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Surface area (2019)
Prism surface area
Is there a general formula? No, in general, no. You just need to look for the areas of the side faces and sum them up.
The formula can be written for straight prism:
Where is the perimeter of the base.
But it’s still much easier to add up all the areas in each specific case than to memorize additional formulas. For example, let's calculate the total surface of a regular hexagonal prism.
All side faces- rectangles. Means.
This was already shown when calculating the volume.
So we get:
Surface area of the pyramid
The general rule also applies to the pyramid:
Now let's calculate the surface area of the most popular pyramids.
Surface area of a regular triangular pyramid
Let the side of the base be equal and the side edge equal. We need to find and.
Let us now remember that
This is the area of a regular triangle.
And let’s remember how to look for this area. We use the area formula:
For us, “ ” is this, and “ ” is also this, eh.
Now let's find it.
Using the basic area formula and the Pythagorean theorem, we find
Attention: if you have a regular tetrahedron (i.e.), then the formula turns out like this:
Surface area of a regular quadrangular pyramid
Let the side of the base be equal and the side edge equal.
The base is a square, and that's why.
It remains to find the area of the side face
Surface area of a regular hexagonal pyramid.
Let the side of the base be equal and the side edge.
How to find? A hexagon consists of exactly six identical regular triangles. We have already looked for the area of a regular triangle when calculating the surface area of a regular triangle. triangular pyramid, here we use the found formula.
Well, we’ve already looked for the area of the side face twice.
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The area of the lateral surface of an arbitrary pyramid is equal to the sum of the areas of its lateral faces. It makes sense to give a special formula for expressing this area in the case of a regular pyramid. So, let us be given a regular pyramid, at the base of which lies a regular n-gon with side equal to a. Let h be the height of the side face, also called apothem pyramids. The area of one side face is equal to 1/2ah, and the entire side surface of the pyramid has an area equal to n/2ha. Since na is the perimeter of the base of the pyramid, we can write the found formula in the form:
Lateral surface area of a regular pyramid is equal to the product of its apothem and half the perimeter of the base.
Regarding area full surface , then we simply add the area of the base to the side one.
Inscribed and circumscribed sphere and sphere. It should be noted that the center of the sphere inscribed in the pyramid lies at the intersection of the bisector planes of the internal dihedral angles of the pyramid. The center of the sphere described near the pyramid lies at the intersection of planes passing through the midpoints of the edges of the pyramid and perpendicular to them.
Truncated pyramid. If a pyramid is cut by a plane parallel to its base, then the part enclosed between the cutting plane and the base is called truncated pyramid. The figure shows a pyramid; discarding its part lying above the cutting plane, we get a truncated pyramid. It is clear that the small discarded pyramid is homothetic to the large pyramid with the center of homothety at the apex. Similarity coefficient equal to the ratio heights: k=h 2 /h 1, or side edges, or other corresponding linear dimensions of both pyramids. We know that the areas of similar figures are related like squares of linear dimensions; so the areas of the bases of both pyramids (i.e. the area of the bases of the truncated pyramid) are related as
Here S 1 is the area of the lower base, and S 2 is the area of the upper base of the truncated pyramid. The lateral surfaces of the pyramids are in the same relationship. A similar rule exists for volumes.
Volumes of similar bodies are related like cubes of their linear dimensions; for example, the volumes of pyramids are related as the product of their heights and the area of the bases, from which our rule is immediately obtained. It has absolutely general character and it directly follows from the fact that volume always has a dimension of the third power of length. Using this rule, we derive a formula expressing the volume of a truncated pyramid through the height and area of the bases.
Let a truncated pyramid with height h and base areas S 1 and S 2 be given. If we imagine that it is continued to full pyramid, then the similarity coefficient between the full pyramid and the small pyramid is easy to find as the root of the ratio S 2 /S 1 . The height of a truncated pyramid is expressed as h = h 1 - h 2 = h 1 (1 - k). Now we have for the volume of a truncated pyramid (V 1 and V 2 denote the volumes of the full and small pyramids)
formula for the volume of a truncated pyramid
Let us derive the formula for the area S of the lateral surface of a regular truncated pyramid through the perimeters P 1 and P 2 of the bases and the length of the apothem a. We reason in exactly the same way as when deriving the formula for volume. Complementing the pyramid top part, we have P 2 = kP 1, S 2 =k 2 S 1, where k is the similarity coefficient, P 1 and P 2 are the perimeters of the bases, and S 1 and S 2 are the areas of the lateral surfaces of the entire resulting pyramid and its upper part, respectively. For the lateral surface we find (a 1 and a 2 are apothems of the pyramids, a = a 1 - a 2 = a 1 (1-k))
formula for the lateral surface area of a regular truncated pyramid
In this lesson:
- Problem 1. Find the total surface area of the pyramid
- Problem 2. Find the lateral surface area of a regular triangular pyramid
.
