00:01
Okay, so we have two cubes.
00:03
One of them is open, one of them is closed, and we're looking at the surface area compared for them.
00:09
The other piece of information is that the closed cube's edge is one centimeter less than the open one.
00:17
So we're going to kind of represent all this information in a moment.
00:22
The first thing we need to know is that for a cube, we have six square faces.
00:27
So if we call this edge s, then for a normal cube, each face would be s times s, which is s squared.
00:38
And since there's six of them, the area of a normal cube is 6s squared.
00:43
For this, we have two kind of cases we have to deal with.
00:47
For the open one, that's if we took the top off the cube.
00:51
We would only have five faces to account for it.
00:54
So for the surface area of the top one, it's 5s squared.
01:01
Now for the bottom, we still have six faces.
01:06
But the catch is we know that the edge is one centimeter less.
01:11
So instead of s, i'm going to call this s minus 1, because i don't know what it is, but i know it's one less than the other one.
01:18
So it's 6 times s minus 1 squared.
01:22
And we know these are equal to each other.
01:24
We're told that the surface area is the same.
01:27
So we're going to set these values equal to each other, and we're going to solve.
01:31
So 5s squared equals 6 times s minus 1 squared.
01:37
Now, in order for us to actually solve this, we're going to have to multiply this out.
01:43
So this is 6 times s minus 1 times s minus 1.
01:50
And you can use foiling, you can use whatever method you want to multiply these together.
01:57
So we have 6.
01:59
What we would get is s squared minus 2s plus 1.
02:05
So we have s squared minus 1s minus another s plus 1 in the end.
02:10
So we get 5s squared equals 6.
02:15
And mentioning here, we're going to distribute the six.
02:17
So 6x squared minus 12s plus 1.
02:26
In order for us to solve this, we want to move all the variables to one side.
02:29
So we're going to subtract 5s squared from both sides.
02:37
So 0 equals s squared minus 12s.
02:43
And this is a case where we could use factoring, we could use quadratic formula.
02:48
We have a lot of options here as far as solving this.
02:52
Factoring, there isn't really much option for this because we have one as the last number, so it would have to be one in one or negative one and negative one, and that doesn't seem like it really works.
03:01
So we're going to use quadratic formula to do the job.
03:07
Quadratic formula is always a method that you can use to solve a quadratic.
03:10
So we're going to try it out.
03:14
So in this case, s is our variable.
03:18
Negative b, plus or minus, i actually erase that smudge because that shouldn't be there.
03:22
Negative b plus or minus square root of b squared minus 4a c all over 2a so our b value is negative 12 so negative negative 12 is positive 12 plus or minus b squared so negative 12 squared is 144 minus 4 a is 1 is 1 is 1 all over 2 times 1 so you get 12 plus or minus the square root.
04:00
So this is 144 minus 4, which is 140, all over 2.
04:07
Squared of 140, we can simplify.
04:09
So i'm going to erase some of this work that we did earlier because we need some space to do this.
04:16
So square root of 140, we'd have to break down 140 here.
04:22
It's 14 times 10.
04:24
14, 7, and 2, 10 is 5 and 2.
04:28
Since we have a pair of numbers that are the same, we can bring that out to the outside of the square.
04:34
Root.
04:36
Sorry about that...