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

| September 16, 2021 | education

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

Since the lengths of all edges are 18 cm, the side faces of the pyramid are equilateral triangles, and its base is square.

The height BH of an equilateral triangle is equal to: RH = CD * √3 / 2 = 18 * √3 / 2 = 9 * √3 cm.

The diagonals AC and BD at the base of the pyramid at point O are divided in half, point H is the middle of the side CD, then the segment OH is the middle line of the triangle ACD, and then OH = AD / 2 = 18/2 = 9 cm.

In a right-angled triangle RON, according to the Pythagorean theorem, PO ^ 2 = PH ^ 2 – OH ^ 2 = 243 – 81 = 162. PO = 9 * √2 cm.

Since the section passes through the middle of the lateral ribs, HH1 = PH / 2 = 9 * √3 / 2 = 4.5 * √3 cm, OO1 = PO / 2 = 9 * √2 / 2 = 4.5 * √2 cm.

Answer: Apothem of the pyramid is 4.5 * √3 cm, the height of the pyramid is 4.5 * √2 cm.



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