How to calculate the area of a triangle
The triangle is a well-known figure. And this, despite the rich variety of its forms. Rectangular, equilateral, acute-angled, isosceles, obtuse-angled. Each one is different. But for anyone it is required
Common for all triangles formulas that use the length of the sides or heights
The designations adopted in them: the parties - a, b, c; heights on corresponding sidesa, nat, nfrom.
1. The area of a triangle is calculated as the product of ½, the sides and the height lowered on it. S = ½ * a * na. Similarly, you should write the formulas for the other two sides.
2. Heron's formula, in which the semi-perimeter appears (it is usually denoted by the small letter p, in contrast to the full perimeter). The semi-perimeter must be counted as follows: put all sides together and divide them by 2. The semi-perimeter formula: p = (a + b + c) / 2. Then the equality for the area of the figure looks like this: S = √ (p * (p - a) * ( p - c) * (p - c)).
3. If you do not want to use a semi-perimeter, then a formula is useful in which only the lengths of the sides are present: S = ¼ * √ (((a + b + c) * (b + c - a) * (a + c - b) * (a + b - c)).It is somewhat longer than the previous one, but it will help out, if you forgot how to find a semi-perimeter.
General formulas in which the angles of a triangle appear
The designations that are required to read the formulas: α, β, γ - angles. They lie opposite the sides a, b, c, respectively.
1. On it, half the product of two sides and the sine of the angle between them is equal to the area of the triangle. That is: S = ½ a * b * sin γ. Similarly, formulas should be written for the other two cases.
2. The area of a triangle can be calculated by one side and three known angles. S = (a2* sin β * sin γ) / (2 sin α).
3. There is also a formula with one well-known side and two corners adjacent to it. It looks like this: s = with2/ (2 (ctg α + ctg β)).
The last two formulas are not the easiest. Remember them quite difficult.
General formulas for the situation when the radii of inscribed or circumscribed circles are known.
Additional designations: r, R - radii. The first is used for the radius of the inscribed circle. The second is for the described.
1. The first formula by which the area of a triangle is calculated is connected with a semi-perimeter. S = p * r. In another way, it can be written as follows: S = ½ r * (a + b + c).
2. In the second case, you will need to multiply all sides of the triangle and divide them by the quadruple radius of the circumcircle.In literal terms, it looks like this: S = (a * c * s) / (4R).
3. The third situation allows you to do without knowing the sides, but the values of all three angles will be required. S = 2 R2* sin α * sin β * sin γ.
Special case: right triangle
This is the simplest situation, since it requires knowledge of only the length of both legs. They are denoted by the Latin letters a and c. The area of a right triangle is equal to half the area of the rectangle completed to it.
Mathematically, it looks like this: S = ½ a * c. It is remembered most easily. Because it looks like a formula for the area of a rectangle, only another fraction appears, meaning half.
Special case: isosceles triangle
Since it has two equal sides, some formulas for its area look somewhat simplified. For example, Heron’s formula, which calculates the area of an isosceles triangle, takes the following form:
S = ½ in √ (((a + ½ in) * (a - ½ in)).
If you convert it, it will become shorter. In this case, Heron’s formula for an isosceles triangle is written as:
S = in √ (4 * a2- b2).
Somewhat simpler than for an arbitrary triangle, the square formula looks like if the sides and the angle between them are known. S = ½ a2* sin β.
Special case: equilateral triangle
Usually in problems about him is known the side or you can somehow find out. Then the formula by which the area of such a triangle is found is as follows:
S = (a2√3) / 4.
Tasks on finding the area if the triangle is depicted on checkered paper
The simplest is the situation when the right triangle is drawn so that its legs coincide with the lines of the paper. Then you just need to count the number of cells that fit into the legs. Then multiply them and divide by two.
When the triangle is acute or obtuse, it must be drawn to the rectangle. Then in the resulting shape there will be 3 triangles. One is the one given in the problem. And the other two are auxiliary and rectangular. Determine the area of the last two you need by the method described above. Then count the area of the rectangle and subtract from it those that are calculated for the auxiliary. The area of the triangle is defined.
Much more difficult is the situation in which neither side of the triangle coincides with the lines of the paper. Then it needs to be inscribed in a rectangle so that the vertices of the original figure lie on its sides. In this case, there will be three auxiliary right triangles.
An example of the problem on the formula of Gerona
Condition. Some triangles have sides. They are equal to 3, 5 and 6 cm. You need to know its area.
Decision. First of all it is necessary to count the semi-perimeter of a triangle. Make the sum of all three, given in the problem, numbers and divide it into two. Simple calculations lead to the number 7. This is the value of the half-meter.
You can now calculate the area of a triangle using the above formula. Under the square root is the product of four numbers: 7, 4, 2 and 1. That is, the area is √ (4 * 14) = 2 √ (14).
If greater precision is not required, then the square root of 14 can be extracted. It is 3.74. Then the area will be equal to 7.48.
Answer. S = 2 √14 cm2or 7.48 cm2.
Example of a problem with a right triangle
Condition. One leg of a right triangle is 31 cm longer than the second one. It is required to know their lengths if the area of the triangle is 180 cm.2.
Decision. We'll have to solve a system of two equations. The first is related to the square. The second is with the attitude of the legs, which is given in the problem.
180 = ½ a * c;
a = in + 31.
First, the value of "a" must be substituted into the first equation. It turns out: 180 = ½ (in + 31) * c. It has only one unknown quantity, so it is easy to solve. After opening the parentheses, we get a quadratic equation:2+ 31 in - 360 = 0. It gives two values for "in": 9 and - 40. The second number is not suitable as an answer, since the length of the side of a triangle cannot be a negative value.
It remains to calculate the second leg: add 31 to the number obtained. It turns out 40. These are the quantities sought for in the problem.
Answer. The legs of the triangle are 9 and 40 cm.
The task of finding the side through the area, side and angle of a triangle
Condition. The area of a triangle is 60 cm2. It is necessary to calculate one of its sides, if the second side is 15 cm, and the angle between them is 30º.
Decision. Based on the accepted notation, the desired side "a", known "in", is a given angle "γ". Then the area formula can be rewritten as:
60 = ½ a * 15 * sin 30º. Here the sine of 30 degrees is 0.5.
After the transformations, “a” is equal to 60 / (0.5 * 0.5 * 15). That is 16.
Answer. The required side is 16 cm.
The problem of a square inscribed in a right triangle
Condition. The top of the square with a side of 24 cm coincides with the right angle of the triangle. The other two are on the legs. The third belongs to the hypotenuse. The length of one of the legs is 42 cm. What is the area of a right-angled triangle?
Decision. Consider two right triangles.The first one is given in the task. The second is based on the known leg of the original triangle. They are similar because they have a common angle and are formed by parallel lines.
Then the relations of their legs are equal. The legs of the smaller triangle are 24 cm (side of the square) and 18 cm (the given leg 42 cm subtract the side of the square 24 cm). The corresponding legs of the large triangle are 42 cm and x cm. It is this “x” that is needed in order to calculate the area of the triangle.
18/42 = 24 / x, i.e. x = 24 * 42/18 = 56 (cm).
Then the area is equal to the product of 56 and 42, divided into two, that is 1176 cm2.
Answer. The required area is 1176 cm2.