Construct point "O" on QC such that PO = OC. This gives angle POQ = 2Theta, thus angle QPC is 90 degrees. Also PQ is sqrt(5) and tan(theta) = 1/2 so PC is 2sqrt(5) and Pythagoras yields X = 5.
Once Known that PQ=sqrt5 you can find sin theta = sqrt5/5 and Cos theta=2sqrt5/5 . Setting CQ=x and CP=y with cosine law we can write: 5=x^2 + y^2 - 2xy*2sqrt5/5 Setting s=side of the square X^2 = s^2 + (s-1)^2 Y^2 = s^2 + (s-2)^2 With sines area formula : Area(CPQ)=1/2*x*y*sqrt5/5 With difference between areas: Area(CPQ)= s^2 - (s*(s-2)*1/2) - (s*(s-1)*1/2) - 2*1*1/2 Comparing 1/2*xy*sqrt5/5=3/2s - 1 xy*sqrt5/5=3s - 2 Then substituting in the cosine law equation: 5=2s^2 - 2S + 1 + 2s^2 - 4s + 4 - 4*(3s - 2) 4s^2 - 18s + 8 = 0 S = 4 X^2 = s^2 +(s-1)^2=4^2 +((4-1)^2= 16+9 X^2=25 X=5
Cut CN=2 and draw a perpendicular from N on CP that cut CQ at M. So, from similar triangle CNM and triangle CPQ X/CM=PQ/NM = > X=CM*(PQ/NM)=sqrt5*(sqrt5/1)=5.
Law of Tangents and using on theta and angles DCP and BCQ gives tan 90 = infinity. This happens when side length is 4. Have to use LoT twice and be careful.
The only possibility for θ = θ, is that triangle CPQ is a right triangle. For any other position of points P and Q, angle θ of right triangle APQ never can be equal to θ of triangle CPQ, and this triangle can't be a right triangle. Right triangle APQ: c² = 1²+2² = 5² -> c=√5 cm Similarly of triangles: x/c= c/1 ; x/√5=√5/1 x = 5 cm ( Solved √ )
@@3r4kl3s If you draw a circumference with diameter x=CQ, this circumference will cross point B, because CBQ is a right triangle. In the same way, will cross point "P", demonstrating that CPQ is a right triangle, and CPQB is a cyclic quadrilateral !!!
@@3r4kl3s Once done all calculations, with any method, and being that triangle inscribed in a square, then if quadrilateral BCPQ meets Ptolomy's theorem, then CPQ is a right triangle. Ptolemy's theorem: a.c + b.d = e.f √5*4+3*2√5 = 5*2√5 (Done √)
Go Pythagorean as to triangle PAQ. Angle at point P then becomes 26.56 degrees, as does the angle at C. Going Pythagorean again as to line PQ, gives a distance of square root of 5. Line PQ is the sine of angle C, of triangle CPQ Sine of an angle of 26.56 is .4471. Simply divide the square root of 5, by the sine of 26.56 degrees and this will equal 5 for line CQ. Short and sweet.
I preferred a faster way. Firstly, I found the hypotenuse of the smaller triangle, which is the opposite side of the bigger one. Considering that it is half of the adjacent side as it was with the smaller triangle, it is easy to find x , using only Pythagorean Theorem.
@@d-8664 Si no fuese triángulo rectángulo, entonces el trazado geométrico propuesto sería imposible. De todas formas, aúnque no veas con claridad esa evidencia, de la figura propuesta y de la hipótesis del triángulo rectángulo se deducen inmediatamente los valores siguientes: Ángulo DCP=θ ; PQ=√5 ; PC=2√5 ; CQ=5 ; PD=2 ; DC=4 ; CD=4 ; QB=3 → Todos esos valores son coherentes con el trazado propuesto y con la hipótesis inicial, lo cual indica que ésta era correcta. Un saludo cordial.
@@d-8664 Ptolemy's theorem: e.f = a.c + b.d 5*2√5 = √5*4+2√5*3 The quadrilatelai BCPQ meets Ptolemy's theorem. This means that is a cyclic quadrilateral. And triangle CBQ is a right triangle, with its hypotenuse CQ as diameter of circumference that holds the mentioned cyclic quadrilateral. Therefore, CPQ also has to be a right triangle !!!
