Question

Найдите периметр прямоугольника, если его диагональ равна 13, а его площадь 60 см в квадрате - задача на систему уравнения

Answer 1

Диагональ делит прямоугольник на 2 одинаковых прямоугольных треугольника, гипотенуза которого равна 13 см;

Пусть одна сторона прямоугольника, а также прямоугольного треугольника -  х, а другая сторона прямоугольника, а также прямоугольного треугольника -  у.
Составим и решим систему уравнений:
х * у = 60 (формула нахождения площади прямоугольника)
х² + у² = 13² (теорема Пифагора)
Выразим из первого уравнения х и подставим во второе:
х = $\frac{60}{y}$
$( \frac{60}{y} )^{2}$+ у² = 169

х = $\frac{60}{y}$
$\frac{3600}{ y^{2} } + y^{2} = 169$

Решаем второе уравнение:
$\frac{3600}{ y^{2} } + \frac{ y^{4} }{ y^{2} }$
$\frac{3600 + y^{4} }{ y^{2} } = 169$
3600 + $y^{4}$ = 169y²
y⁴ - 169y² + 3600 = 0

По теореме Виетта:
у₁² = 25
у₁ = 5;

у₂² = 144
у₂ = 12;

Находим $x_{1} и x_{2}$ путем подстановки у в первое уравнение:
х₁ = $\frac{60}{5}$
х₁ = 12; 

х₂ = $\frac{60}{12}$
х₂ = 5.

Следовательно, стороны прямоугольника равны 5 см и 12 см.
Теперь находим периметр прямоугольника:
P = (5 + 12) * 2 = 34 см.

Ответ: Периметр прямоугольника равен 34 см.

Answer 2

Perimeter of a Rectangle with Diagonal and Area

To find the perimeter of a rectangle when given its diagonal and area, we can use the following steps:

1. Let's assume the length of the rectangle is a and the width is b. 2. We know that the diagonal of a rectangle forms a right triangle with the sides of the rectangle. Therefore, we can use the Pythagorean theorem to relate the length, width, and diagonal of the rectangle. 3. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (diagonal) is equal to the sum of the squares of the other two sides (length and width). - Using the Pythagorean theorem, we have: a^2 + b^2 = diagonal^2. - In this case, the diagonal is given as 13, so we have: a^2 + b^2 = 13^2. 4. We also know that the area of a rectangle is given by the product of its length and width. - In this case, the area is given as 60, so we have: a * b = 60. 5. We now have a system of equations with two variables (a and b) that we can solve to find the values of a and b. 6. Once we have the values of a and b, we can calculate the perimeter of the rectangle using the formula: perimeter = 2 * (a + b).

Let's solve the system of equations to find the values of a and b and then calculate the perimeter.

Solving the System of Equations

Using the given information, we have the following equations:

1. a^2 + b^2 = 13^2 (Equation 1) 2. a * b = 60 (Equation 2)

To solve this system of equations, we can use substitution or elimination. Let's use substitution:

From Equation 2, we can express a in terms of b:

a = 60 / b

Substituting this value of a into Equation 1, we get:

(60 / b)^2 + b^2 = 13^2

Simplifying this equation, we have:

3600 / b^2 + b^2 = 169

Multiplying through by b^2, we get:

3600 + b^4 = 169b^2

Rearranging the equation, we have:

b^4 - 169b^2 + 3600 = 0

This is a quadratic equation in terms of b^2. Let's solve it using factoring:

(b^2 - 25)(b^2 - 144) = 0

This equation has two possible solutions:

1. b^2 - 25 = 0 - Solving for b, we get: b = ±5 2. b^2 - 144 = 0 - Solving for b, we get: b = ±12

Since the width of a rectangle cannot be negative, we discard the negative values. Therefore, we have two possible values for b: 5 and 12.

Let's calculate the corresponding values of a using Equation 2:

1. For b = 5, we have: a = 60 / 5 = 12 2. For b = 12, we have: a = 60 / 12 = 5

Calculating the Perimeter

Now that we have the values of a and b, we can calculate the perimeter of the rectangle using the formula: perimeter = 2 * (a + b).

1. For a = 12 and b = 5, the perimeter is: perimeter = 2 * (12 + 5) = 34 2. For a = 5 and b = 12, the perimeter is: perimeter = 2 * (5 + 12) = 34

Therefore, the perimeter of the rectangle is 34 cm.

Conclusion

The perimeter of the rectangle, given that its diagonal is 13 and its area is 60 cm², is 34 cm. This was calculated by solving the system of equations formed by the Pythagorean theorem and the formula for the area of a rectangle.