Can You Solve This Shaded Area? | Mystery of 2 Overlapping Squares
ฝัง
- เผยแพร่เมื่อ 11 ธ.ค. 2024
- 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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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! 👍
You're like a teacher 😜
@@ZackJENKINS-yp3db I guess I'll take that as a compliment... 😂😂😂
1st 😂
👍🏼 Good problem - nice solution. 😉
@@ubncgexam ❤️
This one was fun to solve. Thanks for sharing.
Glad you like it! Thanks for watching!
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!
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
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
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!!!
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 √)