In a regular quadrangular pyramid MABCD, the side edge is 18 and the height of the pyramid is 8 roots of 5.

| August 9, 2021 | education

In a regular quadrangular pyramid MABCD, the side edge is 18 and the height of the pyramid is 8 roots of 5. Find the cross-sectional area of this pyramid by a plane passing through line AC and midpoint L of edge MB.

Let’s draw a diagonal BD at the base of the pyramid. Since the MO is the height of the pyramid, the triangle of the MOF is rectangular.

By the Pythagorean theorem, in the right-angled triangle OBM, we determine the length of the leg OB.

ОВ ^ 2 = ВМ ^ 2 – ОМ ^ 2 = 324 – 320 = 4.

OB = 2 cm.

Since there is a square at the base of the pyramid, point O divides its diagonals in half, then AC = BD = 2 * OB = 2 * 2 = 4 cm.

Consider a triangle DМВ. Point L is the midpoint of the edge BM, point O is the midpoint of side BD, then the segment OL is the midline of the triangle DMB. Then OL = DМ / 2 = 18/2 = 9 cm.

Then the cross-sectional area of ​​the ACL will be equal to: Sacl = AC * OL / 2 = 4 * 9/2 = 18 cm2.

Answer: The cross-sectional area is 18 cm2.



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