The total surface area of a regular quadrangular prism is 130 cm2, and the base area is 25 cm2. Find the diagonal of the prism.

Let there be given a regular quadrangular prism ABCDA₁B₁C₁D₁: S.e. = 130 cm², Sb. = 25 cm². The diagonal of the prism is the line segment AC₁, which can be found from the rectangular △ ACC₁.

1. The bases of ABCDA₁B₁C₁D₁ contain squares. The area of ​​the square is:

S = a²,

where a is the side length.

In this way:

Sosn. = a²;

a² = 25;

a = √25;

a = 5 cm.

From △ ACD we find AC by the Pythagorean theorem:

AC = √ (AD² + CD²) = √ (5² + 5²) = √ (25 + 25) = √ (2 * 25) = 5√2 (cm).

1. The total surface area is equal to:

Sp.p. = S side. + 2 * S main.

Let’s find the lateral surface area:

S side. = S.p. – 2 * S main. = 130 – 2 * 25 = 130 – 50 = 80 (cm²).

The lateral surface area is equal to the sum of the areas of all four prism faces, then the area of ​​one DD₁C₁C face is equal to:

S = S side. / 4 = 80/4 = 20 (cm²).

The face DD₁C₁C is a rectangle:

S = C₁C * CD;

5 * C₁C = 20;

C₁C = 20/5;

C₁C = 4 (cm).

1. From △ ACC₁ by the Pythagorean theorem we find AC₁:

AC₁ = √ (AC² + C₁C²) = √ ((5√2) ² + 4²) = √ (50 + 16) = √66 (cm).

Answer: AC₁ = √66 cm.