The large diagonal section of a regular hexagonal pyramid is an equilateral triangle

| September 7, 2021 | education

The large diagonal section of a regular hexagonal pyramid is an equilateral triangle, the side of which is 18m. Calculate the volume of the pyramid.

ABC – axial section of a regular hexagonal pyramid. Since, according to the condition ABC is equilateral, its height CE is also the bisector and median of the triangle ABC, then AH = BH = AB / 2 = 18/2 = 9 cm.

In a right-angled triangle ACH, we define the CH leg.

CH ^ 2 = AC ^ 2 – AH ^ 2 = 18 ^ 2 – 9 ^ 2 = 324 – 81 = 243.

CH = 9 * √3 cm.

A regular hexagon lies at the base of the pyramid, then its large diagonals divide it into six equilateral triangles, the side of which is half the larger diagonal. AH = 9 cm.

Then the area of ​​ADH is equal to Sadn = AH ^ 2 * √3 / 4 = 81 * √3 / 4.

Then the area of ​​the hexagon is equal to: Sb = 6 * Sb = 243 * √3 / 2 cm2.

Let’s determine the volume of the pyramid:

V = Sbas * CH / 3 = (243 * √3 / 2) * (9 * √3) / 3 = 2187/2 = 1093.5 cm3.

Answer: The volume of the pyramid is 1093.5 cm3.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.