Hmm, yoda_daro claims that this is a test to determine their sexuality...
Instruction The frames of the film merge for us into continuous movement due to ...
Consider a circle inscribed in a triangle (Fig. 302). Recall that its center O is placed at the intersection of the bisectors of the interior angles of the triangle. The segments OA, OB, OS, connecting O with the vertices of the triangle ABC, will divide the triangle into three triangles:
AOB, BOS, SOA. The height of each of these triangles is equal to the radius, and therefore their areas are expressed as
The area of the whole triangle S is equal to the sum of these three areas:
where is the semiperimeter of the triangle. From here
The radius of the inscribed circle is equal to the ratio of the area of the triangle to its half-perimeter.
To obtain a formula for the radius of the circumscribed circle of a triangle, we prove the following proposition.
Theorem a: In any triangle, the side is equal to the diameter of the circumscribed circle multiplied by the sine of the opposite angle.
Proof. Consider an arbitrary triangle ABC and a circle circumscribed around it, the radius of which will be denoted by R (Fig. 303). Let A- sharp corner triangle. Let's draw the radii OB, OS of the circle and drop the perpendicular OK from its center O to the side BC of the triangle. Note that the angle a of a triangle is measured by half the arc BC, for which the angle BOC is central corner. From here it is clear that . Therefore, from a right-angled triangle SOK we find , or , which was required to be proved.
The given fig. 303 and the argument refer to the case of an acute angle of a triangle; it would not be difficult to carry out a proof for the cases of direct and obtuse angle(the reader will do this on his own), but you can use the sine theorem (218.3). Since it must be where
The sine theorem is also written in. form
and comparison with the notation (218.3) gives for
The radius of the circumscribed circle is equal to the ratio of the product of the three sides of the triangle to its quadruple area.
Task. Find the sides of an isosceles triangle if its inscribed and circumscribed circles have radii, respectively
Decision. Let's write the formulas expressing the radii of the inscribed and circumscribed circles of the triangle:
For an isosceles triangle with a side and a base, the area is expressed by the formula
or, reducing the fraction by a non-zero factor , we have
that leads to quadratic equation relatively
It has two solutions:
Substituting instead of its expression into any of the equations for or R, we finally find two answers to our problem:
Exercises
1. Height of a right triangle drawn from a vertex right angle, delnt the hypotenuse in relation Find the ratio of each of the legs to the hypotenuse.
2. The bases of an isosceles trapezoid inscribed about a circle are equal to a and b. Find the radius of the circle.
3. Two circles touch externally. Their common tangents are inclined to the line of centers at an angle of 30°. The length of the tangent segment between the points of contact is 108 cm. Find the radii of the circles.
4. The legs of a right triangle are equal to a and b. Find the area of a triangle whose sides are the height and median of the given triangle, drawn from the vertex of the right angle, and the segment of the hypotenuse between the points of their intersection with the hypotenuse.
5. The sides of the triangle are 13, 14, 15. Find the projection of each of them onto the other two.
6. In a triangle, the side and heights are known. Find the sides b and c.
7. Two sides of the triangle and the median are known. Find the third side of the triangle.
8. Given two sides of a triangle and an angle a between them: Find the radii of the inscribed and circumscribed circles.
9. The sides of the triangle a, b, c are known. What are the segments into which they are divided by the points of contact of the inscribed circle with the sides of the triangle?
How to find the radius of a circle? This question is always relevant for schoolchildren studying planimetry. Below we will look at a few examples of how you can cope with the task.
Depending on the condition of the problem, you can find the radius of the circle like this.
Formula 1: R \u003d L / 2π, where L is and π is a constant equal to 3.141 ...
Formula 2: R = √(S / π), where S is the area of the circle.
Formula 1: R = B/2, where B is the hypotenuse.
Formula 2: R \u003d M * B, where B is the hypotenuse, and M is the median drawn to it.
How to find the radius of a circle if it is circumscribed around a regular polygon
Formula: R \u003d A / (2 * sin (360 / (2 * n))), where A is the length of one of the sides of the figure, and n is the number of sides in this geometric figure.
How to find the radius of an inscribed circle
An inscribed circle is called when it touches all sides of the polygon. Let's look at a few examples.
Formula 1: R \u003d S / (P / 2), where - S and P are the area and perimeter of the figure, respectively.
Formula 2: R \u003d (P / 2 - A) * tg (a / 2), where P is the perimeter, A is the length of one of the sides, and is the angle opposite this side.
