The height of a regular quadrangular pyramid is 80 cm, the side of the base is 120 cm. Calculate the cross-sectional

| May 17, 2021 | education

The height of a regular quadrangular pyramid is 80 cm, the side of the base is 120 cm. Calculate the cross-sectional area passing through the center of the base parallel to the side face of the pyramid.

Since the pyramid is correct, ABCD is a square, and its side is 120 cm. Let us determine the length of the diagonal of the square. AC = AB * √2 = 120 * √2, then half of the diagonal O1A = AC / 2 = 60 * √2 cm.

Their right-angled triangle OO1A, by the Pythagorean theorem, we determine the length OA.

OA ^ 2 = OO1 ^ 2 + O1A ^ 2 = 80 ^ 2 + (60 * √2) ^ 2 = 6400 + 7200 = 13600.

ОА = ОВ = ОВ = ОD = 20 * √34 cm.

In the triangle AOD, the segment KM is its middle line, since it is parallel to OD and point K is the middle of AD, then KM = PH = OD / 2 = 10 * √34 cm.

In triangle AOB, segment MH is also the middle line of this triangle. MH = AB / 2 = 120/2 = 60 cm.

The section of the CMNR is an isosceles trapezoid. Let’s draw the height MM1 in it. The М1К segment is equal to the half-difference of the trapezoid bases.

М1К = (КР – МН) / 2 = (120 – 60) / 2 = 30 cm.

By the Pythagorean theorem, from the triangle KMM1 we determine the height of MM1.

MM1 ^ 2 = KM ^ 2 – M1K ^ 2 = (10 * √34) ^ 2 – 30 ^ 2 = 3400 – 900 = 2500.

MM1 = 50 cm.

Let’s determine the cross-sectional area of ​​the CMNR.

S = (KR + MN) * MM1 / 2 = (120 + 60) * 50/2 = 4500 cm2.

Answer: The cross-sectional area is 4500 cm2.



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