The height of a regular quadrangular pyramid is equal to the root of six cm, and the lateral rib is inclined to the plane

| February 11, 2021 | education

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

Consider a right-angled triangle AOO1 and through the angle A and the height OO1 define the segments OA and O1A.

ОА = ОО1 / Sin600 = √6 / (√3 / 2) = 2 * √2 cm.

O1A = AO / Cos600 = 2 * √2 / 2 = √2 cm.

Since the pyramid is correct, then at its base there is a square, and its diagonals at the point of intersection are divided in half, then AC = 2 * AO1 = 2 * √2 cm.

From the right-angled triangle of ACD, we determine the legs of BP and CD.

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

8 = 2 * AD ^ 2.

BP ^ 2 = 8/2 = 4.

BP = 2 cm.

Then О1Н = AD / 2 = 2/2 = 1 cm.

In a right-angled triangle OO1H, according to the Pythagorean theorem, we define the apothem of the pyramid OH.

OH ^ 2 = OH1 ^ 2 + O1H ^ 2 = (√6) ^ 2 + 1 ^ 2 = 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 edge is 2 * √2 cm, the lateral surface area is 4 * √7 cm2.



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