# In a rectangular sheet of cardboard with dimensions of 8dm x 6dm, a rectangular hole with an area of 24

In a rectangular sheet of cardboard with dimensions of 8dm x 6dm, a rectangular hole with an area of 24 dm must be cut out so that the sides of the hole are equidistant from the ends of the cardboard sheet. What should be the dimensions of this hole?

Suppose that the length of the hole is x dm, then the distance to the edges of the cardboard sheet will be equal to (8 – x) / 2 dm.

Suppose that the width of the hole is equal to y dm, then the distance to the edges of the cardboard sheet will be equal to (6 – y) / 2 dm.

According to the condition of the problem, we can compose the following system of equations:

(8 – x) / 2 = (6 – y) / 2,

x * y = 24.

From the first equation we obtain that

8 – x = 6 – y,

-y = 2 – x,

y = x – 2.

Substitute the resulting value for y into the second equation:

x * (x – 2) = 24,

x² – 2 * x – 24 = 0.

The discriminant of this equation is:

(-2) ² – 4 * 1 * (-24) = 4 + 96 = 100.

Since x can only be a positive number, the problem has a unique solution:

x = (2 + 10) / 2 = 6 (dm) – hole length.

y = 6 – 2 = 4 (dm) – hole width.

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