Can you find the area of the Blue shaded region? | (Four quadrilaterals in a squares) |

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PreMath

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@UnknownUser.x04 9 ай бұрын

I did it in the first method, but the second method is more interesting! ✨

@PreMath 9 ай бұрын

Nice work! Thanks ❤️

@jimlocke9320 9 ай бұрын

In the second method, we are told that A₁ + A₂ = B₁ + B₂ without being provided with a proof. One way of proving would be to go back to method 1. The yellow and green combined areas is composed of 4 triangles, ΔPGB + ΔPBF, + ΔPED + ΔPDH. However, the following pairs of triangles have equal area: ΔPGB = ΔPGC, ΔPBF = ΔPAF, ΔPED = ΔPEA and ΔPDH = PCH. If we replace each of the triangles in the sum with the triangle of equal area, we get the combined area of the pink and blue regions. Another proof is to connect F, G. H and E to form a square. The sides are the hypotenuses of 4 congruent isosceles right triangles. The square is composed of 4 triangles, ΔPEF, ΔPFG, ΔPGH and ΔPHE. If the sides which do not include point P are the bases, the bases have equal length, side of square FGHE. A line segment from EH to FG through point P and perpendicular to both has length equal to a side of square FGHE, and is also the sum of heights of ΔPFG and ΔPHE. A line segment from EF to HG through point P and perpendicular to both has length equal to a side of square FGHE, and is also the sum of heights of ΔPEF and ΔPGH. So, combined area ΔPFG + ΔPHE = ΔPEF + ΔPGH. Add 2 of the 4 congruent triangles to each side: (ΔPFG + ΔBFG) + (ΔPHE + ΔDHE) = (ΔPEF + ΔAEF) + (ΔPGH + ΔCGH). -> yellow + green = pink + blue.

@flesby 9 ай бұрын

Hi one questions regrading Method 2: You are stating that A1+A2 = B1+B2, what is that theorem called, I saw other math-channels proceeding in this order as well, however I was never taught why that is the case. ... An explanation would be much appreciated.

@UnknownUser.x04 9 ай бұрын

∆APB=∆BPF=a(let) [same height and base] Similarly, ∆BPG=∆CPG=b ∆CPH=∆DPH=c ∆DPE=∆APE=d Now, FPGB+EDHP=(a+b)+(c+d) Again, CGPH+AEPF=(b+c)+(a+d)

@sandanadurair5862 9 ай бұрын

I too had this doubt.😂

@flesby 9 ай бұрын

@@UnknownUser.x04 Thx for the answer, I think you meant to say "∆APF=∆BPF=a(let) [same height and base]" (first line of your comment). So if I am not completely mistaken this property holds true for all squares that are dissected in a similar fashion as the one show in the video, that is the dissecting lines start in the middle of each side and unite in a common vertex that can be located anywhere within the square. Is this observation correct? - And if so is there a name to this property such as "intersecting chords theorem" in circles or does it occur to seldom to have a name of its own?

@UnknownUser.x04 9 ай бұрын

@@flesby yea, sorry for the mistake. And your observation is completely correct, but the interesting fact is that it works with all kinds of Quadrilaterals and not just with squares. You can draw and try to understand.

@flesby 9 ай бұрын

@@UnknownUser.x04 Thank you for expanding upon the topic and no need to be sorry, after all you took the time to explain something to me and typos simply do happen. :) - I tried it with a drawing .... simply amazing this fact never occurred to me.

@phungpham1725 9 ай бұрын

Blue area + pink area= green area +yellow area----->Blue area+ 15 sqx -9x+5= 15 sqx-9x+12-----> Blue area= 12-5= 7 sq units

@PreMath 9 ай бұрын

Thanks ❤️

@LuciMama-iy9ij 9 ай бұрын

Áreas: CPG=g=GPB ; BPF=b-g=FPA ; APE=a-b+g=EPD ; DPH=d-a+b-g=HPC 》CHPG=HPC+CPG=d-a+b = (10XX-6X+8)-(15XX-9X+5)+(5XX-3X+4)=8-5+4=7 Interesante acertijo. Gracias y saludos.

@raulalbertoreyesmartinez8847 Ай бұрын

Great video. Just a question, can you provide proof for method 2? Why is the sum of the areas you gave equal?

@jpopelish 9 ай бұрын

In second method, you did not justify why A1+A2=B1+B2. Recently you solved a problem that proved that the areas of opposite triangles, inside a square, (that all share a common, arbitrary point, inside the square, formed if you connect the point to each of the corners of the square) add up to the same total area. But, here, you skipped the step that forms such a square by connecting the mid points of the given square, and all the 4 triangles outside that inner square are equal in area.

