In preparing for the exam in mathematics, students have to systematize knowledge of algebra and geometry. I would like to combine all the known information, for example, on how to calculate the area of the pyramid. And starting from the base and side faces to the area of the entire surface. If the situation with the side faces is clear, since they are triangles, then the base is always different.
What to do when finding the area of the base of the pyramid?
It can be absolutely any figure: from an arbitrary triangle to an n-gon. And this basis, in addition to the difference in the number of angles, can be the right figure or the wrong one. In tasks of interest for students on the exam, there are only tasks with the correct figures at the base. Therefore, we will only talk about them.
Right triangle
That is equilateral. One in which all sides are equal and marked with the letter “a”. In this case, the base area 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:
S = a 2 .
Arbitrary regular n-gon
The side of the polygon has the same notation. The latin letter n is used for the number of angles.
S = (n * a 2 ) / (4 * tg (180º / n)).
What to do when calculating the area of the lateral and full surface?
Since the base is the correct figure, then all the faces of the pyramid are equal. Moreover, each of them is an isosceles triangle, since the side ribs are equal. Then, in order to calculate the lateral area of the pyramid, you 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 the apothem. Its designation is “A”. The general formula for the lateral surface area is as follows:
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 side ribs (c) and a flat angle at its apex (α) are given. Then it is supposed to use such a formula to calculate the lateral area of the pyramid:
S = n / 2 * in 2 sin α .
Task number 1
Condition. Find the total area of the pyramid if it is based on an equilateral triangle with a side of 4 cm, and the apothem has a value of √3 cm.
Decision. It must begin with the calculation of the perimeter of the base. Since this is a regular triangle, 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, the following area value is obtained: (4 2 * √3) / 4 = 4√3 cm 2 .
To determine the entire area, you need to add two resulting values: 6√3 + 4√3 = 10√3 cm 2 .
Answer. 10√3 cm 2 .
Task number 2
Condition . There is a regular quadrangular pyramid. The length of the side of the base is 7 mm, the side rib is 16 mm. It is necessary to find out its surface area.
Decision. Since the polyhedron is quadrangular and regular, then at its base lies a square. Having learned the area of the base and the side faces, it will be possible to calculate the area of the pyramid. The formula for the square is given above. And at the side faces, all sides of the triangle are known. Therefore, you can use the Heron formula to calculate their area.
The first calculations are simple and lead to such a number: 49 mm 2 . For the second value, you need to calculate the half-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 .
The answer . The desired value is 267.576 mm 2 .
Task number 3
Condition . The correct quadrangular pyramid needs to calculate the area. The side of the square is known in it - 6 cm and height - 4 cm.
Decision. The easiest way is to use the formula with the product of the perimeter and apothem. The first value is easy to find. The second is a bit more complicated.
We will have to recall the Pythagorean theorem and consider a right triangle. It is formed by the height of the pyramid and apothema, which is the hypotenuse. The second leg is equal to half the side of the square, since the height of the polyhedron falls in its middle.
The desired apothem (hypotenuse of a right triangle) is √ (3 2 + 4 2 ) = 5 (cm).
Now you can calculate the desired value: ½ * (4 * 6) * 5 + 6 2 = 96 (cm 2 ).
Answer. 96 cm 2 .
Task number 4
Condition. Given a regular hexagonal pyramid. The sides of its base are 22 mm, the side ribs are 61 mm. What is the lateral surface area of this polyhedron?
Decision. The reasoning in it is the same as was described in task No. 2. Only there was given a pyramid with a square at the base, and now it is a hexagon.
The first step is to calculate the base area using the above formula: (6 * 22 2 ) / (4 * tg (180º / 6)) = 726 / (tg30º) = 726√3 cm 2 .
Now you need to find out the half-perimeter of an isosceles triangle, which is a side face. (22 + 61 * 2): 2 = 72 cm. It remains 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 according to the Heron formula: √ (72 * (72-22) * (72-61) 2 ) = √435600 = 660 cm 2 . Calculations that give the area of the lateral surface: 660 * 6 = 3960 cm 2 . It remains to fold them to find out the entire surface: 5217.47≈5217 cm 2 .
Answer. The bases are 726√3 cm 2 , the lateral surface is 3960 cm 2 , the entire area is 5217 cm 2 .