Home Forums Hunting & Shooting Hunter's Campfire More Caveman math...

Thanks man... That was fun...

Keeping the old Caveman on his toes.

How do you figure 2.24 as the short sides? If BC is 5 those short sides are 2.5.
Gotta eat lunch and test some loads.

BC is not 5... that was the trap I missed.

AD is 10... thus making two hippopotami... from there that thar jus math stuff.

As we both agree, point E is at the intersection of AD and BC. We also agree that we have a triangle of interest with three sides AE, BE, and AB. Pythagoras requires AB^2 + BE^2 = AE^2. Where we disagree is in the equation for BE. You state that it is AE/2 or half the hypotenuse. I state that it is BC/2 which is equal to AB/2. My equation becomes AB^2 + (AB/2)^2 = AE^2 which is solved above.

With your definition for BE, the Pythagorean equation to be solved is: AB^2 + (AE/2)^2 = AE^2 which reduces to AB^2 = 3/4 * AE^2, which reduces AB to sqrt(18.75), and finally your 4.33. But the problem with this solution is that the BC segment is too long, as CK now recognizes. Your BC segment is 5 and not equal to AB and CD. BE must be constrained to half of BC (which equals AB) and not half of the hypotenuse AE.

If you thought that was fun, check this out...

Problem 3.

Pythagoras tells us that

AB^2 + BC^2 = AC^2

Now multiply both sides by π/72.

π/72 * AB^2 + π/72 * BC^2 = π/72 * AC^2

But isn't π/72 * AB^2 = 4 * π / 2 * (AB/12)^2 which is the area of four half circles with a radius 1/12 of AB. The very same area we need in our calculation.

Hmm, so the Pythagorean equation tells us that the four half circles along AB plus the four along BC equal the four along AC!

I think Cash is trying to save some money and get the 'Far to fill in the blanks on his plans for the WVa. estate.
Some of those questions look awful familiar.

LOL... Not really... Those takesoffs are easy peasy... even the roof on Papa Bear is simple math.

These problems come from a Fakebook Group of International geometry geeks... I typically only do the ones that require "No Math".

Cause all that Pythagoras stuff retards my hillybilly ways...

Exactly... I just did it without all that complicated math stuff.

No kidding. When we did PersianDog's homework, he'd at least occasionally send out some virtual ammo.

Mr Hoosier...

Since ya like all that Pythagoras stuff...

Here is an easy one fo ya...

It's not very pretty. I solved with brute force mathematics. I solved for the (x,y) coordinate where the two right angles of the two triangles meet. There are two solutions: (5.79, 10.93) and (2.31, 13.61). The total area for the two shaded triangles are 131.29 and 80.09 corresponding to the two solutions respectively. Note that the former coordinate and area more closely matches the figure. If there is a more elegant solution, it has not occurred to me.

And now I must yield to you. Please, a respite! Thanks for the exercises.

"Cipherin'!!!"

[i][/i]

Not really... ya still gotta do da math.



BTW... Pythagoras was an ascetic and a vegan.

I am more akin to Socrates and his asebian manner... and expect to meet a similar fate... like many of us.

Damn, I see it now. Take the red chord from lower left, and pivot it clockwise along its arc into the lower right quadrant at B.

Now take the top triangular-ish shape CD, and flip it on the AC line so now it is on the edge CB.

Now all the red area is in the right triangular 'quadrant' of a square divided diagonally into four.

But I confirmed the math first.

The light when on when figuring out area of the lower-left chord. Rotate it to the right and then subtract the area of a triangle from 1/2 the area of the half circle. That's when the imagery kicked in.

Yeah, while I slept the brain found this more elegant solution using two intersecting circles of known centers and radii. I did the math this morning and went from yesterday's page and a half spent resolving multivariate quadratic equations to a few lines of trig. Same answer. Thanks for your help training the brain.