The height of a regular triangular pyramid is 8 cm. The radius of a circle circumscribed about its base is (8√3)

| September 14, 2021 | education

The height of a regular triangular pyramid is 8 cm. The radius of a circle circumscribed about its base is (8√3) cm. Calculate: a) The length of the side edge of the pyramid. b) the area of the lateral surface of the pyramid.

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 = 8 * √3 cm.

In a right-angled triangle AOD, we determine the length of the hypotenuse AD.

AD ^ 2 = AO ^ 2 + DO ^ 2 = 192 + 64 = 256.

AD = 16 cm.

Since the point O, by the property of the medians, divides it in the ratio of 2/1, then OH = AO / 2 = 8 * √3 / 2 = 4 * √3 cm.

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

DH ^ 2 = DO ^ 2 + OH ^ 2 = 64 + 48 = 112.

DН = 4 * √7 cm.

The height of the ABC triangle is equal to: AH = AO + OH = 8 * √3 + 4 * √3 = 12 * √3 cm.

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

BC = 2 * AH / √3 = 2 * 12 * √3 / √3 = 24 cm.

The area of ​​the triangle ВСD is equal to: Sвсд = ВС * DН / 2 = 24 * 4 * √7 / 2 = 48 * √7 cm2.

Then S side = 3 * Svsd = 3 * 48 * √7 = 144 * √7 cm2.

Answer: The lateral rib is 16 cm, the lateral surface area is 144 * √7 cm2.



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