You will need

- The triangle with the given parameters
- The compass
- Line
- Gon
- Table of sines and cosines
- Mathematical concepts
- Determination of the height of the triangle
- Formulas of sines and cosines
- Formula of area of triangle

Instruction

1

Draw a triangle with the desired settings. The triangle can be build either on three sides or by two sides and angle between them, or a side and two adjacent angles thereto. Label the vertices of the triangle as A, b and C, the angles as α, β, and γ, and opposite the vertex angle side as a, b and c.

2

Guide height to all sides of the triangle and find the point of intersection. Label the height as h with the corresponding sides of indices. Find the point of intersection and label it O. She will be the center of the circle. Thus, the radius of this circle will be the segments OA, Ob and OC.

3

The radius of the circumscribed circle can be found in two formulas. For one you must first calculate the area of a triangle. It is equal to the product of all the sides of a triangle the sine of any of the angles divided by 2.

S=abc*sinα

In this case, the radius of the circumscribed circle is calculated by the formula

R=a*b*c/4S

For other formulas, it is sufficient to know the length of one side and the sine of its opposite angle.

R=a/2sinα

Calculate the radius and describe a triangle around the circumference.

S=abc*sinα

In this case, the radius of the circumscribed circle is calculated by the formula

R=a*b*c/4S

For other formulas, it is sufficient to know the length of one side and the sine of its opposite angle.

R=a/2sinα

Calculate the radius and describe a triangle around the circumference.

Useful advice

Remember, what is the height of the triangle. It is the perpendicular drawn from angle to opposite side.

The area of a triangle can be represented as the product of the square of one of the parties to the sines of the two adjacent angles, divided by the doubled sine of the sum of these angles.

S=A2*sinβ*sinγ/2sinγ

The area of a triangle can be represented as the product of the square of one of the parties to the sines of the two adjacent angles, divided by the doubled sine of the sum of these angles.

S=A2*sinβ*sinγ/2sinγ

# Advice 2: How to find radius of circumscribed circle

A circle is considered circumscribed about a polygon into if it covers all its vertices. Remarkably, the center of this

**circle**coincides with the point of intersection of the perpendiculars drawn from the midpoints of the sides of the polygon.**The radius of the**described**circle**depends entirely on the polygon around which it is described.You will need

- Know of sides of the polygon, its area/perimeter.

Instruction

1

The calculation of the radius of the circumscribed around the triangle

If a circle circumscribed around the triangle with sides a, b, c, area S and angle ?, lying up against the side a, the radius R can be calculated by the following formulas:

1) R = (a*b*c)/4S;

2) R = a/2sin?.

**of the circle**.If a circle circumscribed around the triangle with sides a, b, c, area S and angle ?, lying up against the side a, the radius R can be calculated by the following formulas:

1) R = (a*b*c)/4S;

2) R = a/2sin?.

2

The calculation of the radius

To calculate the radius

R = a/(2 x sin (360 / (2 x n))) where

a - side of a regular polygon;

n is the number of sides.

**of a circle**circumscribed around a regular polygon.To calculate the radius

**of a circle**circumscribed around a regular polygon, you need to use the following formula:R = a/(2 x sin (360 / (2 x n))) where

a - side of a regular polygon;

n is the number of sides.

Note

Around a polygon, you can describe a circle only if it is regular, i.e. all its sides equal and all its angles equal.

The thesis, which States that the center of the circumscribed around the circumference of the polygon is the intersection of its middle perpendiculars are valid for all regular polygons.

The thesis, which States that the center of the circumscribed around the circumference of the polygon is the intersection of its middle perpendiculars are valid for all regular polygons.

# Advice 3: How to find on three sides the area of a triangle

The search area of the triangle is one of the most common tasks in school geometry. Knowledge of the three sides of the triangle sufficient to determine the area of any triangle. In special cases of isosceles and equilateral triangles, it is sufficient to know the lengths of two and one side, respectively.

You will need

- the lengths of the sides of triangles, Heron's formula, the theorem of cosines

Instruction

1

Imagine you are given a triangle ABC with sides AB = c, AC = b, BC = a. The area of this triangle can be found by Heron's formula.

The perimeter P is the sum of the lengths of its three sides: P = a+b+c. We denote its properiter over p. It will be equal to p = (a+b+c)/2.

The perimeter P is the sum of the lengths of its three sides: P = a+b+c. We denote its properiter over p. It will be equal to p = (a+b+c)/2.

2

Heron's formula for area of a triangle as follows: S = sqrt(p(p-a)(p-b)(p-c)). If you paint properiter p, we get: S = sqrt(((a+b+c)/2)((b+c-a)/2)((a+c-b)/2)((a+b-c)/2)) = (sqrt((a+b+c)(a+b-c)(a+c-b)(b+c-a)))/4.

3

It is possible to derive a formula for the area of a triangle and for other reasons, for example, applying the theorem of cosines.

