Hi Mike! Can you go over dSAT Test 4, Module 2, #26? Thanks!

Bear with me here because normally a question like this would be a little easier to explain with some drawings, but I’m not able to make good drawings right now. So we’ll have to use a little imagination.

First important concept: remember that when you glue two prisms together as described in this question, you don’t just add their surface areas together. This is because when you glue, you are actually ERASING some surface area from each of the original prisms. The square faces that get glued together here become part of the inside of the new bigger prism, so they’re not part of the surface area anymore.

Now, we know the height of the prisms we’re starting with, 90 cm, but we don’t know the dimensions of the square base. If we say those are x, then we can create an expression for the surface area of a prism. The square bases will each have an area of x\times x=x^2, and there are two of those, so 2x^2. Then each of the four sides will have an area of 90x, so 4(90x)=360x.  Putting it all together, each prism before the gluing will have a surface area of 2x^2+360x. That’s what the question is calling K.

When we glue the prisms together, we lose two of the square base faces. Each of those was x^2. So the way I’d describe the gluing algebraically is:

\left(2x^2+360x\right)+\left(2x^2+360x\right)-\left(2x^2\right)

That’s adding the whole surface areas of each prism together, then subtracting out the two faces that  become internal. Let’s simplify:

\left(2x^2+360x\right)+\left(2x^2+360x\right)-\left(2x^2\right)\\=\left(4x^2+720x\right)-\left(2x^2\right)\\=2x^2+720x

Now we need to figure out this \dfrac{92}{47}K thing.

We figured out before that 2x^2+360x=K. And now we just figured out that the surface area of the new, glued prism is 2x^2+720x…that’s what is going to equal \dfrac{92}{47}K.

We’ve come all this way algebraically so I’m going to finish the job, but at this point I want to pause and say, I kinda wish I had just started trying answer choices a while ago! We could have figured out the surface area of a prism with square base 4 and height 90 pretty easily, right? And then calculated the area of the new prism when we glue two of those together? And if those didn’t work out to the right ratio we could have moved on to the next choice? I say all this, I guess, just to admit that even the guy who always tells everyone else to backsolve sometimes forgets to backsolve and that’s OK. People are complicated.

Right, back to algebra.

\dfrac{92}{47}\left(2x^2+360x\right)=2x^2+720x\\\\92\left(2x^2+360x\right)=47\left(2x^2+720x\right)\\\\184x^2+33120x=94x^2+33840x\\\\90x^2=720x\\\\90x=720\\\\x=8

Phew!

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