Can you find area of the Yellow shaded region in the square?

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Can you find area of the Yellow shaded region in the square? | (Trapezoid) |

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7 ай бұрын

Learn how to find the area of the Yellow shaded region in the square. Important Geometry skills are also explained: area of the trapezoid formula; Pythagorean theorem; isosceles triangles. Step-by-step tutorial by PreMath.com
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Пікірлер: 81

@MultiYesindeed 7 ай бұрын

Thanks for your continual & amazing work over 2023…wishing you health & happiness in 2024

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@andreasproteus1465 7 ай бұрын

By symmetry the yellow area is half that of the whole square. There is no need to consider the trapezium.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@phungpham1725 7 ай бұрын

1/ The two right triangles ECF and GBF are congruent so FE=FG ------>the triangle EFG is a right isosceles -------> EF=20 sqrt2-----> sq EC= sqEF- sq4-----> EC=28 2/ The side of the square = 32. The area of the trapezium=( DE+AG)/2 x AD=512 sq units

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@phungpham1725 7 ай бұрын

@@PreMath Happy new year to you from Texas!

@giuseppemalaguti435 7 ай бұрын

Dato l il lato del quadrato risulta EG=√(l-8)^2+l^2=a+b...ma a^2+20^2=4^2+(l-4)^2=b^2+20^2,per cui a=b..moda cui risulta l=32..Ay=(4+28)*32/2=32*16=512

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@quigonkenny 7 ай бұрын

By observation there is symmetry between the lines HE HF and HG, so all are equal to 20 and by extension of HF to a spot HI 4 units above A, we can aee that H is the center of the square. Therefore, by symmetry, the yellow trapezoid is 1/2 the area of the square. Add a line EF and this creates right isoceles triangle ∆EFH, as HE and HF are both 20. By the Pythagorean formula: HF² + HE² = EF² 20² + 20² = EF² EF² = (400)2 EF = 20√2 EC² + CF² = EF² EC² + 4² = 800 EC² = 800 - 16 = 784 EC = √784 = √(16)(49) = 28 DC = DE + EC = 4 + 28 = 32 Area of ADEG is ABCD/2 = 32²/2 = 1024/2 = 512

@DB-lg5sq 5 ай бұрын

شكرا لكم على المجهودات يمكن استعمال x=AD مع x>4 ECF=FBG قياس GEF هو 45 درجة EH=20 EG^2=x^2+(x-8)^2 EH=1/2 EG ..... x=32 S=32^2 /2=512

@m.h.6470 7 ай бұрын

Solution: Since it is a square, you can easily see, that yellow = white, which means, that yellow = half of the square area. Due to 90° point symmetry, the line FH = EH = GH. Therefore there is a isosceles triangle EFH with the Hypotenuse of 20√2 (from Pythagoras a² + b² = c² → 20² + 20² = c² → c =√(2 * 20²) → c = 20√2) This means that the right-angle triangle CEF has the side CE of 28 (from Pythagoras a² + b² = c² → 4² + (CE)² = (20√2)² → CE = √(2*20² - 16) → CE = √784 → CE = 28) So the side length of the square is 28 + 4 = 32. This means that the yellow area is 1/2 * 32² = 512

@PreMath 7 ай бұрын

Great! Thanks ❤️ Happy, safe, and prosperous New Year!

@christianaxel9719 3 ай бұрын

We only need to trace EF and FG to realize that CEF is congruent to BFG (sides 4,x, 90º angle, side-angle-side case), then ∠CEF=∠BFG, ∠CFE+∠BFG=∠CFE+∠CEF=90º so ∠EFG=90º. EFG is isosceles with ∠EFG=90º so ∠EH=∠FGE=45º, then ∠EFH=∠FHG=45º and EH=HG=20. Finally EF=20sqrt(2), x²=800-16=784, x=28, yellow area=(x+4)²/2=32²/2=512.

@neilmccafferty5886 6 ай бұрын

these problems are reasonably straightforward IF YOU HAVE LOTS OF TIME. the skill is in working out the methodology within limited time constraint, as in an exam. Very good puzzle! I enjoy them!

@Waldlaeufer70 7 ай бұрын

I like your solution with the rotated square inside square ABCD! 😊 My solution was: The yellow area is half of the square ABCD. A = s²/2 Pythagorean theorem: (s/2)² + (s/2 - 4)² = 20² s²/4 + s²/4 - 4s + 16 = 400 s²/2 - 4s + 16 = 400 s² - 8s + 32 = 800 s² - 8s + 16 + 16 = 800 (s - 4)² + 16 = 800 (s - 4)² = 784 s - 4 = +- 28 s = - 24 (doesn't work) s = 32 (works!) A = s²/2 = 32²/2 = 512 square units Happy New Year! 😊🙏😊

@PreMath 7 ай бұрын

Bravo! Thanks ❤️ Happy, safe, and prosperous New Year!