Cosine theorem AC^2 = (AB^2)+(BC^2)-2*AB*BC*cos(ABC). Using the introduced notation, these expressions can also be written in the form: b^2 = (a^2)+(c^2)-2a*c*cos(ABC). Hence, cos(ABC) = ((a^2)+(c^2)-(b^2))/(2*a*c)

Cosine theorem AC^2 = (AB^2)+(BC^2)-2*AB*BC*cos(ABC). Using the introduced notation, these expressions can also be written in the form: b^2 = (a^2)+(c^2)-2a*c*cos(ABC). Hence, cos(ABC) = ((a^2)+(c^2)-(b^2))/(2*a*c)

4

The area of the triangle is also the formula S = a*c*sin(ABC)/2 two sides and the angle between them. The sine of the angle ABC can be expressed via its cosine by using the basic trigonometric identities: sin(ABC) = sqrt(1-((cos(ABC))^2). Substituting the sine in the formula for the area and describing it, you can come to the formula for the area of triangle ABC.

# Advice 4: How to find the area of a triangle by three points

Three points uniquely define a triangle in a Cartesian coordinate system is a vertex. Knowing their position relative to each of the coordinate axes it is possible to calculate any of the parameters of this plane figure including and limited to its perimeter

**area**. This can be done in several ways.Instruction

1

Use Heron's formula to calculate the area

**of a triangle**. It involves the dimensions of three faces, so the calculations start with their definition. The length of each side must be equal to the square root of the sum of the squares of the lengths of its projections on the coordinate axes. If the coordinates of the vertices are A(X₁,Y₁,Z₁), B(X₂,Y₂,Z₂) and C(X₃,Y₃,Z₃), the lengths of their sides can be expressed as: AB = √((X₁-X₂)2 + (Y₁-Y₂)2 + (Z₁-Z₂)2), BC = √((X₂-X₃)2 + (Y₂-Y₃)2 + (Z₂-Z₃)2) AC = √((X₁-X₃)2 + (Y₁-Y₃)2 + (Z₁-Z₃)2).2

To simplify the calculations, enter the auxiliary variable properiter (R). As the name suggests, which is half the sum of the lengths of all sides: P = ½*(AB+BC+AC) = ½*(√((X₁-X₂)2 + (Y₁-Y₂)2 + (Z₁-Z₂)2) + √((X₂-X₃)2 + (Y₂-Y₃)2 + (Z₂-Z₃)2) + √((X₁-X₃)2 + (Y₁-Y₃)2 + (Z₁-Z₃)2).

3

Calculate

**the area**(S) by Heron's formula - extract the root of the product of properiety on the difference between him and the length of each of the parties. In General it can be written as: S = √(P*(P-AB)*(P-BC)*(P-AC)) = √( P*(P-√((X₁-X₂)2 + (Y₁-Y₂)2 + (Z₁-Z₂)2))*(P-√((X₂-X₃)2 + (Y₂-Y₃)2 + (Z₂-Z₃)2))*(P-√((X₁-X₃)2 + (Y₁-Y₃)2 + (Z₁-Z₃)2)).4

For practical calculations it is convenient to use specialized online calculators. These are the scripts that are hosted on servers of some sites that will do all the necessary calculations based on the coordinates you entered in the appropriate form. The only disadvantage of this service - it gives no explanations and justifications for each step of the calculation. Therefore, if you are only interested in the end result, not the calculation in a General view, go, for example, on the page http://planetcalc.ru/218/.

5

In the form fields separately enter each coordinate of each vertex

**of the triangle**- they are denoted here as Ax, Ay, Az, etc. If a triangle is specified two-dimensional coordinates, in the field of applicator that con - Az, Bz and Cz - write zero. In the "calculation Precision" set the desired number of decimal places by clicking the mouse icon to a plus or minus. Placed under the form the orange "Calculate" button to press is not necessary, the calculations will be made without it. You will find the answer next to "Square**the triangle**" - it is placed just under the orange button.# Advice 5: How to find the circumcenter

Sometimes around a convex polygon you can draw a circle so that the tops of all the angles lying on it. Such a circle relative to the polygon should be called described. Her

**center**does not have to be inside the perimeter of the inscribed figure, but using the described properties**of the circle**, to find this point is usually not very difficult.You will need

- Ruler, pencil, protractor or straight edge, a compass.

Instruction

1

If the polygon about which it is necessary to describe a circle drawn on paper, for finding

**the centre**of the circle and quite a ruler, pencil and protractor or angle. Measure the length of any of sides of a shape, determine its middle and put in this part of the drawing auxiliary point. With the help of the set square or the protractor guide on the inside of the polygon perpendicular to the side segment to the intersection with the opposite side.2

Do the same thing with any side of the polygon. The intersection of the two constructed segments is the required point. This follows from basic properties are described

**of a circle**- its**center**in a convex polygon with any number of sides, always lies at the intersection of middle perpendiculars drawn to these parties.3

For the regular polygons defining

**the center of**a inscribed**circle**can be much easier. For example, if it is a square, then draw two diagonals and their intersection will be**the center**om of the inscribed**circle**. The regular polygon with any even number of sides, it is sufficient to connect the auxiliary segments two pairs lying opposite each other at the corners -**center**of the described**circle**must coincide with the point of their intersection. In a right triangle to solve the problem just define the middle of the longest side of the figure is the hypotenuse.4

If conditions are unknown, whether it is possible in principle to draw a circumscribed circle for that polygon, after determining the proposed point

**center**and any of the described ways you can find out. Mark on the compass the distance between the found point and any vertices, set the compass at the intended**center****of the circle**and draw a circle, every vertex must lie on this**circle**. If not, then not running one of the basic properties and describe a circle about a given polygon is impossible.