The lesson of knowledge control on the topic. The test work is designed to test students’ ability to use formulas for solving problems of the surface areas of bodies of revolution.

**Option 1.**

1. The radius of the base of a cone is 10 cm, and the formant is inclined to the plane of the base at an angle of 45°. Find the area of the section passing through the two formations, the angle between which is 30°, and the area of the total surface of the cone.

The axial section of a cylinder is a square with a diagonal of 8 cm. Find the total surface area of the cylinder.

3. The height of the cone is 8 and the diameter of the base is 20. Find the area of the lateral surface of the cone.

4. The radius of the sphere is 11 cm. Find the area of the cross section of the sphere removed from its center by 8 cm.

**Option 2.**

1. The height of the cone is 12 cm and the diameter of the base is 14 cm. Find the area of the total surface of the cone.

2. The radius of the ball is 12 cm. Find the area of the cross section of the sphere removed from its center by 9 cm.

3. The radius of the base of the cone is 6 cm, and the formant is inclined to the plane of the base at an angle of 60°. Find the area of the section passing through the two formations, the angle between which is 45°, and the area of the lateral surface of the cone.

4. The axial section of the cylinder is a square with a diagonal of 4 cm. Find the area of the total surface of the cylinder.

**Option 3.**

1. A rectangle rotates around a larger side. The lengths of the sides of the rectangle are 10 cm and 4 cm. Find the area of the total surface of the cylinder.

2. Draw a truncated cone. On the drawing, note the radii of the bases OC and O1M, the formation AB and the height KS.

3. The height of the cone is 8 cm and the radius of the base is 6 cm. Find the area of the lateral surface of the cone.