The height of a regular triangular pyramid is equal to a√3, the radius of a circle circumscribed about its base is 2a.

| April 30, 2021 | education

The height of a regular triangular pyramid is equal to a√3, the radius of a circle circumscribed about its base is 2a. Find: a) the apothem of the pyramid b) the angle between the lateral face and the base c) the area of the lateral surface

Point O, the point of intersection of the medians of the triangle ABC, is the center of the circumcircle and the inscribed circle. Then R = AO = 2 * a cm.

Since the point O, by the property of the medians, divides it in the ratio 2/1, then OH = AO / 2 = 2 * a / 2 = a cm.

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

DH ^ 2 = OH ^ 2 + DO ^ 2 = a^2 + a^2 * 3 = 4 * a2.

DН = 2 * a cm.

In a right-angled triangle DOH, the length of the hypotenuse DН is twice as long as the leg OH, then the angle ОDН = 30, and therefore the angle DHO = 30.

The height of the triangle ABC is equal to: AH = AO + OH = 2 * a + a = 3 * a cm.

AH = b * √3 / 2, where b is the side of an equilateral triangle.

BC = 2 * AH / √3 = 2 * 3 * a / √3 = 6 * a / √3 cm = 2 * a * √3 cm.

The area of ​​the triangle ВСD is equal to: Sвсд = ВС * DН / 2 = 2 * a * √3 * 2 * a / 2 = 2 * a^2 * √3 cm2.

Then S side = 3 * Svsd = 3 * 2 * √3 * a^2 = 6 * √3 * a2 cm2.

Answer: The length of the apothem is 2 * a cm, the angle is 60, the lateral surface area is 6 * √3 * a2 cm2.



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