The height of a regular quadrangular pyramid is equal to the root of 6 cm, and the lateral edge is inclined to the base

The height of a regular quadrangular pyramid is equal to the root of 6 cm, and the lateral edge is inclined to the base plane at an angle of 60 degrees. a) find the side edge of the pyramid b) find the area of the side surface of the pyramid

From the right-angled triangle OO1A, we determine the length of the hypotenuse OA and the leg O1A.

ОА = ОО1 / Sin60 = √6 / (√3 / 2) = 2 * √2 cm.

O1A = AO / Cos60 = 2 * √2 / 2 = √2 cm.

Since there is a square at the base of the pyramid, AC = 2 * AO1 = 2 * √2 cm.

From the right-angled triangle ACD we define the legs AD and CD.

AC ^ 2 = AD ^ 2 + CD ^ 2 = 2 * AD ^ 2.

8 = 2 * AD ^ 2.

АD ^ 2 = 8/2 = 4.

AD = 2 cm.

Then О1Н = АD / 2 = 2/2 = 1 cm.

Consider a right-angled triangle OO1H, and by the Pythagorean theorem, define the apothem OH.

OH ^ 2 = OH1 ^ 2 + O1H ^ 2 = (√6) 2 + 12 = 7.

OH = √7 cm.

Let us determine the area of ​​the side face of the ODC.

Sodc = DC * OH / 2 = 2 * √7 / 2 = √7 cm2.

Determine the area of ​​the side surface of the pyramid.

Sside = 4 * Sods = 4 * √7 cm2.

Answer: The lateral rib is 2 * √2 cm, the lateral surface area is 4 * √7 cm2.




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