Note . If you need to solve a geometry problem that is not here, write about it in the forum. In tasks, instead of the "square root" symbol, the sqrt() function is used, in which sqrt is the symbol square root, and the radical expression is indicated in brackets. For simple radical expressions, the sign "√" can be used.
Problem 1. Find the total surface area of a regular pyramid
The height of the base of a regular triangular pyramid is 3 cm, and the angle between the side face and the base of the pyramid is 45 degrees.Find the total surface area of the pyramid
Solution.
At the base of a regular triangular pyramid lies equilateral triangle.
Therefore, to solve the problem, we will use the properties of a regular triangle: 
We know the height of the triangle, from where we can find its area.
h = √3/2a
a = h / (√3/2)
a = 3 / (√3/2)
a = 6 / √3
Whence the area of the base will be equal to:
S = √3/4 a 2
S = √3/4 (6 / √3) 2
S = 3√3
In order to find the area of the side face, we calculate the height KM. According to the problem, the angle OKM is 45 degrees.
Thus:
OK / MK = cos 45
Let's use the table of values of trigonometric functions and substitute known values.
OK / MK = √2/2
Let's take into account that OK is equal to the radius of the inscribed circle. Then
OK = √3/6a
OK = √3/6 * 6/√3 = 1
Then
OK / MK = √2/2
1/MK = √2/2
MK = 2/√2
The area of the side face is then equal to half the product of the height and the base of the triangle.
Sside = 1/2 (6 / √3) (2/√2) = 6/√6
Thus, the total surface area of the pyramid will be equal to
S = 3√3 + 3 * 6/√6
S = 3√3 + 18/√6
Answer: 3√3 + 18/√6
Problem 2. Find the lateral surface area of a regular pyramid
In a regular triangular pyramid, the height is 10 cm and the side of the base is 16 cm . Find the lateral surface area .Solution.
Since the base of a regular triangular pyramid is an equilateral triangle, AO is the radius of the circle circumscribed around the base.
(This follows from)
We find the radius of a circle circumscribed around an equilateral triangle from its properties
Whence the length of the edges of a regular triangular pyramid will be equal to:
AM 2 = MO 2 + AO 2
the height of the pyramid is known by condition (10 cm), AO = 16√3/3
AM 2 = 100 + 256/3
AM = √(556/3)
Each side of the pyramid is an isosceles triangle. We find the area of an isosceles triangle from the first formula presented below 
S = 1/2 * 16 sqrt((√(556/3) + 8) (√(556/3) - 8))
S = 8 sqrt((556/3) - 64)
S = 8 sqrt(364/3)
S = 16 sqrt(91/3)
Since all three faces of a regular pyramid are equal, the lateral surface area will be equal to
3S = 48 √(91/3)
Answer: 48 √(91/3)
Problem 3. Find the total surface area of a regular pyramid
The side of a regular triangular pyramid is 3 cm and the angle between the side face and the base of the pyramid is 45 degrees. Find the total surface area of the pyramid.
Solution.
Since the pyramid is regular, there is an equilateral triangle at its base. Therefore the area of the base is 
So = 9 * √3/4
In order to find the area of the side face, we calculate the height KM. According to the problem, the angle OKM is 45 degrees.
Thus:
OK / MK = cos 45
Let's take advantage
When preparing for the Unified State Exam in mathematics, students have to systematize their knowledge of algebra and geometry. I would like to combine all known information, for example, on how to calculate the area of a pyramid. Moreover, starting from the base and side edges to the entire surface area. If the situation with the side faces is clear, since they are triangles, then the base is always different.
How to find the area of the base of the pyramid?
It can be absolutely any figure: from an arbitrary triangle to an n-gon. And this base, in addition to the difference in the number of angles, can be a regular figure or an irregular one. In the Unified State Exam tasks that interest schoolchildren, there are only tasks with correct figures at the base. Therefore, we will talk only about them.
Regular triangle
That is, equilateral. The one in which all sides are equal and are designated by the letter “a”. In this case, the area of the base of the pyramid is calculated by the formula:
S = (a 2 * √3) / 4.
Square
The formula for calculating its area is the simplest, here “a” is again the side:
Arbitrary regular n-gon
The side of a polygon has the same notation. For the number of angles, the Latin letter n is used.
S = (n * a 2) / (4 * tg (180º/n)).
What to do when calculating the lateral and total surface area?
Because at the base lies correct figure, then all faces of the pyramid turn out to be equal. Moreover, each of them is an isosceles triangle, since lateral ribs are equal. Then in order to calculate lateral area pyramid, you will need a formula consisting of the sum of identical monomials. The number of terms is determined by the number of sides of the base.
The area of an isosceles triangle is calculated by the formula in which half the product of the base is multiplied by the height. This height in the pyramid is called apothem. Its designation is “A”. The general formula for lateral surface area is:
S = ½ P*A, where P is the perimeter of the base of the pyramid.
