Can You Solve This Shaded Area? | Mystery of 2 Overlapping Squares
ฝัง
- เผยแพร่เมื่อ 8 ก.พ. 2025
- Welcome to an exciting geometry challenge! In this video, we explore the intriguing mystery of two overlapping squares. Your task is to calculate the shaded area formed by the overlapping sections. While it might seem straightforward at first, this puzzle will test your problem-solving skills and knowledge of geometry.
Do you think you can solve it? Give it a try before I reveal the solution, and feel free to share your answer in the comments! This challenge is perfect for anyone who loves puzzles, geometry enthusiasts, or students preparing for math competitions.
Ready to dive into this mystery? Watch the full video to uncover the solution and boost your geometry skills!
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Join the Chanel 👉 / @thephantomofthemath
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#GeometryChallenge #ShadedArea #MathPuzzle #OverlappingSquares #MathProblemSolving #MysteryPuzzle #Geometry
Interesting Problem! I generalized it somewhat: Same configuration, but let the small square be split up in two parts A1 and A2, both positive, A1+A2 = 4 and let 0
You're like a teacher 😜
@@ZackJENKINS-yp3db I guess I'll take that as a compliment... 😂😂😂
Note that FCBQ is a trapezoid (also called a trapezium) and we can solve for its area. At 6:55, we have enough information to determine the lengths of both bases FC and QB as well as height CB. CB = CD = AB = AD = AP + PD = √5 + 1/(√5) = (6√5)/5. FC = CD - FD = (6√5)/5 - 2/(√5) = (4√5)/5. AQ = AP = √5 by corresponding sides of congruent triangles, so AQ = √5. QB = AB - AQ = (6√5)/5 - √5 = (√5)/5. Area of FCBQ = (1/2)(FC + QB)(CB) = (1/2)((4√5)/5 + (√5)/5)((6√5)/5) = (1/2)(√5)((6√5)/5) = 3, as The Phantom of the Math also found.
Yes, absolutely! FCBQ is a trapezoid, so you can do it that way also. Great job btw! 👍
Picking up from GP = FP = 1 and PA = sqrt(5)
Area(PGA) / Area(FDP) = (PA /FP)^2 = 5
=> Area(FDP) = 1/5
Now, Extend line FQ to intersect CB at T such that
FCT is similar to PGA and similar to QBT
Also note that DP = QB => QT = PF = 1 and Area(QBT) = Area(FDP) = 1/5
So FT = 2+1+1 = 4
Area(FCT)/Area(FDP) = (FT/FP)^2 = (4/1)^2 = 16
Area(FCT) - 16/5
Area(FQBC) = Area(FCT) - Area(QBT) = 16/5-1/5 - 3
It is not so complicated as it sounds in my writing :)
Thanks for sharing interesting problems that trigger the thought process
Thank you so much for your kind words! I’m glad you find the problems thought-provoking. It’s always a joy to share challenges that inspire curiosity and creativity. 😊
This one was fun to solve. Thanks for sharing.
Glad you like it! Thanks for watching!
FC=4/√5
CB=6√5
QB=1/√5
area=3
1st 😂
👍🏼 Good problem - nice solution. 😉
@@ubncgexam ❤️
3
1) If we make the height of trapezoid we have 4 similar right triagles 2) From the left one of them, we khow area ( 1 ) and one side (2) and we can find the other sides (type of area and PY.TH.)
3) With the theorems of similar triagles we can find the bases of trapezoid ( 4*sq(5)/5 and sq(5)/5 and the height 6*sq(5) and the area 3) It's easy. Thank's From Greece "No good English".
Very nice! I will add translation in Greek for next video. I promise!
s² = (1+3) = 4 cm² --> s= 2cm
A₁=A₃= ½.s.h₁ = 1cm² --> h₁ = 1 cm
tan α = h₁/s = 1/2 --> α = 26,565°
β = 45° - α = 18,435°
S² = (d cosβ)² = (2√2 cosβ)² = 7,2 cm²
A₄ = ½ s/2 .cosα. s/2 .sinα = 0,2cm²
A = S² - s² - A₄ = 7,2 - 4 - 0,2
A = 3 cm² ( Solved √)
Nice!!!