@Waldlaeufer70 7 ай бұрын

@@PreMath Same to you, and everyone else! 😊

@unknownidentity2846 7 ай бұрын

The inscribed square is indeed very nice. I did it essentially the same way you did after getting that not only the triangles BFG and CEF are congruent, but the triangles FGH and EFH are congruent as well. So H represents exactly the middle point of the square. Happy new year and best regards from the baltic sea.

@Math25362 7 ай бұрын

@@unknownidentity2846Explain further please. Yes, H is the midpoint of EG. But how does that make it the centre of the sqaure

@unknownidentity2846 7 ай бұрын

@@Math25362 With s being the side length of the square we can assume the following coordinates for the corners of the square: A: ( 0 ; 0 ) B: ( s ; 0 ) C: ( s ; s ) D: ( 0 ; s ) If H is really the center of the square, its coordinates should be ( s/2 ; s/2 ). This can be proven by calculating the midpoint of EG: E: ( 4 ; s ) G: ( s−4 ; 0 ) xH = (xE + xG)/2 = (4 + s−4)/2 = s/2 yH = (yE + yG)/2 = (s + 0)/2 = s/2 Best regards from Germany

@soli9mana-soli4953 7 ай бұрын

Triangle EFG is isosceles because EF=FG=4²+(S-4)² with S=side of the square. Being FH perpendicular to EG H is the midpoint of EG. WE can find EG with: EG² = S² + (s-8)² EG = sqrt(2S²-16S+64) EH = sqrt(2S²-16S+64)/2 EF² - EH² = HF² (4²+(S-4)²) - (sqrt(2S²-16S+64)/2)² = 20² S² - 8S + 32 - (2S² - 16S + 64)/4 = 400 S² - 8S - 768 = 0 S = 32

@robertlynch7520 7 ай бұрын

Way more complicated (my way) than needed to be, but here was the idea. I divided the diagonal in half which by symmetry is the midpoint of the square on all sides. Figuring 𝒒 = EH, 𝒒² = (𝒔 - 4)² ⊕ 4² - 20²) 𝒒² = 𝒔² - 8𝒔 - 368 And noting at the bottom of the square that one can also find the length of the diagonal as 𝒕² = (𝒔 - 8)² - 𝒔² 𝒕² = 𝒔² - 16𝒔 + 64 - 𝒔² 𝒕² = 2𝒔² - 16𝒔 + 64 Now just set 'em equal, and figure it out 𝒒² = ½² 𝒕² 𝒔² - 8𝒔 - 368 = ½𝒔² - 4𝒔 + 16 … rearrange ½𝒔 - 4𝒔 - 384 = 0 … being quadratic has roots 𝒔 = [ 32, -24 ] with -24 not possible so 𝒔 = 32; Area of trapezoid remains ½ the square (½𝒔²) = (½ 32²) = (½ 1024) = 512 u² Done!

@hcgreier6037 7 ай бұрын

So you call this less complicated? OK 🤔

@juanalfaro7522 7 ай бұрын

I extended the 20-unit diagonal to P (because of symmetry of the EG transversal) to make it 40 units; then found X by Pythagoras -----> 40^2 = X^2 + (X-4-4)^2 = X^2 + (L-8) ^2 ----> 2 X^2 - 16X + 64 = 1600 ---> X=32 and then calculated the area of the trapezoid just like you did.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@howardaltman7212 7 ай бұрын

Nice simple solution. I wish that I had your confident intuition in making symmetry conclusions.

@georgebliss964 7 ай бұрын

Good👍

@xsilata 7 ай бұрын

Point H is the center of square ABCD. Draw a perpendicular from H to BC at point Q. FQ^2+HQ^2=400,FQ=HQ-4 => (HQ-4)^2+HQ^2=400, HQ=1/2 AB

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@user-sw9lb2zs6e 12 күн бұрын

Symmetry is like a cross rotating in the middle of the big square

@marcelowanderleycorreia8876 7 ай бұрын

Very good!!

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@marcgriselhubert3915 7 ай бұрын

Let's use an adapted orthonormal. A(0;0) B(c;0) C(c;c) D(0;c) E(4;0) F(c;c-4) G(c-4;0) It's easy to obtain the equation of (EG) which is cx +(c-8)y +4c - c^2 = 0, Then we use the formula giving the distance from a point to a straight line. Distance from F to (EG) is: abs(c^2 + (c-8).(c-4) +4c -c^2) / sqrt (c^2 + (c-8)^2), which is equal to ( c^2 -8c +32) / sqrt (2 (c^2 -8c + 32)) when simplified. (c^2 -8c +32) is positive. We can now simplify by sqrt (c^2 -8c +32). We obtain: sqrt(c^2 -8c +32) / sqrt (2), which is equal to 20. Taking the square, we have then c^2 -8c + 32 = (20. sqrt(2))^2 = 800, or c^2 -8c - 768 = 0. Delta' = 16 + 768 = 784 = (28)^2 . The solutions are c = 4 + 28 = 32 or c= 4 -28 = -24 which is rejected as beeing negative. Finally c = 32 and the unknown area is ((4 + 28) / 2). (32) = 512.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@marcgriselhubert3915 7 ай бұрын

@@PreMath Thanks and same to you and everyone else.