How to find the radius of a circle if it is inscribed in a right triangle
Formula 1:
Radius of a circle inscribed in a rhombus
A circle can be inscribed in any rhombus, both equilateral and inequilateral.
Formula 1: R \u003d 2 * H, where H is the height of the geometric figure.
Formula 2: R \u003d S / (A * 2), where S is and A is the length of its side.
Formula 3: R \u003d √ ((S * sin A) / 4), where S is the area of \u200b\u200bthe rhombus, and sin A is the sine of the acute angle of this geometric figure.
Formula 4: R \u003d V * G / (√ (V² + G²), where V and G are the lengths of the diagonals of a geometric figure.
Formula 5: R = B * sin (A / 2), where B is the diagonal of the rhombus, and A is the angle at the vertices connecting the diagonal.
Radius of a circle that is inscribed in a triangle
In the event that in the condition of the problem you are given the lengths of all sides of the figure, then first calculate (P), and then the semi-perimeter (p):
P \u003d A + B + C, where A, B, C are the lengths of the sides of the geometric figure.
Formula 1: R = √((p-A)*(p-B)*(p-B)/p).
And if, knowing all the same three sides, you are also given, then you can calculate the required radius as follows.
Formula 2: R = S * 2(A + B + C)
Formula 3: R \u003d S / p \u003d S / (A + B + C) / 2), where - p is the semi-perimeter of the geometric figure.
Formula 4: R \u003d (n - A) * tg (A / 2), where n is the half-perimeter of the triangle, A is one of its sides, and tg (A / 2) is the tangent of half the angle opposite this side.
And the formula below will help you find the radius of the circle that is inscribed in
Formula 5: R \u003d A * √3/6.
Radius of a circle that is inscribed in a right triangle
If the problem is given the lengths of the legs, as well as the hypotenuse, then the radius of the inscribed circle is found out as follows.
Formula 1: R \u003d (A + B-C) / 2, where A, B are legs, C is the hypotenuse.
In the event that you are given only two legs, it's time to remember the Pythagorean theorem in order to find the hypotenuse and use the above formula.
C \u003d √ (A² + B²).
Radius of a circle that is inscribed in a square
The circle, which is inscribed in the square, divides all its 4 sides exactly in half at the points of contact.
Formula 1: R \u003d A / 2, where A is the length of the side of the square.
Formula 2: R \u003d S / (P / 2), where S and P are the area and perimeter of the square, respectively.
A circle is considered to be inscribed in the boundaries of a regular polygon if it lies inside it, while touching the lines that pass through all sides. Consider how to find the center and radius of a circle. The center of the circle will be the point where the bisectors of the corners of the polygon intersect. Radius is calculated: R=S/P; S is the area of the polygon, P is the semiperimeter of the circle.
In a triangle
Only one circle is inscribed in a regular triangle, the center of which is called the incenter; it is the same distance from all sides and is the intersection of the bisectors.
In a quadrilateral
Often you have to decide how to find the radius of the inscribed circle in this geometric figure. It must be convex (if there are no self-intersections). A circle can be inscribed in it only if the sums of opposite sides are equal: AB+CD=BC+AD.
In this case, the center of the inscribed circle, the midpoints of the diagonals, are located on one straight line (according to Newton's theorem). A segment whose ends are where they intersect opposite sides regular quadrilateral lies on the same line, called the Gauss line. The center of the circle will be the point at which the heights of the triangle intersect with the vertices, the diagonals (according to Brocard's theorem).
In a rhombus
It is considered a parallelogram with the same side length. The radius of a circle inscribed in it can be calculated in several ways.
- To do this correctly, find the radius of the inscribed circle of the rhombus, if the area of the rhombus is known, the length of its side. The formula r=S/(2Xa) is applied. For example, if the area of a rhombus is 200 mm square, the side length is 20 mm, then R = 200 / (2X20), that is, 5 mm.
- An acute angle of one of the vertices is known. Then it is necessary to use the formula r=v(S*sin(α)/4). For example, with an area of 150 mm and a known angle of 25 degrees, R= v(150*sin(25°)/4) ≈ v(150*0.423/4) ≈ v15.8625 ≈ 3.983 mm.
- All angles in a rhombus are equal. In this situation, the radius of a circle inscribed in a rhombus will be equal to half the length of one side of this figure. If we argue according to Euclid, who claims that the sum of the angles of any quadrilateral is 360 degrees, then one angle will be equal to 90 degrees; those. get a square.
Circle inscribed in a triangle
Existence of a circle inscribed in a triangle
Recall the definition angle bisector .
Definition 1 .Angle bisector called a ray that divides an angle into two equal parts.