Lösung:
A = linke untere Ecke des kleinen Quadrates,
B = rechte untere Ecke des kleinen Quadrates,
C = rechte obere Ecke des kleinen Quadrates,
D = linke obere Ecke des kleinen Quadrates,
E = rechte untere Ecke des großen Quadrates,
F = rechte obere Ecke des großen Quadrates,
G = linke obere Ecke des großen Quadrates,
H = Schnittpunkt beider Quadrate,
I = unterer Schnittpunkt der Verlängerung von BC mit dem großen Quadrat.
Das ganze schwarze, kleine Quadrat hat die Fläche 3+1 = 4, also ist die Kante vom schwarzen, kleinen Quadrat a = 2. Die linke Dreiecksfläche hat die Fläche 1, daraus folgt:
a*DH/2 = 2*DH/2 = 1 ⟹
DH = 1 ⟹ HC = 2-DH = 2-1 = 1
Pythagoras:
AH = √(AD²+DH²) = √(2²+1²) = √5
Ähnlichkeit: HG/HC = DH/AH ⟹ HG/1 = 1/√5 ⟹ HG = 1/√5 ⟹
AG = AH+HG = √5+1/√5 = (5+1)/√5 = 6/√5 ⟹
Fläche des großen Quadrates = AG² = (6/√5)² = 36/5 = 7,2
Ähnlichkeit: GC/HC = AD/AH ⟹ GC/1 = 2/√5 ⟹ GC = 2/√5 ⟹
Ähnlichkeit: AI/AB = AH/AD ⟹ AI/2 = √5/2 ⟹ AI = √5 ⟹
Fläche des schwarzen Trapezes = (GC+AI)/2*AG = (2/√5+√5)/2*6/√5
= (2+5)/√5*3/√5 = 7*3/5 = 21/5 = 4,2
Rote Fläche = Fläche des großen Quadrates - Fläche des schwarzen Trapezes
= 7,2-4,2 = 3
AGFE se puede dividir en cuatro triángulos rectángulos congruentes de superficie unitaria y lados 1/2/√5→ QEA es también congruente con esos triángulos y PDF es semejante a ellos→ Razón de semejanza s=1/√5→ s²=1/5→ Área de PDF =1*1/5=1/5 → PD=GP*s=1*1/√5 =√5/5→ AD=AP+PD =6√5/5 ---> Área sombreada BCFQ =ABCD-AQFD =(6√5/5)² -(1/5)-3-1=15/5 =3 ud².
Buen rompecabezas. Gracias y un saludo cordial.
Good job! Gracias por ver
CB = 1/(√5) + (√5) = (√5)/5 + (5√5)/5 = (6√5)/5,
QB is 1/√5,
FC = CB - 2√5 = (6√5)/5 - 2/√5 = (6√5)/5 - (2√5)/5 = (4√5)/5
Red area = (CB x (FC + QB) )/2
=(6√5)/5 x ((4√5)/5 + 1/√5) /2
=(3√5)/5 x ((4√5)/5 + √5/5)
=(3√5)/5 x (5√5)/5 = 15 * 5 / 25 = 75 / 25 = 3
The right 6*sq(5)/5 for height. I am sorry.
I know what you meant
s = √(1+3) = 2 cm
A₁ = ½.s.h₁ = 1 cm² --> h₁ = 1 cm
tan α = 1/2. --> α = 26,565°
β = 45° - α = 18,435°
S = d cosβ = 2√2 cosβ = 2,6834 cm
A₃ = A₁ = 1cm²
A₄ = ½ s/2 .cosα. s/2 .sinα = 0,2cm²
A = S² - A₂ - A₃ - A₄ = 7,2 - 3 - 1 - 0,2
A = 3 cm² ( Solved √)