23MT

This is hard, dgfgrafgdfge111 hang on for a ride that is very confusing and is probably wrong.

FIND \(SQ\)

Based on pythagorean thoerem, we can see that it is \(25\)

Next, we find the altitude of \(\Delta{SPQ}\).

We FIRST do that by finding the area, which is 20*15 = 300 / 2 = 150

Using the triangle area formula \(\frac{1}{2}ab=150\), with A being the altitude, and B being the base, we can plug in and solve for altitude.

The base is 25, which is SQ.

So plugging in:

\(\frac{1}{2}a(25)=150\)

\(\frac{1}{2}a=6\)

\(a=12\)

So we find the altitude is 12. SRQ is congruent to SPQ, so SRQ also has an altitude of 12.

To help us visualize better, we unfold the two visible faces of the cube with triangles SPQ and SRQ from three dimensions into 2 dimensions.

If folded back into 3 dimensions, the length of the orange line will be the answer. (try to imagine it, I know its hard)

The orange, red, and half the top blue line form a right triangle in 3 dimensions.

While the red, green, and half the bottom blue form a right triangle in 2 dimensions.

The length of the blue line is 24, because we know the altitude is 12, because 12 * 2 = 24.

We need to know the length of the green line. So we can use pythagorean theorem to find the length of the red line.

We can use some geoemtric proofs to prove that SQ Is trisected. And the green part is 1/3 of SQ. (ask for the proof if you need it there is a lot to write.)

So since it is proved that SQ Is trisected, we know that the green length is \(\frac{25}{3}\).

So, after millions of years of calculations, I think i got (c) as the answer

I am not so sure, so please check!

Oof my brain got fried