Super easy peasy...

4 (1/4 of the total area).

Exactly...

Easy peasy.

-8πa

If there's a Logical thinking way of solving this please walk us through it. The only red area my pea-brain saw an easy way to calculate is the upper-most one.
I'm thinking that one has to be 16 (the area of the entire square)
Minus
16 x Pi / 4 (the area of the quarter circle who's center is at "A"
divided by 2. I get ~ 1.717.

The rest of them have my brain in a cramp.

Rex

Yeah, it's so easy . . .

Nah, just an arrogant know it all .

Damit Cash, letters are fer spellin, not cypherin !

I am a math lover, a logical person and REALLY appreciate geometry.

Forgive me, but this looks like a timesink, I'll leave it for more practical topics. I love your enthusiasm, however.

Just fold all the red parts to the upper quadrants. The square is split on the 45. No math needed.

Fold the A over to B. Then AB up to D. It fills up quarter of the box abcd. 4 x 4=16 /4= 4

🤔

BINGO...

6

Oh, just like a Mad Magazine fold-in. This Caveman could handle that . . .

prove the area is 4.
start with assumptions-- validate the assumptions

No... I refuse.

Aaah. How about that?
Rex

It's assumed the upper arc is a quarter circle centered at A
It's assumed that the lower arc is a half circle with the line A-B going through the center.

Really?

If you printed it you could mostly verify it with a compass but it's not stated.

A four sided square with obvious radii...

Dude... ease back on the over-thinking.

Next thing you will say is that Big Mike is a gal... unless... I prove otherwise.

Which I ain't.

IF you assume what appears to be obvious.

How in the fugg is a radius that is based on a perfect square... somehow NOT A RADIUS?

Seriously...

Good Lord.

Poindexter.

I apologize. This is a game that isn't intended to be pure math. Math is a pure science that doesn't allow for guesses and that's not the intent of a game. I got a little carried away with the math part and left out the game part.

Cool RC... No apology needed.

I treat math... like I treat science, like medicine, like the law... engineering and architecture... hunting and fishing...

Nothing is fact... just practical "Practice".

Only a dog's love is Fact...

To me.

I sent the original problem to my friend, a PhD Optical Physicist that worked on the James Webb Space Telescope. He sent me back:

8-2π + π-2+ π-2= 4

BTW, π = Pi, if it doesn't show well in our text.
And on international Pi Day at that.
Just like me, he didn't "see" the caveman-level solution. Didn't keep him from the answer though.

Rex

I think you transposed the formula a bit. I believe it should be:

8-2n - n-2 - n-2 = 4

25 % of total area of the square......4 sq whatevers

Longbob
What you typed as "n" is supposed to be Pi. It didn't endure the font transposition as well as I'd have liked when copy/pasted.

As I presented it (copy/pasted from my friend's reply to me, whose name is also Bob, BTW) it = 1.7168 + 1.1416 + 1.1416 = 4.0000
As you're suggesting it = 1.7168 - 1.1416 - 1.1416 = (-0.5664)

One thing I've learnt is not to mess with Bob, when it comes to cypherin'

Cheers,
Rex

I don't do hyperbola or parabola.

You are right. I was driving, figuring, and typing at that same time.

Being able to do hard math like that is very impressive to me. I can not.

I just work out stuff like this in "my head"... visually folding paper (in this case)...

Projecting lines and angles in other problems... kinda thing.

Mostly it is just a silly hobby... but it kinda helps me think out other stuff sometimes.

Two part problem...

What is the Area?

What is the length of AB?

Fitty%/2

area = 60

AB = sq root of 5, whatever that is.

60 and 5.

Looking at it again: 60 and 2.5 x square root of 3 which is 1.732 = 4.33

Area is 60 and length of AB is sqrt of 20.

Let’s see if I can explain my logic..

Draw a line from A to D.
Call the point where the that line intersects line BC point E
That line bisects line BC which makes BE = 2.5 and also makes line AE = 5.
If you rotate the triangle CDE counter clockwise you will see the two triangles are congruent and then form a 10 x 6 rectangle.
The area is 10 x 6 or 60

Now to solve for AB:
The hypotenuse of triangle ABE is 5 (half of ten)
Side BE is 2.5
That makes triangle ABE a 30/60 right triangle, where the sides are 1, 2 and square root of 3 (1.732)
Side AB is 2.5 x 1.732 = 4.33

No, BE is not 2.5. BE is half of BC or AB which you assert is 4.33.

