SABCD is a regular quadrangular pyramid, the length of each edge is 2 cm, point O is the midpoint

| July 14, 2021 | education

SABCD is a regular quadrangular pyramid, the length of each edge is 2 cm, point O is the midpoint of the edge DC. Construct a section of the pyramid with a plane passing through point O perpendicular to line SC. Calculate the perimeter of this section.

From point K, draw a straight line KM to the intersection with the BC. Let’s connect points O and M. Triangle OCM is our required section.

At the base of the pyramid lies a square with a side of 2 cm.

Let us determine the length of the diagonal BD of the square. ВD = ВС * √2 cm = 2 * √2 cm.

The segment ОМ is the middle line of the triangle ВСD, then ОМ = ВD / 2 = 2 * √2 / 2 = √2 cm.

The side faces of the pyramid are equilateral triangles. We will construct the height DH of the triangle SСD and determine its length.

DН = SC * √3 / 2 = 2 * √3 / 2 = √3 cm.

The segments DH and OK are perpendicular to SC, which means they are parallel to each other. Point O is the middle of CD, then point K is the middle of CH, and the segment OK is the middle line of the triangle CDH.

OK = DH / 2 = √3 / 2.

MK = OK = √3 / 2.

Then Psech = √2 + √3 / 2 + √3 / 2 = √2 + √3 cm.

Answer: The perimeter of the section is √2 + √3 cm.



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