Cómo la mitad del cuadrado es 2 .este tiene 4 unidades y la hipotenusa del triángulo pequeño resulta de aplicar teorema de Pitágoras que da raíz cuadrada de 5 que es un cateto del triángulo que averigua el valor de x y por lógica el otro cateto pide 4 unidades .Al aplicar la suma de las raíces en su cuadrado da raíz cuadrado de 26 igual a 5 ..más fácil no ?
cevabı bu kadar uzatıp karmaşıklaştırmaya hiç gerek yok benzer açı kenar kenar benzerliği yakaladığımız 2 üçgen için; küçük üçgende hipotenüsü buluruz (karekök5)kısa kenar kök 5 ise uzun kenar 2 kök 5 olur sonrası tekrardan hipotenüs yapmakta 2kök5+kök5 in karelerini aldığımızda kök 25 buluyoruz sadeleştirdiğimizde X=5 olduğu ortaya çıkıyor
Let
Хорошее решение!
Very good
Nice method
Опишем окружность с центром O вокруг треугольника PCQ.
Construct point "O" on QC such that PO = OC. This gives angle POQ = 2Theta, thus angle QPC is 90 degrees.
Also PQ is sqrt(5) and tan(theta) = 1/2 so PC is 2sqrt(5) and Pythagoras yields X = 5.
Once Known that PQ=sqrt5 you can find sin theta = sqrt5/5 and Cos theta=2sqrt5/5 . Setting CQ=x and CP=y with cosine law we can write:
5=x^2 + y^2 - 2xy*2sqrt5/5
Setting s=side of the square
X^2 = s^2 + (s-1)^2
Y^2 = s^2 + (s-2)^2
With sines area formula :
Area(CPQ)=1/2*x*y*sqrt5/5
With difference between areas:
Area(CPQ)= s^2 - (s*(s-2)*1/2) - (s*(s-1)*1/2) - 2*1*1/2
Comparing
1/2*xy*sqrt5/5=3/2s - 1
xy*sqrt5/5=3s - 2
Then substituting in the cosine law equation:
5=2s^2 - 2S + 1 + 2s^2 - 4s + 4 - 4*(3s - 2)
4s^2 - 18s + 8 = 0
S = 4
X^2 = s^2 +(s-1)^2=4^2 +((4-1)^2= 16+9
X^2=25
X=5
Cut CN=2 and draw a perpendicular from N on CP that cut CQ at M. So, from similar triangle CNM and triangle CPQ X/CM=PQ/NM = > X=CM*(PQ/NM)=sqrt5*(sqrt5/1)=5.
Law of Tangents and using on theta and angles DCP and BCQ gives tan 90 = infinity. This happens when side length is 4.
Have to use LoT twice and be careful.
The only possibility for θ = θ, is that triangle CPQ is a right triangle.
For any other position of points P and Q, angle θ of right triangle APQ never can be equal to θ of triangle CPQ, and this triangle can't be a right triangle.
Right triangle APQ:
c² = 1²+2² = 5² -> c=√5 cm
Similarly of triangles:
x/c= c/1 ; x/√5=√5/1
x = 5 cm ( Solved √ )
Altoughmy solution as well. My way: Triangel qap und Triangel cda are Seminar.
Can you please explain why CPQ must be a right triangle if θ = θ ?
@@3r4kl3s
If you draw a circumference with diameter x=CQ, this circumference will cross point B, because CBQ is a right triangle. In the same way, will cross point "P", demonstrating that CPQ is a right triangle, and CPQB is a cyclic quadrilateral !!!
@@3r4kl3s
Once done all calculations, with any method, and being that triangle inscribed in a square, then if quadrilateral BCPQ meets Ptolomy's theorem, then CPQ is a right triangle.
Ptolemy's theorem:
a.c + b.d = e.f
√5*4+3*2√5 = 5*2√5 (Done √)
Go Pythagorean as to triangle PAQ. Angle at point P then becomes 26.56 degrees, as does the angle at C. Going Pythagorean again as to line PQ, gives a distance of square root of 5. Line PQ is the sine of angle C, of triangle CPQ Sine of an angle of 26.56 is .4471. Simply divide the square root of 5, by the sine of 26.56 degrees and this will equal 5 for line CQ. Short and sweet.
This is wrong you haven't proven that the triangle is a right triangle.
*_Solução:_*
Seja ∠DPC = α. Daí,
∠CPQ= 180° - α - θ. No triângulo ∆CQP, ∠PQC= α.
