In a regular quadrangular pyramid PABCD, the base side is 10, the height is PH 5√6.

| September 26, 2021 | education

In a regular quadrangular pyramid PABCD, the base side is 10, the height is PH 5√6. a) find the angle of inclination of the side edge of the pyramid to the plane of its base b) Construct a section of the pyramid with a plane parallel to the base plane ABCD and passing through point M at the height PH (PM: MH = 4: 1) and find the area of this section

At the base of the pyramid lies the square ABCD.

By the Pythagorean theorem, we determine the length of the hypotenuse AC in a right-angled triangle ACD. AC ^ 2 = AD ^ 2 + CD ^ 2 = 100 + 100 = 2 * 100.

AC = 10 * √2 cm, then AH = AC / 2 = 5 * √2 cm.

In a right-angled triangle ARN, tgPAH = PH / AH = 5 * √6 / 5 * √2 = √3.

Angle PAH = 60.

Triangles APH and A1PM are rectangular and similar in acute angle, then PH / PM = AH / A1M.

А1М = РМ * АН / РН = 4 * 5 * √2 / 5 = 4 * √2 cm.

In a right-angled triangle A1MD1, A1M = D1M = 4 * √2, then A1D1 ^ 2 = A1M ^ 2 + D1M ^ 2 = 3 ^ 2 + 3 ^ 2 = 64.

A1D1 = 8 cm.

Then Ssection = 8 * 8 = 64 cm2.

Answer: The angle of inclination is 60, the cross-sectional area is 64 cm2.



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