In a regular quadrangular pyramid, the side faces are inclined to the base at an angle of 60.

| August 17, 2021 | education

In a regular quadrangular pyramid, the side faces are inclined to the base at an angle of 60. The distance from the middle of the base side to the opposite face is 4√3. Find the area of the side surface of the pyramid.

Since MK is the shortest distance from point M to the lateral face of PBC, the triangle MKH is rectangular. Then MH = MK / Sin60 = 4 * √3 / (√3 / 2) = 8 cm.

The MPH triangle is isosceles, since MP = HP as the apothem of a regular pyramid, and since one of its angles is 60, the triangle is equilateral, MP = HP = MH = 8 cm.

Quadrangle AMHB is a rectangle, then AB = MH = 8 cm.

Let us determine the area of the side face of the PBC. Srvs = BC * HP / 2 = 8 * 8/2 = 32 cm2.

Then S side = 4 * Srvs = 4 * 32 = 128 cm2.

Answer: The lateral surface area is 128 cm2.



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