Algebraic Approach to this one

I wanted to try this one with this algebraic approach. It was fun to go through.

[johnsanko]

Using the geometric shapes as placeholders for the numbers in the equation is a really great way to help people visualize what the numbers mean.

[janda]

Another method could be to ”move” the x-areas into the ”citcular hole” in the z-area. Doing this on both x-areas creates a non-shaded triangle of b=h=4 which means an area of 4x4/2=8. The area of the quarter circle is 4pi meaning the shaded area is 4pi - 8

[NEDinACTION]

nothing has ever blown my mind harder than the sentence "we can subtract triangle from both sides"! I absolutely love the way you think about these problems and these videos always make me smile.

[ratzou]

both white areas can be cut and moved around to create simple 2x2 squares.. so the white area is 8 sq units, and the whole quarter circle is 4pi.
There you have it, the blue area is 4pi-8.

[thibault]

Not as straight forward as I thought it will be but still exciting

[euum]

Since both 2x + y and x + y + z were equal to 2pi it was probably easier to just determine that x and z were equal to begin with, and that 2x + 2z can be renamed to 4x equals the area.
Then once getting that pi - 2 is x, you can just multiply out to get 4pi - 8 is the area.

[zaidq]

I never visualized the equation like this ever in my life. This way of teaching makes it so simple and interesting.

[CuriousCyclist]

These problems are indeed fun and your enthusiasm is infectious. ❤

[doesnotexist]

It's faster to recognize that the two x regions can be moved to mate with the z regions, creating a single circular segment of radius 4 and angle 90. Then it's just the area of the full quarter circle minus the area of the right equilateral triangle.

[Anonymous-ptue]

Lame Question: At 1:31, how do we know that the blue areas, x and z exactly meet at the centre? Isn't that an assumption critical to make y equals 2?

[YardyHardy-zh]

I love how you don't waste anytime over explaining things. Subscribed

[bregenoranthoran]

I was so excited around 1:40 when I saw Z and X were the same. I was like "I KNEW IT!"

[Masfugo]

I think the most important think to solve this kind of question is knowing how to breakdown the shape into more workable form

how exciting

[icarus-wings]

What’s most fascinating about this is that z and x are the same area, despite not looking anywhere near the same area visually!

[Falcon]

How exciting! I did it slightly differently, I took the bottom left part to be a square with sides 2. With that area being 4 square units, if I took away 1/4 of the circle with radius 2 from the 4 I would have a variable I called x. I saw that if you had a square of area 4 and you took away 2 of those x’s you’d be left with the blue value between the smaller circles. If you then took the whole quarter circle, took away 2 quarter circles of radius 2, the square of 4 and added back on the value found earlier you’d arrive at the same conclusion.

[Formatsli]

Andy your the greatest teacher at math in youtube, please teach us how to make it simpler because alot of people are having a hard time, only few can do this like you.

[VideoFusco]

If you rotate the two blue circular segments until they are adjacent to the two uppermost blue pieces, the white and blue areas retain the values of their areas but with this maneuver the white area becomes an isosceles right triangle with legs equal to 4 (therefore area 8) and the blue area is a circular segment that has an area equal to the difference between the area of the quarter circle you calculated at the beginning (4pi) and the area of the white triangle (8). So the final area is 4pi-8.

[Oricorio-Pom-Pom]

I did it as the following: label the top blue part “A”, the bottom blue part “B”; take the 2x2 square around B. Area of B is the overlap between the two white parts in that square, since there is no empty parts, its just the combined area of the white parts in the square, mod the area of the square. A_whiteareas is just pi x d^2/8 = 2pi. Now we have area of B as 2pi mod 4 or 2pi-4. Now for the area of A. This is the area of the whole quarter circle, - the areas of the white + back the overlap. A_qcircle is pi x r^2 /4 = 4pi. Substituting with all the values we have gives 4pi - 4pi + 2pi - 4, canceling out the 4pi’s, we have area A as 2pi-4. Adding this with the other one gives us the final area of the blue as 4pi-8 u^2. Quite lengthy but thats just what came to mind when i saw it.

[bastianrevazov]

please solve more stuff with algebra
i get so excited when algebra is in play

[theduckthatsees]

I did this one a bit differently. Rather than focusing on the blue area, I found it by working out the white area and subtracting it from the area of the total shape.

If we break it down, the white area is made up of two quarter circles plus an extra bit that we will get to later. To find the area of the quarter circles, each one has a radius of 2. Since the area of a quarter circle is πr²/4, we substitute in 2 to get:
2²π/4 = 4π/4 = π
Due to there being two of these quarter circles their total area would be 2π.

If we drew a square of radius 2 around the pointed oval type shape in the bottom left, it becomes apparent that the white area within that square is equal to two of the remainding area when you draw a quarter circle of radius 2 within the square. Using the same formula, πr²/4, we can work out:
2 * (4 - 2²π/4) = 2 * (4 - 4π/4) = 2 * (4 - π) = 8 - 2π

Thus, the total white area is equal to:
2π + 8 - 2π = 8

Using πr²/4 once again, we can work out that the area of the total shape, as it has a radius of 4, is equal to:
4²π/4 = 16π/4 = 4π

Therefore, as it is equal to the total area minus the white area, the blue area is equal to 4π - 8 square units.