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.
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".
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.
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! 👍
This one was fun to solve. Thanks for sharing.
Glad you like it! Thanks for watching!
1st 😂
👍🏼 Good problem - nice solution. 😉
@@ubncgexam ❤️
You're like a teacher 😜
@@ZackJENKINS-yp3db I guess I'll take that as a compliment... 😂😂😂
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!
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!!!
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
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 √)