Find the area of the diagonal section of a quadrangular pyramid if the side of the base is 8√2 and the side is 17 cm.

The diagonal section of a regular quadrangular pyramid is an isosceles triangle, the sides of which are equal to the side faces of the pyramid, and the base is equal to the diagonal of the base.

Let’s find the diagonal of the base of this pyramid.

The diagonal of the base is the hypotenuse of a right-angled triangle, the legs of which are equal to the sides of the base of the pyramid.

Suppose that the diagonal of the base is equal to x, then by the Pythagorean theorem we obtain:

(8√2) ² + (8√2) ² = x²,

64 * 2 + 64 * 2 = x²,

x² = 256,

x = 16.

Now we can find the height of the diagonal section.

The height of the section is the leg of a right-angled triangle, the second leg of which is half the diagonal, and the hypotenuse is equal to the lateral rib.

Let the height be equal to y. We get:

y² + (16/2) ² = 17²,

y² = 289 – 64,

y² = 225,

y = 15.

Therefore, the area of ​​the diagonal section is equal to:

S = (16 * 15) / 2 = 120 cm².