For the record, I didn't solve the problem graphically either. Like you guys I solved it mathematically but, even though it was pi-day, I withheld the introduction of pi hoping that it would cancel out. It did. Let f(r) be the area of a one-eighth pie shaped portion of a circle with radius r. If needed, this is πr^2/8. The area that we are to find is the sum of three pieces which, in turn, can be expressed as differences in areas of circle sections and triangles. In total:

A = [2*f(2) - 2] (The lower left section being a quarter circle of radius two minus a triangle of area 2)
+ [f(4) - 2*f(2) - 2] (The middle section being an eighth circle of radius four minus the same triangle and quarter circle from above)
+ [8 - f(4)] (The final section being the large top left triangle minus an eighth circle of radius four)

All the circle sections cancel, leaving simply 4 as already shown.

In the second problem:

BE is 2.5. AB, BC and CD are all 5. Line AD bisects line BC so BE and CE are both 2.5. The hypotenuse of both triangles is 5.

ABE is a right triangle. If one side of a right triangle is 2.5 and the hypotenuse is 5 (given) the triangle is a 30/60 right triangle. In a 30/60 right triangle if the short side is 1 and the hypotenuse 2, then the other side is 1.732 or the square root of 3. A squared plus B squared = C squared

2.5 x 1.732 = 4.33 for AB.

Along the same lines...

This is either visually obvious...

Or... WAY OVER MY HEAD mathematically. i.e. I should have paid attention (and not been throwing spitballs) back in Caveman Math 101...

AB can't be both 5 and 4.33. We have a right triangle but not a 30/60 right triangle

Letting AD and BC both be bisected at E. Then BE is one half AB since AB equal BC. By the Pythagorean equation AE^2 = AB^2 + BE^2 = AB^2 + AB^2/4. Solving for AB^2 yields 4/5*AE^2...

Problem 3.

36 - Neat!

Hint, consider the square of the lengths AB, BC, and AC. 4 AB^2 = BC^2 = 4/5 AC^2.

I said:

AB, BC and CD are all 5.
BC is bisected by line AD and therefore BE is 2.5. So the two sides are 2.5, 4.33 and hypotenuse is 5. If you make the short side 1 and the hypotenuse 2, the longer side is the square root of three which is1.732. By definition, if the hypotenuse is double the short side it is a 30/60 right triangle.

To check:
5 squared is 25
2.5 squared is 6.25
4.33 squared is 18.75

6.25+18.75=25

Note the two places I emboldened from your earlier reply. In the first, you assert that AB is 5. In the second, you've calculated AB to be 4.33.

I don't do metric conversion.......................

Nor should you.

This is America!

Yep, should have read AE, DE and BC are all 5. I confoozed myself. 😊

FIFY - Cheers! The 4.33 is still incorrect.

The hippopotamus of the triangle be 5.

Half if the hippopotamus be 2.5...

Thus the AB etc. is 4.33.

Alert!!! The teacher got it wrong! Sure, half the hyp (AE) is 2.5, but who said BE is half the hyp (AE)? BE is half BC. If BE is also half the hyp, then BC is 5 and then NOT equal to AB which is a precondition! Get away from the assumption we are dialing with a 30/60/90 triangle and use Pythagoras.

CK, Navlav8r,

You both assume that the double ticked segment (BE) is 2.5. This segment is half of BC which equals AB. Therefore, BE must equal half of AB. Your math does not check since 2.5 is greater than half of 4.33. Your assumption is wrong.

AB is the square root of twenty or 4.47. BE is half that at the square root of five or 2.23. Adding the square of the sides yields 25 which is the square of the hypotenuse (Pythagoras). The hypotenuse is thus calculated to be 5 which verifies.

Are you trolling me?

Any right triangle:
Height square + length square = hypotenuse square

this example: 2 height equals length, and hypotenuse equals 5

height equals sq root of (25/3), length equals sq root [2x(25/3)]

19 (339) 6346^ 3387649> ( (6#3) 44/332 345.7665x0

Maybe you missed this in the OP?