@alster724 7 ай бұрын

Okay, got it! Have a blessed 2024 Premath friends! Happy New Year from The Philippines 🇵🇭 🎇🎆🎉🎇🎆🎉

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@georgebliss964 7 ай бұрын

I noticed point H was the centre of the square, and that the yellow area was equal to the white area by symmetry. I drew a horizontal line from point H to the right to meet line FB at point P, making a triangle HPF. Then HP = S/2, where S = square side length. PF = (S/2 - 4). Using Pythagoras, (S/2)^2 + (S/2 - 4) = 20^2. Let S/2 = X..............(1) Then X^2 + (X-4)^2 = 20^2.. X^2 + X^2 - 8X +16 = 400. X^2 -4X + 8 = 200. X^2 - 4X - 192 = 0. (X - 16) (X +12) = 0. X = 16. Thus S = 32 from (1) Then Area of square = 32 x 32. Area of yellow trapezium = half of that, 32 x 16.

@misterenter-iz7rz 7 ай бұрын

Your method is smartest. 😮

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@tasfiandnafi6359 Ай бұрын

"Mathematics is easy if you understand it, but difficult if you don't."

@dantallman5345 7 ай бұрын

Fun problem. Happy New Year.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@cyruschang1904 7 ай бұрын

x is the length of each side of the square x^2 + (x - 8)^2 = 40^2 2x^2 - 16x + 64 = 1600 x^2 - 8x + 32 = 800 x^2 - 8x - 768 = 0 (x - 32)(x + 24) = 0 x = 32 yellow area = (32)(32)/2 = 512

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@cyruschang1904 7 ай бұрын

@@PreMath Wishing you a year of happiness and success 🎉🥂🎆

@howardaltman7212 7 ай бұрын

After my terrible first solution that involved extending CB and EG to form similar triangles that resulted in solving a quartic equation. Got the correct answer, but ugh! Looked for a simpler approach by drawing diagonal DB to prove ΔDEH≅ΔBGH by ASA. Hence DH=BH which makes H the center of the square and CH=s/√2. Since CH is part of diagonal CA, ∠BCH=45°. Applying the Law of Cosines to ΔCFH gives s=-24 or s=32. Hence [ADEG]=s^2/2=512. I love questions like this with so many different valid ways to solve.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@murdock5537 7 ай бұрын

CF = a-4; CF = 4; EF = 20√2; sin⁡(FCE) = 1 → a - 4 = √(800 - 16) = 28 → a = 32 → area AGEF = 4(32) + (1/2)(24)(32) = 2(16)^2

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@murdock5537 7 ай бұрын

Dear Sir, thanks, happy New Year too!@@PreMath

@nunoalexandre6408 7 ай бұрын

Love 🎉it

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@AmirgabYT2185 Ай бұрын

S=512

@user-sk9oi9jl2g 7 ай бұрын

Спасибо за интересные задачи. Жду новых в Новом году. Поздравляю с Новым годом.❤

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@wackojacko3962 7 ай бұрын

🙂

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@Jack_Callcott_AU 7 ай бұрын

Happy New Year to all the PreMath people.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@Copernicusfreud 7 ай бұрын

Yay! I solved the problem.

@PreMath 7 ай бұрын

Bravo! ❤️ Happy, safe, and prosperous New Year!

@jans1616 7 ай бұрын

h^2 + (h-8)^2 = (20+20)^2 h=32 P=4*32 + 1/2*(32*28) = 512

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@hermannschachner977 7 ай бұрын

32^2:2

@jaiprashanth2005 7 ай бұрын

Happy new year😊

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@JSSTyger 7 ай бұрын

OK so I'm not sure this worked out but the side length of the square to be 67.5 and the area of the trapezoid to be 2415.1.

@JSSTyger 7 ай бұрын

I made a simple mistake. I ended up using the quadratic formula but instead of "-4AC" I used "+4AC". Now I get the side length equal to 32 and area equal to 512.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@prossvay8744 7 ай бұрын

Area of the yellow shaded region=1/2(4+28)(32)=512 Square units

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!

@prossvay8744 7 ай бұрын

@@PreMath Happy New Year Teacher

@user-id9cp4rm3u 3 ай бұрын

the yellow area=470

@LuisdeBritoCamacho 7 ай бұрын

In my Geometric Calculations; given the Point H as the Center (0 , 0) of a Circle with Radius = 10 ; (x^2 + y^2 = 400), the only Possible and Unique Solution is that Yellow Area (Half of the Square Area) = 512 su. Trapezoid Area = (B + b) * h / 2 B = 28 b = 4 h = 32 So: (4 + 28) * (32/2) = 32 * 16 = 512 Square Area = 32 * 32 =1.024 su. I didn't prove it, Algebraically speaking.

@PreMath 7 ай бұрын

Thanks ❤️ Happy, safe, and prosperous New Year!