Let AB=a and BC=b, from which FC=10/b and AE=12/a, then CD=a-10/b and ED=b-12/a, from which (a-10/b)(b-12/a)=12, and from which (ab-4)(ab-30)=0, so ab=30, which is the area of the rectangle, so the area of the triangle is equal to 30-6-6-5=13
Проведем через точку E прямую параллельно АВ. Проведем через точку F прямую параллельно AD. Эти прямые разделят наш прямоугольник на четыре части. Пусть площадь верхней правой части равна x. x/(10-x)=(12-x)/12 x=4 S(ABCD)=12+12+10-4=30 S(EBF)=30-6-6-5=13
As Math Booster solves it, he ends up with one equation and 2 unknowns. However, he doesn't need to solve for the 2 unknowns to get the area of the square. Instead, he must solve for the product of the 2 unknowns, which he does. There are an infinite number of possible pairs (a, b) which will solve the problem. Another way to solve: let AB = CD = a and AD = BD = b as PreMath did, but let AE = c and CF = d, then ED = b - c and DF = a - d. We get one equation from A = (1/2)bh for each triangle. ΔABE: (1/2)(AB)(AE) = (1/2)(a)(c) = 6 and ac = 12. ΔBCF: (1/2)(CF)(BC) = (1/2)(d)(b) = 5 and bd = 10. ΔDEF: (1/2)(DF)(DE) = (1/2)(a - d)(b - c) = 6 and ab - bd - ac + cd = 12. We substitute 10 for bd and 12 for ac: ab - 12 - 10 + cd = 12, so ab + cd = 34. We note that ac = 12 and bd = 10, so abcd = (12)(10) = 120. and cd = 120/(ab). So, ab + 120/(ab) = 34. This is the same equation that Math Booster got at 6:24, so we complete the solution the same way he did.
Trazamos una recta horizontal por E y una vertical por F y el rectángulo queda dividido en cuatro celdas rectangulares que denominaremos por la letra de su vérice exterior→ Áreas de celdas: [A]=2*6-b=12-b; [B]=b; [C]=2*5-b=10-b; [D]=2*6=12 → (12-b)/12=b/(10-b)→ b²-34b+120=0→ b=4→ Área ABCD=12-4+4+12+10-4=30→ Área BEF=30-6-6-5=13. Gracias y saludos.
@@ДмитрийИвашкевич-я8т Este problema lo proponen con mucha frecuencia, y el método que usamos los dos es el más sencillo y directo, aunque lo usa poca gente. Enhorabuena. Un saludo cordial.
Let AB=a and BC=b, from which FC=10/b and AE=12/a, then CD=a-10/b and ED=b-12/a, from which (a-10/b)(b-12/a)=12, and from which (ab-4)(ab-30)=0, so ab=30, which is the area of the rectangle, so the area of the triangle is equal to 30-6-6-5=13
Проведем через точку E прямую параллельно АВ.
Проведем через точку F прямую параллельно AD.
Эти прямые разделят наш прямоугольник на четыре части. Пусть площадь верхней правой части равна x.
x/(10-x)=(12-x)/12
x=4
S(ABCD)=12+12+10-4=30
S(EBF)=30-6-6-5=13
As Math Booster solves it, he ends up with one equation and 2 unknowns. However, he doesn't need to solve for the 2 unknowns to get the area of the square. Instead, he must solve for the product of the 2 unknowns, which he does. There are an infinite number of possible pairs (a, b) which will solve the problem.
Another way to solve: let AB = CD = a and AD = BD = b as PreMath did, but let AE = c and CF = d, then ED = b - c and DF = a - d. We get one equation from A = (1/2)bh for each triangle. ΔABE: (1/2)(AB)(AE) = (1/2)(a)(c) = 6 and ac = 12. ΔBCF: (1/2)(CF)(BC) = (1/2)(d)(b) = 5 and bd = 10. ΔDEF: (1/2)(DF)(DE) = (1/2)(a - d)(b - c) = 6 and ab - bd - ac + cd = 12. We substitute 10 for bd and 12 for ac: ab - 12 - 10 + cd = 12, so ab + cd = 34. We note that ac = 12 and bd = 10, so abcd = (12)(10) = 120. and cd = 120/(ab). So, ab + 120/(ab) = 34. This is the same equation that Math Booster got at 6:24, so we complete the solution the same way he did.
Trazamos una recta horizontal por E y una vertical por F y el rectángulo queda dividido en cuatro celdas rectangulares que denominaremos por la letra de su vérice exterior→ Áreas de celdas: [A]=2*6-b=12-b; [B]=b; [C]=2*5-b=10-b; [D]=2*6=12 → (12-b)/12=b/(10-b)→ b²-34b+120=0→ b=4→ Área ABCD=12-4+4+12+10-4=30→ Área BEF=30-6-6-5=13.
Gracias y saludos.
Вот и я про тоже😊
@@ДмитрийИвашкевич-я8т Este problema lo proponen con mucha frecuencia, y el método que usamos los dos es el más sencillo y directo, aunque lo usa poca gente. Enhorabuena. Un saludo cordial.
Si arriva facilmente all'equazione (bh)^2-34(bh)+120=0...bh=17+13=30...da cui risulta A=bh-6-6-5=30-17=13