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# Painted cube and cuboid

Yahooooooooooooo) For math class=)
by

## K-k* =)

on 9 January 2011

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#### Transcript of Painted cube and cuboid

painted Cube investigation Let's imagine that we droped cube into the paint...So...how many painted faces are we going to have?o_O Another important thing - there are some more cubes inside!=) Size of cube 3 painted faces 2 painted faces 1 painted face 0 painted faces Total painted faces
2x2x2 8 0 0 0 24
3x3x3 8 12 6 1 54
4x4x4 8 24 24 8 96
5x5x5 8 36 54 27 150
Let's take different sizes of cube and work out painted faces and put all results into the table. So...now we can see some patterns! For the '3 painted faces' we can say that it's '8' everytime, because there are 4 angles in the cube everytime. a(n) = 8 For '2 painted faces' we have to add '12' to the previous answer if the size of cube is increasing by '1'. a(n) = 12(n-2) For '1 painted faces' formula will be: a(n)=6(n-2)(n-2) For the '0 painted faces' formula is going to be: a(n)=(n-2)³ And now, by using these formulas, let's find out painted faces for the 6x6x6 cube. '3 painted faces' '2 painted faces' '1 painted face' '0 painted faces' Total painted faces And to calculate total painted faces: a(n)=6n² 8 12(6-2) = 48 6(6-2)(6-2) = 96 (6-2) = 64 ³ 6 x 36 = 216 Painted cuboid investigation And now let's imagine that we have cuboid... Now we have to choose some different sizes and write results into the table: Size of cuboid 3 painted faces 2 painted faces 1 painted faces 0 painted faces Total painted faces

2*3*4 8 12 4 0 52
2*3*5 8 16 6 0 62
3*4*5 8 24 22 6 78 Here we can see that all cuboids have 8 '3 painted faces'. '3 painted faces' '2 painted faces' '1 painted faces' '0 painted faces' a(n) = 8 a(n) = 4(A+B+C)-24 You can see this formula if you'll look at the diagram above - all the cubes with two painted (purple) are one in from either side so the formula must have a-2, b-2 and c-2 in it which is the cuboid lengths minus 2. Also as each of the cuboid lengths 'appear' 4 times on the cuboid...so we have to multiply it by '4'. a(n) = 2((A-2)(B-2)+(B-2)(C-2)+(A-2)(C-2)) One painted face - all in the middle - pink-> a-2, b-2 and c-2. Also - all of the shapes are cuboids, the cubes with one painted face are always going to be in a rectangle so we can say that the formula will also be to do with multiplying the side lengths together for each different size of rectangle and as each different rectangle occurs twice the formula will also have multiplying by 2 in it. a(n)= (A-2)(B-2)(C-2) All of the cubes with no painted faces are inside of the cuboid-> you have to minus 1 from each side of the length, therefore a-2, b-2 and c-2 (are going to be in the formula). Also know that the cube or cubes that have no painted faces - >will also form a cuboid=> you have to multiply all of the new lengths together. Total painted faces a(n) = 2ab+2ac+ABC-4B-4C I got this formula by adding products of 'painted faces' and 'nth term' formulas. Painted cube and cuboid investigations
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