In a regular quadrangular pyramid SABCD with vertex S, the edges are known: AB = 1, SD = 2.

| August 13, 2021 | education

In a regular quadrangular pyramid SABCD with vertex S, the edges are known: AB = 1, SD = 2. Point M is the midpoint of edge SC. Find the cross-sectional area of this pyramid passing through points M, A and D

The section of the pyramid will be an isosceles trapezoid SCMV, in which the larger base is equal to the side of the base of the square, AD = AB = 1 cm. The smaller base of the KM section is the middle line of the triangle SBC, then KM = BC / 2 = 1/2 cm.

The lateral face of the DM section is the median of the isosceles triangle SСD.

Then МD = (√2 * SD ^ 2 + 2 * SD ^ 2 – SC ^ 2) / 2 = (√2 * 2 + 2 * 1 – 4) / 2 = √6 / 2 cm.

Let’s draw the height of the MH section.

The length of the segment is DH = (AD – KM) / 2 = (1 – 0.5) / 2 = 1/4 cm.

Then, by the Pythagorean theorem, МН ^ 2 = МD ^ 2 – DH ^ 2 = (6/4) – (1/16) = 23/16.

MH = √23 / 4 cm.

Then Ssection = (AD + KM) * MH / 2 = ((1 + 1/2) * √23 / 4) / 2 = (3/4) * (√23) / 4 = 3 * √23 / 16 cm2 …

Answer: The cross-sectional area is 3 * √23 / 16 cm2.



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