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

| September 10, 2021 | education

In a regular triangular pyramid, the side faces are inclined to the base plane at an angle of 60 °. The distance from the center of the base to the side face is 2 cm. Find the area of the side surface of the pyramid.

In a right-angled triangle НOK, the length of the segment OК, by condition, is 2 cm.

Then Sin60 = OK / OH.

OH = OK / Sin60 = 2 / (√3 / 2) = 4 / √3 cm.

In a right-angled triangle DOH, we determine the length of the hypotenuse DН.

Cos60 = OH / DH.

DN = OH / Cos60 = (4 / √3) / (1/2) = 8 / √3 cm.

In triangle ABC, the length of the segment BO, by the property of intersecting medians, BO = 2 * OH = 2 * 4 / √3 cm.Then BH = BO + OH = 8 / √3 + 4 / √3 = 12 / √3 = 4 * √3 cm.

Height BH = AC * √3 / 2.

AC = 2 * BH / √3 = 2 * 4 * √3 / √3 = 8 cm.

Sac = AC * DH / 2 = 8 * (8 / √3) / 2 = (64 / √3) / 2 = 32 / √3 cm2.

Sside = 3 * Sacd = 3 * 32 / √3 = 32 * √3 cm2.

Answer: The lateral surface area is 32 * √3 cm2.



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