The height of the cylinder is 160 m, the radius is 208 m. The plane parallel to the axis of the cylinder is 192 m away from it.

The height of the cylinder is 160 m, the radius is 208 m. The plane parallel to the axis of the cylinder is 192 m away from it. Find the cross-sectional area of the cylinder with this plane.

A cylinder is a geometric body created by rotating a rectangle around its side.

Since the cross-section by a plane has the shape of a rectangle, its cross-sectional area is equal to the product of the height by the width of the cross-section:

SABCD = AB · BC.

To do this, you need to calculate the width of the given section.

Consider the base of the cylinder.

Point O is its center, segments BО and OC are radii, segment OH is the distance from the axis to the section, segment BC is the width of the secant plane.

The triangle ΔBОС is isosceles, since its lateral sides are radii. In this way:

BH = HC;

ВС = ВН + НС.

The height of an isosceles triangle divides it into two equal right-angled triangles.

Consider, for example, ΔBOН.

To calculate BH, we apply the Pythagorean theorem:

BO ^ 2 = BH ^ 2 + OH ^ 2;

BH ^ 2 = BO ^ 2 – OH ^ 2;

BH ^ 2 = 208 ^ 2 – 192 ^ 2 = 43264 – 36864 = 6400;

BH = √6400 = 80 cm.

BC = 80 + 80 = 160 cm.

SABCD = 160 160 = 25600 cm2.

Answer: the area of ​​the cutting plane is 25600 cm2.