A regular quadrangular pyramid, all edges of which are equal to 18 cm, is intersected by a plane parallel

A regular quadrangular pyramid, all edges of which are equal to 18 cm, is intersected by a plane parallel to the base of the pyramid and passing through the middle of the lateral rib. Find the height and apothem of the resulting truncated pyramid.

Let’s start with a non-truncated pyramid. Let’s draw the diagonals AC and BD at the base of the pyramid, which is a square with a side of 18 cm.

Let us determine the length of the AS.

AC = AD * √2 = 18 * √2 cm.

The diagonals of the square, at the point of intersection, are halved. ОА = АС / 2 = 18 * √2 / 2 = 9 * √2 cm.

From the right-angled triangle AOS, by the Pythagorean theorem, we define the leg SO.

SO ^ 2 = SA ^ 2 – AO ^ 2 = 18 ^ 2 – (9 * √2) ^ 2 = 324 – 162.

SO = 9 * √2 cm.

Let us draw the apothem SH and, by the Pythagorean theorem from the triangle ASH, determine its length.

SH ^ 2 = AS ^ 2 – AH ^ 2 = 18 ^ 2 – 9 ^ 2 = 324 – 91 = 243.

SH = 9 * √3 cm.

Since the section of the pyramid is parallel to the base, then it cuts off a similar pyramid from the main pyramid, and so it passes through the middle of the lateral edge, the coefficient of similarity is two.

Then the height OO1 of the truncated pyramid is equal to half the height of SO, and the apothem HH1 is equal to half of SH.

OO1 = 9 * √2 / 2 = 4.5 * √2 cm.

НН1 = 9 * √3 / 2 = 4.5 * √3 cm.

Answer: The height of the truncated pyramid is 4.5 * √2 cm, the apothem is 4.5 * √3 cm.