In a regular quadrangular pyramid SABC with base ABC, a section is drawn through the midpoints

| June 15, 2021 | education

In a regular quadrangular pyramid SABC with base ABC, a section is drawn through the midpoints of edges AB and BC and vertex S. Find the area of this section if the side edge of the pyramid is 7 and the side of the base is 8.

Let’s draw the diagonals AC and BD at the base of the pyramid and determine their length.

BD ^ 2 = AC ^ 2 = AB ^ 2 + BC ^ 2 = 2 * AB ^ 2 = 2 * 64.

AC = BD = 8 * √2 cm.

Since points K and M are the middle of the edges AB and BC, then the segment KM is the middle line of the triangle ABC, then KM = AC / 2 = 8 * √2 / 2 = 4 * √2 cm.

The segments SK and SM are the medians of the side faces, and since the side faces are isosceles triangles, SK and SM are their heights. Then SM ^ 2 = SK ^ 2 = SB ^ 2 – BM ^ 2 = 49 – 16 = 33.

SM = SK = √33 cm.

Section SKM is an isosceles triangle, then its height SH divides the CM in half.

NM = KM / 2 = 2 * √2 cm.

Then SH ^ 2 = SM ^ 2 – HM ^ 2 = 33 – 8 = 25

SH = 5 cm.

Determine the cross-sectional area. Ssection = SH * KM / 2 = 5 * 2 * √2 / 2 = 5 * √2 cm2.

Answer: The cross-sectional area is 5 * √2 cm2.



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