NotThatSmart

We don't actually have to find the area of the polyhedron itself. 

We know how to find the area of the cube and the area of the pramids cut off, so we just have to subtract them. 

Each of the pyramids cut off must have a right iscocoles base with sidelengths 3 and height 3. 

From this, we can write

\(\text{Volume} = 6^3 - 8 \cdot \dfrac13 \left(\dfrac{3 \cdot 3}2\right)\cdot 3 = 180\)

So the volume is 180. 

Thanks! :)

Well, I don't reccomend it, but I used a calculator because I'm too lazy. 

\(729y^{12}+2916y^{11}-2430y^{10}-19980y^9+135y^8+59976y^7+6364y^6-99960y^5+375y^4+92500y^3-18750y^2-37500y+15625\)

As you can see, the coefficient for y^4 is 375. 

Thanks! :)

There's a pretty handy formula to find the circumradius of a triangle. 

We have \(\frac{(abc)}{\sqrt{(a + b + c)(b + c - a)(c + a - b)(a + b - c)}}\) where a, b, and c are the sidelegnths. 

Here, we just have \(\frac{a^3}{\sqrt{3a^4}} = \frac{a^3}{a^2 \cdot \sqrt3}= \frac{a}{\sqrt{3}}\). 

So \(\frac{a}{\sqrt3}\) is our answer. 

Thank You! :)

This was the post that got me to 300 points LOL :)

Let x be the length from the right edge to the strip. 

We have

\([rectangle]=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ [strip]=1\cdot \sqrt{8.5375^2+8^2}\\ [strip]=11.7\)

Thanks! :)

First, from the first part of the problem, we know that

\(2x + 2y = 40 \\ x + y = 20 \\ y = 20 - x\)

Now, we use the fact that the diaganol is 10 sqrt2, we have

\(x^2 + y^2 = (10 sqrt 2) ^2 \\ x^2 + ( 20 - x)^2 = 200 \\ x^2 + x^2 - 40x + 400 = 200 \\ 2x^2 - 40x + 200 = 0 \\ x^2 - 20x + 100 = 0\)

\((x -10)^2 = 0 \\ x - 10 = 0 \\ x = 10 \\ y = 20 - 10 = 10\)

The area is just 10*10 = 20. 

Thanks! :)

I believe the rhombus is a square for main reasons. 

For one thing, the opposite angles of a rhombus are 90 degrees, so we have two 90 degree angles.  

The adjacent angle also should be 90 degrees, so all 4 angles are 90 degrees. 

Therefore, we just have a square. 

We have \(12^2 = 144\)

So the area is 144. 

Thanks! :)

To start, if all sides of a quadrilateral are equal and the diaganols are equal, then we defintely have a square. 

The square of diagonal of a square is equal to 2 times the square of the sidelegnth. We have

\(2PQ^2 = (\sqrt2)^2 \\ 2PQ^2 = 2 \\ PQ^2 = 1 \\ PQ = 1\)

The perimeter is just \(1+1+1+1 = 4\)

So 4 is the answer. 

(We could check. The diagonal of a square with sidelength 1 is actually sqrt2, so we are clear!)

Thanks! :)