Had to take a break. Momma sat a plate of bacon, eggs, and a toasted English muffin in front of me. Then I had to get the HP 35 S from the desk drawer.

25/ 3= 8.33 sq root is 2.88 equals height of triangle.

2x25/3 = 16.66 sq root is 4.08 equals base of triangle

check the math:
4.08 sq = 16.64
2.88 sq = 8.29
8.29 + 16.44 = 24.73 because I only carried two decimal places.

It can not be a true 3-4-5 triangle and maintain the 2/1 ratio of base to height.

Correct.

Never said it was...

Us Cavemen only use 3-4-5 triangles when laying out our cave entrances... or doing Caveman Woolly Mammoth cave art...

Why is our math different?

ETA: at 24.98 for hyp square, yours is obviously more acccurate.

No you did not. But some seemed to be working under that assumption.

Ain't math fun...

It’s not a 3,4,5 right triangle. It’s a 30, 60 right triangle where, if the short side is 1, the hypotenuse is 2 and the longer side is 1.732. So the hypotenuse is double the short side. The other, longer side is 1.732 times the short side, 1.732 being the square root of 3.

Why did you sort out the angles?

And are you sure it is a 30-60-90?

I most certainly did not miss it! It is your undoing! When you state that the two double ticked portions of BC (shown below in a later image) are each 2.5, then BC as well as AB and CD are 5. They cannot then also be 4.33.

Note that none of AB, BC, or CD is the hypotenuse. As determined by my and N8r's point E, AE and DE are.

Except the drawing specifically shows the base as twice the height, with the hypotenuse as 5.

If the base is twice the height, the hypotenuse must be longer than that.

FYI...

And Ya done made me pull out my Caveman calculator for the angles...

Geez.

Pythagoras states that a^2 + b^2 = c^2. The problem further constrains b:a be 2:1. CK and N8r mistakenly impose c:a to be 2:1 as a condition. It is not. There is no 30-60-90 triangle.

If you want to specify the short side as the base, the hypotenuse is double the base and the vertical side is 1.732 times the base.
Just do a quick search for 30, 60 right triangle.

In this drawing AB = CD <> BC! Look at it! 4.33 = 4.33 <> 5!

I yield... You are correct.

BC is indeed 4.33... NOT 5.

The 2.5 shown above should be 2.17

The angles are 25.7, 64.3 and 90.

Look at the drawing again. It is specifically NOT a 30-60-90 triangle.

The hypotenuse is identified as 5.
The base is identified as 2 x height.

The longest side of a right triangle is, by definition, the hypotenuse.
The shortest side is commonly referred to as height.
The base is commonly referred to as the longer leg attached to the right triangle.

Here, the base is 2 X height. Hypotenuse sq equals base sq plus height sq.

25 equals base sq plus height sq.

NO!!! 2.17, 4.33, 5 violates Pythagoras for a right triangle. Sqrt(5), Sqrt(20), and 5 make up the triangle. AB is Sqrt(20).

The work:
AE^2 = AB^2 + (AB/2)^2
4*AE^2 = 4*AB^2 + AB^2
4*AE^2 = 5*AB^2
AB^2 = 4/5*AE^2
AB^2 = 4/5*5^2
AB^2 = 20
AB = Sqrt(20)

The base of the large rectangle is 10 (a given).
The two sides of the rectangle are 6 (a given)
A line drawn from the highest point of both sides therefore bisects line BC so each half is 2.5. That gives two congruent triangles with two known sides…2.5 and 5.
A right triangle with one aide = 2.5 and the hypotenuse is 5 defines a 30. 60 right triangle.
In a 30, 60 right triangle the longer side is 1.732 times as long as the shorter side.
1.732 X 2.5 = 4.33.

By definition , the hypotenuse is the side opposite the right angle. If the hypotenuse is 5 and one of the sides adjacent to the right angle is 2.5, the other side adjacent to the right angle is 1.732 times the short side (1.732 is the square root of 3).
So A (2.5) squared plus B (4.33)squared does indeed = C (5) squared or 25

In my math, I assumed the square of base would be a 2/1 ratio with sq of height. Obviously not true.

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.