An isosceles right-angled triangle with a 2 cm leg rotates around one of the legs

An isosceles right-angled triangle with a 2 cm leg rotates around one of the legs. Find the axial cross-sectional area of the resulting body of revolution.

In a right-angled isosceles triangle, the legs are equal. When rotating around one of them, a cone 2 cm high with a base radius of 2 cm is formed.

The axial section of the cone Sk will be equal to the area of the doubled area of the triangle forming the cone.

The area Sт of the triangle is equal to half the product of the legs:

St = (2 cm * 2 cm) / 2 = 2 cm ^ 2;

Sk = 2St = 2 cm ^ 2 * 2 = 4 cm ^ 2.

Answer: 4 cm ^ 2.