There are situations when the sides of the base are not known, but the side edges (c) and the flat angle at its apex (α) are given. Then you need to use the following formula to calculate the lateral area of the pyramid:
S = n/2 * in 2 sin α .
Task No. 1
Condition. Find the total area of the pyramid if its base has a side of 4 cm and the apothem has a value of √3 cm.
Solution. You need to start by calculating the perimeter of the base. Because this regular triangle, then P = 3*4 = 12 cm. Since the apothem is known, we can immediately calculate the area of the entire lateral surface: ½*12*√3 = 6√3 cm 2.
For the triangle at the base, you get the following area value: (4 2 *√3) / 4 = 4√3 cm 2.
To determine the entire area, you will need to add the two resulting values: 6√3 + 4√3 = 10√3 cm 2.
Answer. 10√3 cm 2.
Problem No. 2
Condition. There is a regular quadrangular pyramid. The length of the base side is 7 mm, the side edge is 16 mm. It is necessary to find out its surface area.
Solution. Since the polyhedron is quadrangular and regular, its base is a square. Once you know the area of the base and side faces, you can calculate the area of the pyramid. The formula for the square is given above. And for the side faces, all sides of the triangle are known. Therefore, you can use Heron's formula to calculate their areas.
The first calculations are simple and lead to the following number: 49 mm 2. For the second value, you will need to calculate the semi-perimeter: (7 + 16*2): 2 = 19.5 mm. Now you can calculate the area of an isosceles triangle: √(19.5*(19.5-7)*(19.5-16) 2) = √2985.9375 = 54.644 mm 2. There are only four such triangles, so when calculating the final number you will need to multiply it by 4.
It turns out: 49 + 4 * 54.644 = 267.576 mm 2.
Answer. The desired value is 267.576 mm 2.
Problem No. 3
Condition. The right one quadrangular pyramid you need to calculate the area. The side of the square is known to be 6 cm and the height is 4 cm.
Solution. The easiest way is to use the formula with the product of perimeter and apothem. The first value is easy to find. The second one is a little more complicated.
We will have to remember the Pythagorean theorem and consider It is formed by the height of the pyramid and the apothem, which is the hypotenuse. The second leg is equal to half the side of the square, since the height of the polyhedron falls into its middle.
The required apothem (hypotenuse of a right triangle) is equal to √(3 2 + 4 2) = 5 (cm).
Now you can calculate the required value: ½*(4*6)*5+6 2 = 96 (cm 2).
Answer. 96 cm 2.
Problem No. 4
Condition. Dana correct side its bases are 22 mm, side ribs are 61 mm. What is the lateral surface area of this polyhedron?
Solution. The reasoning in it is the same as that described in task No. 2. Only there was given a pyramid with a square at the base, and now it is a hexagon.
First of all, the base area is calculated using the above formula: (6*22 2) / (4*tg (180º/6)) = 726/(tg30º) = 726√3 cm 2.
Now you need to find out the semi-perimeter of an isosceles triangle, which is the side face. (22+61*2):2 = 72 cm. All that remains is to use Heron’s formula to calculate the area of each such triangle, and then multiply it by six and add it to the one obtained for the base.
Calculations using Heron's formula: √(72*(72-22)*(72-61) 2)=√435600=660 cm 2. Calculations that will give the lateral surface area: 660 * 6 = 3960 cm 2. It remains to add them up to find out the entire surface: 5217.47≈5217 cm 2.
Answer. The base is 726√3 cm 2, the side surface is 3960 cm 2, the entire area is 5217 cm 2.
Surface area of the pyramid. In this article we will look at problems with regular pyramids. Let me remind you that a regular pyramid is a pyramid whose base is a regular polygon, the top of the pyramid is projected into 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 of finding elements (height, base edge, side edge) was raised.
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 of a regular quadrangular pyramid are 72, the side edges are 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.
The base is known, it is equal to six. Let's find the height. Let's consider right triangle(it's highlighted in yellow):
One leg is equal to 4, since this is the height of the pyramid, the other is equal to 3, since it is equal to half the edge of the base. We can find the hypotenuse using the Pythagorean theorem:
This means that the area of the lateral surface of the pyramid is:
Thus, the surface area of the entire pyramid is:
Answer: 96
27069. The sides of the base of a regular quadrangular pyramid are equal to 10, the side edges are equal to 13. Find the surface area of this pyramid.
27070. The sides of the base of a regular hexagonal pyramid are equal to 10, the side edges are equal to 13. Find the lateral surface area of this pyramid.
There are also formulas for the lateral surface area of a regular pyramid. In a regular pyramid, the base is an orthogonal projection of the lateral surface, therefore:
P- base perimeter, l- apothem of the pyramid
*This formula is based on the formula for the area of a triangle.
If you want to learn more about how these formulas are derived, don’t miss it, follow the publication of articles.That's all. Good luck to you!
Sincerely, Alexander Krutitskikh.
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