so the total surface area of the $12$ small bricks is $12(6)=72$.

The difference of the areas, then, is $72-32=40$. This is likely what the original problem was, as mentioned in the comments.

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    Just changed the image. Sorry for the trouble. It seems you were correct. The actual brick has a volume of $2\times 2\times 3$2012-07-27
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SA of larger bricks is $2LW+2LH+2HW = 2(3)(2)+2(3)(2)+(2)(2)(2)=12+12+8= 32 $ because it is not a perfect cube, but it's a rectangular prism.

SA of smaller bricks $= 6\cdot 12 = 72$. Then, the difference = larger - smaller $ = 72-32 = 40$.