How to find the lateral surface area of a pyramid: formulas, example problem. Find the surface area of a regular triangular pyramid S side 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
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Note . If you need to solve a geometry problem that is not here, write about it in the forum. In problems, instead of the "square root" symbol, the sqrt() function is used, in which sqrt is the square root symbol, and the radicand expression is indicated in parentheses. 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 an 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 = √3/2 a
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 the 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
A cylinder is a geometric body bounded by two parallel planes and a cylindrical surface. In the article we will talk about how to find the area of a cylinder and, using the formula, we will solve several problems as an example.
A cylinder has three surfaces: a top, a base, and a side surface.
The top and base of a cylinder are circles and are easy to identify.
It is known that the area of a circle is equal to πr 2. Therefore, the formula for the area of two circles (the top and base of the cylinder) will be πr 2 + πr 2 = 2πr 2.
The third, side surface of the cylinder, is the curved wall of the cylinder. In order to better imagine this surface, let's try to transform it to get a recognizable shape. Imagine that the cylinder is an ordinary tin can that does not have a top lid or bottom. Let's make a vertical cut on the side wall from the top to the bottom of the can (Step 1 in the figure) and try to open (straighten) the resulting figure as much as possible (Step 2).
After the resulting jar is fully opened, we will see a familiar figure (Step 3), this is a rectangle. The area of a rectangle is easy to calculate. But before that, let's return for a moment to the original cylinder. The vertex of the original cylinder is a circle, and we know that the circumference is calculated by the formula: L = 2πr. It is marked in red in the figure.
When the side wall of the cylinder is fully opened, we see that the circumference becomes the length of the resulting rectangle. The sides of this rectangle will be the circumference (L = 2πr) and the height of the cylinder (h). The area of a rectangle is equal to the product of its sides - S = length x width = L x h = 2πr x h = 2πrh. As a result, we received a formula for calculating the area of the lateral surface of the cylinder.
Formula for the lateral surface area of a cylinder
S side = 2πrh
Total surface area of a cylinder
Finally, if we add the area of all three surfaces, we get the formula for the total surface area of a cylinder. The surface area of a cylinder is equal to the area of the top of the cylinder + the area of the base of the cylinder + the area of the side surface of the cylinder or S = πr 2 + πr 2 + 2πrh = 2πr 2 + 2πrh. Sometimes this expression is written identical to the formula 2πr (r + h).
Formula for the total surface area of a cylinder
S = 2πr 2 + 2πrh = 2πr(r + h)
r – radius of the cylinder, h – height of the cylinder
Examples of calculating the surface area of a cylinder
To understand the above formulas, let’s try to calculate the surface area of a cylinder using examples.
1. The radius of the base of the cylinder is 2, the height is 3. Determine the area of the lateral surface of the cylinder.
The total surface area is calculated using the formula: S side. = 2πrh
S side = 2 * 3.14 * 2 * 34.6. Total ratings received: 990.
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?
Since the base is a regular figure, all faces of the pyramid are equal. Moreover, each of them is an isosceles triangle, since the side edges are equal. Then, in order to calculate the lateral area of the 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. Since this is a 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 will be able to 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. For a regular 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. The correct side is given. The sides of its base are 22 mm, the side edges 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 cm2, the side surface is 3960 cm2, the entire area is 5217 cm2.
In a regular triangular pyramid SABC R- middle of the rib AB, S- top.
It is known that SR = 6, and the lateral surface area is equal to 36
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Find the length of the segment B.C..
Let's make a drawing. In a regular pyramid, the side faces are isosceles triangles.
Line segment S.R.- the median lowered to the base, and therefore the height of the side face.
The lateral surface area of a regular triangular pyramid is equal to the sum of the areas
three equal side faces S side = 3 S ABS. From here S ABS = 36: 3 = 12- area of the face.
The area of a triangle is equal to half the product of its base and height
S ABS = 0.5 AB SR. Knowing the area and height, we find the side of the base AB = BC.
12 = 0.5 AB 6
12 = 3 AB
AB = 4
Answer: 4
You can approach the problem from the other end. Let the base side AB = BC = a.
Then the area of the face S ABS = 0.5 AB SR = 0.5 a 6 = 3a.
The area of each of the three faces is equal to 3a, the area of the three faces is equal 9a.
According to the conditions of the problem, the area of the lateral surface of the pyramid is 36.
S side = 9a = 36.
From here a = 4.
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.
Concerning total surface area, 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. The similarity coefficient is equal to the ratio of 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 relation. 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 is of a completely general nature and 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 extended to a full pyramid, then the coefficient of similarity between the full pyramid and the small pyramid can easily be found 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. We supplement the pyramid with the upper 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 the upper part accordingly. 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