@ABhaim 9 ай бұрын

I solved it by applying the shoelace method Let the square side 2a and the dot inside it (b,c). By the shoelaces method and these coordinates, we conclude that 0.5a(2a + c - b) = 5x^2-3x+4 and 0.5a(2a + b - c) = 10x^2-6x+8, therefore, by adding them together, we eliminate b and c, get 2a^2 on the left side, and recall that the side of the square is 2a, so by doubling both sides by 2, we get the whole square area in terms of x and get 30x^2-18x+24, then we subtract the sum of all known areas and get that the remaining area is 7 😊

@yalchingedikgedik8007 9 ай бұрын

Thanks Sir The second method very easier than first method . With glades ❤❤❤❤

@Zero-t3z 9 ай бұрын

Mindblown 😍

@misterenter-iz7rz 9 ай бұрын

By formula, (10x^2-6x+8)+(5x^2-3x+4)=(15x^2-9x+5)(?)+(?), 15x^2-9x+12=15x^2-9x+5+(?), (?)=7.😊

@arnavkange1487 9 ай бұрын

Very nice solution .....

@PreMath 9 ай бұрын

Glad to hear that! Thanks ❤️

@alster724 9 ай бұрын

Easy approach especially method 2

@MrPaulc222 9 ай бұрын

If I've remembered correctly the area is 7, because green and yellow add up to 15x^2 - 12x + 12, and the pink area is 15x^2 - 12x + 5, so add 7 to level them. Yes, I see my way was closest to your second method.

@PreMath 9 ай бұрын

Correct! Thanks ❤️

@wes9627 9 ай бұрын

Blue+Red area = Green+Yellow area, so Blue=Green+Yellow-Red. Proof: Area of square EFGH is 1/2 the area of square ABCD. Why? Red+Blue area inside EFGH = Green+Yellow area inside EFGH and thus, Red+Blue area outside EFGH = Green+Yellow area outside EFGH. Why? Area of a triangle is 1/2 base times height. Look closely at the four triangles inside the square EFGH and figure out their individual areas.

@soli9mana-soli4953 9 ай бұрын

Blu = green + yellow - pink = 7

@jamestalbott4499 9 ай бұрын

Thank you!

@Alishbafamilyvlogs-bm4ip 9 ай бұрын

How do you know the area the green region plus the yellow one it was the

@Waldlaeufer70 9 ай бұрын

Blue + Red = Green + Yellow (B = Blue) B + 15x² - 9x + 5 = 10x² - 6x + 8 + 5x² - 3x + 4 B + 15x² - 9x + 5 = 15x² - 9x + 12 B = 12 - 5 = 7 square units

@seak5988 9 ай бұрын

What is x? length of the one side?

@Unstoppable332 9 ай бұрын

Yes x is the length of one side

@dirklutz2818 9 ай бұрын

x=0

@neilmccafferty5886 9 ай бұрын

Second method so simple, yet I wasn't aware of this quality of a square (or rectangle?) when divided in four with each having a common corner. I suppose if each one was a square it would be obvious!!!

@Copernicusfreud 9 ай бұрын

Yay! I solved the problem. I used method #2 for the solution.

@kennethstevenson976 9 ай бұрын

Great proof, can be applied to any problem with similar construction meeting at one point.

@s.a.s7514 9 ай бұрын

در روش دوم راه حل، چرا آن موازنه و تساوی مساحت بخش‌های گوناگون برقرار است؟

@nunoalexandre6408 9 ай бұрын

Love it!!!!!!!!!!!!!!

@PreMath 9 ай бұрын

Glad to hear that! Thanks ❤️

@RobG1729 9 ай бұрын

It wasn't clear to me in your second method how you could assert that A1 + A2 = B1 + B2.

@jasongoldman3850 9 ай бұрын

How do you know the area of the green region plus the yellow one equals the area of the pink area plus the blue one?

@UnknownUser.x04 9 ай бұрын

∆APF=∆BPF=a(let) [same height and base] Similarly, ∆BPG=∆CPG=b ∆CPH=∆DPH=c ∆DPE=∆APE=d Now, FPGB+EDHP=(a+b)+(c+d) Again, CGPH+AEPF=(b+c)+(a+d)

@geraldgiannotti8364 9 ай бұрын

@@UnknownUser.x04 This was very helpful. Thanks for posting this!

@jasongoldman3850 9 ай бұрын

@@UnknownUser.x04But the polygons aren't triangles and using the second method they weren't divided into triangles. Is there a theorem that says which of the 4 polygons in the square have the same areas?

@jasongoldman3850 9 ай бұрын

@@UnknownUser.x04How do we know A1 + A2 = B1 + B2 without dividing the areas into triangles?

@UnknownUser.x04 9 ай бұрын

@@jasongoldman3850 i divided them into triangles because it's the general way of proving the theorem. Theorems are statements that can be logically proved. The result of the theorem formulates with the quadrilaterals, but the proof includes construction of triangles to get the result.