Usando a Lei do Senos no ∆PCQ:
*sen θ/PQ = sen α /PC*
No ∆PDC, temos: sen α= DC/PC e no triângulo PQA:
PQ² = 2² + 1² (PITÁGORAS)
PQ=√5. Além disso, sen θ = 1/√5.
Assim,
(1/√5)/√5 = (DC/PC) /PC
1/5 = DC/PC² → *PC² = 5DC.*
Seja DP = y, consequentemente, DC= y+2. Sendo ∆DCP um triângulo retângulo, por Pitágoras:
PC² = PD² + DC² (PC² = 5DC)
5DC = PD² + DC²
5(y+2) = y² + (y+2)²
5y + 10 = 2y² + 4y + 4
2y² - y - 6 = 0, com y > 0.
y = (1 ± 7)/4 → y = 2. No triângulo retângulo CBQ, vamos ter BC= 4 e BQ = 3 e, por Pitágoras:
x² = 4² + 3² = 25 → *x = 5.*
I preferred a faster way.
Firstly, I found the hypotenuse of the smaller triangle, which is the opposite side of the bigger one. Considering that it is half of the adjacent side as it was with the smaller triangle, it is easy to find x , using only Pythagorean Theorem.
This is wrong you haven't proven that the triangle is a right triangle.
θ=arctg(1/2)..uso l'equazione arctg(l-2)/l+θ+arctg(l-1)/l=90..svolgo i calcoli risulta l=4..percio x^2=4^2+3^2=25
PQ=√5---> Razón de semejanza entre QAP y QPC, s=√5/1=√5---> X=√5*√5 =5.
Gracias y saludos
Como supiste que QPC era triangulo rectángulo ???
This is wrong you haven't proven that the triangle is a right triangle.
@@d-8664 Si no fuese triángulo rectángulo, entonces el trazado geométrico propuesto sería imposible. De todas formas, aúnque no veas con claridad esa evidencia, de la figura propuesta y de la hipótesis del triángulo rectángulo se deducen inmediatamente los valores siguientes: Ángulo DCP=θ ; PQ=√5 ; PC=2√5 ; CQ=5 ; PD=2 ; DC=4 ; CD=4 ; QB=3 → Todos esos valores son coherentes con el trazado propuesto y con la hipótesis inicial, lo cual indica que ésta era correcta.
Un saludo cordial.
@@d-8664
Ptolemy's theorem:
e.f = a.c + b.d
5*2√5 = √5*4+2√5*3
The quadrilatelai BCPQ meets Ptolemy's theorem. This means that is a cyclic quadrilateral.
And triangle CBQ is a right triangle, with its hypotenuse CQ as diameter of circumference that holds the mentioned cyclic quadrilateral. Therefore, CPQ also has to be a right triangle !!!
@@marioalb9726 The sum of opposite angles of a cyclic quadrilateral is 180°.
Cómo la mitad del cuadrado es 2 .este tiene 4 unidades y la hipotenusa del triángulo pequeño resulta de aplicar teorema de Pitágoras que da raíz cuadrada de 5 que es un cateto del triángulo que averigua el valor de x y por lógica el otro cateto pide 4 unidades .Al aplicar la suma de las raíces en su cuadrado da raíz cuadrado de 26 igual a 5 ..más fácil no ?
This is wrong you haven't proven that the triangle is a right triangle.
@vcVartak, unless you first prove
cevabı bu kadar uzatıp karmaşıklaştırmaya hiç gerek yok benzer açı kenar kenar benzerliği yakaladığımız 2 üçgen için; küçük üçgende hipotenüsü buluruz (karekök5)kısa kenar kök 5 ise uzun kenar 2 kök 5 olur sonrası tekrardan hipotenüs yapmakta 2kök5+kök5 in karelerini aldığımızda kök 25 buluyoruz sadeleştirdiğimizde X=5 olduğu ortaya çıkıyor
(2)^2 (1)^2 ={4+1}=5 {90°A+90°B+90°C+90°D}=360°ABCD/5=70.10ABCD 7^10.10 7^5^5.5^5 3^4^2^3^2^3.2^3^2^3 1^2^2^1^1^1^1.1^1^1^3 1^2.1^3 2.3 (ABCD ➖ 3ABCD+2).
I just solved this with mind calculation , just use similarity
Eu fiz pela tanθ
5
∠CPQ = 90°
5