# Find the total surface area of the cylinder, the axial section diagonal is 13 cm, and the height is 5 cm.

The total surface area of a cylinder is equal to the sum of the areas of its lateral surface and two bases:

Sp.p. = Sb.p. + 2Sn. = 2 * π * r * h + 2 * π * r2.

Consider the axial section of this cylinder. For convenience, we will designate it as ABCD.

The AC diagonal cuts it into two right-angled triangles.

Consider the triangle ∆ABS. BC is the diameter of the base, we find its length using the Pythagorean theorem:

AC ^ 2 = AB ^ 2 + BC ^ 2;

BC ^ 2 = AC ^ 2 – AB ^ 2;

BC ^ 2 = 13 ^ 2 – 5 ^ 2 = 169 – 25 = 44;

BC = √144 = 12 cm.

The base radius is half the diameter:

r = d / 2;

r = 12/2 = 6 cm.

Sp.p. = 2 * 3.14 * 6 * 5 + 2 * 3.14 * 62 = 2 * 3.14 * 6 * 5 + 2 * 3.14 * 36 = 188.4 + 226.08 = 414.48 cm2.

Answer: the total surface area of the cylinder is 414.48 cm2.