The lateral edge of a regular 4-angled pyramid, equal to 12, forms an angle of 60 degrees with

| July 6, 2021 | education

The lateral edge of a regular 4-angled pyramid, equal to 12, forms an angle of 60 degrees with the base plane. Find the Side of the Pyramid.

The MO height and the MB edge form a right-angled triangle, in which we determine the length of the OB leg.

Cos60 = OB / MB.

OB = MB * Cos60 = 12 * 1/2 = 6 cm.

Since ABSD is a square, then OB is half of the diagonal of BD, then BD = OB * 2 = 2 * 6 = 12 cm.

In a right-angled triangle ABD AB = BD, then 2 * AB ^ 2 = BD ^ 2.

AB ^ 2 = BD ^ 2/2 = 144/2.

AB = 12 / √2 = 6 * √2 cm.

Let us draw the height of the MH of the lateral face of the AВM, which is also its median. Then BH = AH = AB / 2 = 3 * √2 cm.

MH ^ 2 = BM ^ 2 – BH ^ 2 = 144 – 18 = 126.

MH = 3 * √14 cm.

Let us determine the area of ​​the lateral face of the ABM.

Savm = AB * MH / 2 = 6 * √2 * 3 * √14 / 2 = 9 * √28 = 18 * √7 cm2.

Then Sside = 4 * Sawm = 4 * 18 * √7 = 72 * √7 cm2.

Answer: The lateral surface area is 72 * √7 cm2.



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