Questions: 다음 그림의 어두운 사각형은 직사각형이고, 나머지 4 개의 사각형은 정사각형이다. 4 개의 정사각형의 둘레의 길이의 합이 80 이고, 4 개의 정사각형과 어두운 직사각형의 넓이의 총합이 140 이다. 어두운 사각형의 이웃한 두 변의 길이를 각각 x, y 라 할 때, x^3+y^3 의 값을 구하시오. [4점]

Solution Steps

The problem involves a cross-shaped figure composed of 5 rectangles: 1 central rectangle and 4 identical surrounding rectangles. We are given:

• The sum of the perimeters of the 4 identical rectangles is 80.
• The total area of the 5 rectangles is 140.
• We need to find $x^3 + y^3$ where $x$ and $y$ are the lengths of the sides of the central rectangle.

Let:

• $a$ and $b$ be the dimensions of the central rectangle.
• $x$ and $y$ be the dimensions of each of the 4 identical rectangles.

1. The perimeter of one identical rectangle is $2(x + y)$. Since the sum of the perimeters of the 4 rectangles is 80: $$4 \times 2(x + y) = 80 \implies 8(x + y) = 80 \implies x + y = 10$$

2. The area of the central rectangle is $ab$, and the area of one identical rectangle is $xy$. The total area of the 5 rectangles is 140: $$ab + 4xy = 140$$

From the equation $x + y = 10$, we can express $y$ in terms of $x$: $$y = 10 - x$$

Substitute $y = 10 - x$ into the area equation: $$ab + 4x(10 - x) = 140 \implies ab + 40x - 4x^2 = 140$$

Using the identity for the sum of cubes: $$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$ Since $x + y = 10$: $$x^3 + y^3 = 10(x^2 - xy + y^2)$$

Using $y = 10 - x$: $$x^2 - x(10 - x) + (10 - x)^2 = x^2 - 10x + x^2 + 100 - 20x + x^2 = 3x^2 - 30x + 100$$

Substitute back into the sum of cubes formula: $$x^3 + y^3 = 10(3x^2 - 30x + 100)$$

Final Answer