The pyramid, at the base of which lies regular hexagon, A sides are formed regular triangles, called hexagonal.
This polyhedron has many properties:
- All sides and angles of the base are equal to each other;
- All edges and dihedral coals of the pyramid are also equal to each other;
- The triangles forming the sides are the same, respectively, they have the same areas, sides and heights.
To calculate the correct area hexagonal pyramid applies standard formula lateral surface area of a hexagonal pyramid:
where P is the perimeter of the base, a is the length of the apothem of the pyramid. In most cases it is possible to calculate lateral area according to this formula, but sometimes you can use another method. Because side faces pyramids formed equal triangles, you can find the area of one triangle, and then multiply it by the number of sides. There are 6 of them in a hexagonal pyramid. But this method can also be used when calculating. Let's consider an example of calculating the lateral surface area of a hexagonal pyramid.
Let a regular hexagonal pyramid be given, in which the apothem is a = 7 cm, the side of the base is b = 3 cm. Calculate the area of the lateral surface of the polyhedron.
First, let's find the perimeter of the base. Since the pyramid is regular, there is a regular hexagon at its base. This means that all its sides are equal, and the perimeter is calculated by the formula:
Substitute the data into the formula:
Now we can easily find the lateral surface area by substituting the found value into the basic formula: 
Also important is the search for the base area. The formula for the area of the base of a hexagonal pyramid is derived from the properties of a regular hexagon: 
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Regular hexagonal pyramid- a pyramid with a regular hexagon at its base.
Designations
- $SABCDEF$ - regular hexagonal pyramid
- $O$ - center of the base of the pyramid
- $a$ - side length of the base of the pyramid
- $h$ - length of the side edge of the pyramid
- $S_(\text(base))$ - area of the base of the pyramid
- $V_(\text(pyramids))$ - volume of the pyramid
Area of the base of the pyramid
At the base of the pyramid there is a regular hexagon with side $a$. According to the properties of a regular hexagon, the area of the base of the pyramid is $$ S_(\text(basic))=\frac(3\sqrt(3))(2)\cdot a^2 $$Regular hexagon at the base of the pyramid
According to the properties of a regular hexagon, triangles AOB, BOC, COD, DOE, EOF, FOA are regular triangles. It follows that $$ AO=OD=EO=OB=CO=OF=a $$ We draw a segment AE intersecting the segment CF at point M. The triangle AEO is isosceles, in it $AO=OE=a,\ \angle EOA =120^(\circ)$. Based on the properties of an isosceles triangle $$ AE=a\cdot\sqrt(2(1-\cos EOA))=\sqrt(3)\cdot a $$ Similarly, we come to the conclusion that $ AC=CE=\sqrt(3 )\cdot a $, $FM=MO=\frac(1)(2)\cdot a$. Finding $SO$
The straight line $SO$ is the height of the pyramid, so $\angle SOF=90^(\circ)$. Triangle $SOF$ is right-angled, with $FO=a,\FS=h$. By properties right triangle$$ SO=\sqrt(FS^2-FO^2)=\sqrt(h^2-a^2) $$Volume of the pyramid
The volume of a pyramid is calculated as one third of the product of the area of its base and its height. The height of a regular pyramid is the segment $SO$. At the base of a regular hexagonal prism there is a regular hexagon, the area of which is known to us. We get $$ V_(\text(pyramids))=\frac(1)(3)\cdot S_(\text(main))\cdot SO=\frac(\sqrt(3))(2)\cdot a ^2 \cdot \sqrt(h^2-a^2) $$Finding $ST$ and $TO$
Let point $T$ be the midpoint of edge $AF$. Triangle $AOF$ is regular, therefore, according to Surface area of the pyramid. In this article we will look at tasks with regular pyramids. Let me remind you that a regular pyramid is a pyramid whose base is regular polygon, the vertex of the pyramid is projected to the center of this polygon.
The side face of such a pyramid is an isosceles triangle.The altitude of this triangle drawn from the vertex of a regular pyramid is called apothem, SF - apothem:
In the type of problem presented below, you need to find the surface area of the entire pyramid or the area of its lateral surface. The blog has already discussed several problems with regular pyramids, where the question was about finding the elements (height, base edge, side edge).
IN Unified State Exam assignments As a rule, regular triangular, quadrangular and hexagonal pyramids are considered. I haven’t seen any problems with regular pentagonal and heptagonal pyramids.
The formula for the area of the entire surface is simple - you need to find the sum of the area of the base of the pyramid and the area of its lateral surface:
Let's consider the tasks:
The sides of the base are correct quadrangular pyramid are equal to 72, lateral ribs are equal to 164. Find the surface area of this pyramid.
The surface area of the pyramid is equal to the sum of the areas of the lateral surface and the base:
*The lateral surface consists of four triangles of equal area. The base of the pyramid is a square.
We can calculate the area of the side of the pyramid using:
Thus, the surface area of the pyramid is:
Answer: 28224
The sides of the base of a regular hexagonal pyramid are equal to 22, the side edges are equal to 61. Find the lateral surface area of this pyramid.
The base of a regular hexagonal pyramid is a regular hexagon.
The lateral surface area of this pyramid consists of six areas of equal triangles with sides 61,61 and 22:
Let's find the area of the triangle using Heron's formula:
Thus, the lateral surface area is:
Answer: 3240
*In the problems presented above, the area of the side face could be found using another triangle formula, but for this you need to calculate the apothem.
27155. Find the surface area of a regular quadrangular pyramid whose base sides are 6 and whose height is 4.
In order to find the surface area of the pyramid, we need to know the area of the base and the area of the lateral surface:
The area of the base is 36 since it is a square with side 6.
The lateral surface consists of four faces, which are equal triangles. In order to find the area of such a triangle, you need to know its base and height (apothem):
*The area of a triangle is equal to half the product of the base and the height drawn to this base.