rpgcodex > the threads can be improved by adding more replies

How advanced are we talking, here?

ItsChon

I solved one of the problems. I missed an obvious right angle marker which made it trivial. Still, not sure how to do this one.
https://i.imgur.com/711eCO6.jpg

Maxwell's Demon

Angle G is a right angle, so it follows that the narrow angle at G is equal to angle F
Sin(X) = opposite/hypotenuse, and Cos(X) is adjacent/hypotenuse

Maxwell's Demon

so Sin(F) = EG/27 and Cos(F) = 4/EG
Sin²(X) + Cos²(X) = 1
(EG/27)² + (2/EG)² - 1 = 0

Maxwell's Demon

EG²/729 - 1 + 4/EG² = 0
EG⁴/729 - EG² + 4 = 0
(1/729)x² - x + 4 = 0 sub into quadratic formula to get EG²
EG² = 729/2 ± 27 √(713)/2
EG = √[729/2 ± 27 √(713)/2]

Maxwell's Demon

FYI: the answer is the smaller root, 4.02. It seems to fit the other numbers. Don't worry about the weird one.

Maxwell's Demon

Now that I think about it, I think I did that wrong.
Opposite/adjacent = tan (X)
so h/25 = 2/h
h²= 50
2² + h² = EG²
4 + 50 = EG²
√(54) = EG

Maxwell's Demon

Derp

Maxwell's Demon

There is a faint right angle marker at H, right?

Syl

Only way I can solve this is if the angles are the same, i.e. if triangle EGH is a scaled version of triangle GFH
It gives: GH / a = b / GH
Along with Pythagoras: a^2 + GH^2 = EG^2
I find EG = square root of 54

Edit: oh, I didn't see someone else answered already.

Maxwell's Demon

It is scaled, see: https://en.wikipedia.org/wiki/Geometric_mean_theorem

Syl

oh, angle EGF is a right angle, I didn't see that.
I thought the figure was missing a piece of information, but that's it.

ItsChon

Ah so I flat out don't remember EVER learning about what you just mentioned in regards to Angle G being equal to Angle F. Why is that the case? Is it due to this Geometric Mean Theorem?

ItsChon

Yeah it appears to be so. Thanks so much. I never took Trigonometry, just Geometry, and for some reason I have no recollection of ever learning this. Wish there was a way to repay y'all with more than just Brofists, but it'll have to do I suppose.

ItsChon

One last thing if either of you wouldn't mind. For something like this, https://i.imgur.com/nc7Y2yp.png what's the plan of action? Do you use this formula, C^2=A^2+ ((2*Area)/A))^2 to find all three sides of the top triangle, to find Angle L, which you can use to get NR and LR for the area? Is there a way other than this to do it?

Maxwell's Demon

I learned the GMT from a Numberphile video, not a classroom.

Maxwell's Demon

This is a simple uniform scaling in Euclidean space. Where s is defined as an arbitrary, closed loop in the plane, the area enclosed by s is always proportional to the length squared (the "length" here can be any arbitrary one so long as it's still the same length before and after the transformation, i.e. compare two circles' radii, or their circumferences, but not one radius to a circumference).

Maxwell's Demon

This works for any arbitrary shape. So in this example, the scale factor is (25/13) and the area is 37 so the new area is 37*(25/13)² or 37 * 625/169 = 136.83...

Maxwell's Demon

The nice thing about this method is that it generalizes to higher dimensions. For a 1-sphere (circle) in 2-space (the plane) it's length squared, for a 2-sphere (sphere) in 3-space (3D space) it's length cubed, for a 3-sphere (etc.) in 4-space it's length raised to the fourth power, etc.

Maxwell's Demon

Also, if you want to compute 2D surface area of higher dimensional figures, you still square the length regardless of the figure, 3D volumes are always length cubed, and so on.