Theorem 1 (Basic property of the angle bisector) . Each point of the bisector of the angle is at the same distance from the sides of the angle (Fig. 1).
Rice. one
Proof D lying on the bisector of the angleBAC , and DE and D.F. on the sides of the corner (Fig. 1).right triangles ADF and ADE equal because they have the same acute anglesDAF and DAE , and the hypotenuse AD - general. Hence,
D.F. = D.E.
Q.E.D.
Theorem 2 (inverse theorem to Theorem 1) . If some , then it lies on the bisector of the angle (Fig. 2).
Rice. 2
Proof . Consider an arbitrary pointD lying inside the cornerBAC and located at the same distance from the sides of the corner. Drop from pointD perpendiculars DE and D.F. on the sides of the corner (Fig. 2).right triangles ADF and ADE equal , since they have equal legsD.F. and DE , and the hypotenuse AD - general. Hence,
Q.E.D.
Definition 2 . The circle is called circle inscribed in an angle if it is the sides of this angle.
Theorem 3 . If a circle is inscribed in an angle, then the distances from the vertex of the angle to the points of contact of the circle with the sides of the angle are equal.
Proof . Let the point D is the center of a circle inscribed in an angleBAC , and the points E and F - points of contact of the circle with the sides of the corner (Fig. 3).
Fig.3
a , b , c - sides of a triangle S -square,
r – radius of the inscribed circle, p - semiperimeter
.
View formula output
a – lateral side of an isosceles triangle , b - base, r – inscribed circle radius
a r – inscribed circle radius
View formula output
,
where
,
then, in the case of an isosceles triangle, when
we get
which is what was required.
Theorem 7 . For the equality
where a - side of an equilateral triangler – radius of the inscribed circle (Fig. 8).
Rice. eight
Proof .
,
then, in the case of an equilateral triangle, when
b=a,
we get
which is what was required.
Comment . I recommend deriving as an exercise the formula for the radius of a circle inscribed in an equilateral triangle directly, i.e. without use general formulas for the radii of circles inscribed in an arbitrary triangle or in an isosceles triangle.
Theorem 8 . For a right triangle, the equality
where a , b - legs of a right triangle, c – hypotenuse , r – radius of the inscribed circle.
Proof . Consider Figure 9.
Rice. nine
Since the quadrilateralCDOF is an , which has adjacent sidesDO and OF are equal, then this rectangle is . Hence,
CB \u003d CF \u003d r,
By virtue of Theorem 3, the equalities
Therefore, taking also into account , we get
which is what was required.
A selection of tasks on the topic "A circle inscribed in a triangle."
1.
A circle inscribed in an isosceles triangle divides one of the sides into two segments at the point of contact, the lengths of which are equal to 5 and 3, counting from the vertex opposite the base. Find the perimeter of the triangle.
2.
3
In triangle ABC AC=4, BC=3, angle C is 90º. Find the radius of the inscribed circle.
4.
The legs of an isosceles right triangle are 2+. Find the radius of the circle inscribed in this triangle.
5.
The radius of a circle inscribed in an isosceles right triangle is 2. Find the hypotenuse c of this triangle. Write c(-1) in your answer.
Here are a number of tasks from the exam with solutions.
The radius of a circle inscribed in an isosceles right triangle is . Find the hypotenuse c of this triangle. Please indicate in your answer.
The triangle is right and isosceles. So his legs are the same. Let each leg be equal. Then the hypotenuse is.
We write the area of triangle ABC in two ways:
Equating these expressions, we get that
. Insofar as, we get that
. Then.
In response, write.
Answer:.
Task 2.
1. In any two sides 10cm and 6cm (AB and BC). Find the radii of the circumscribed and inscribed circles
The problem is solved independently with commenting.
Decision:
AT.
1) Find: 
2) Prove:
and find CK
3) Find: the radii of the circumscribed and inscribed circles
Decision:
Task 6.
R the radius of a circle inscribed in a square is. Find the radius of the circle circumscribed about this square.Given :
To find: OS=?
Decision: in this case the problem can be solved using either the Pythagorean theorem or the formula for R. The second case is simpler, since the formula for R is derived from the theorem. 
Task 7.
The radius of a circle inscribed in an isosceles right triangle is 2. Find the hypotenusewith this triangle. Please indicate in your answer.
S is the area of the triangle
We do not know either the sides of the triangle or its area. Let's denote the legs as x, then the hypotenuse will be equal to:
The area of the triangle will be 0.5x 2 .
Means
So the hypotenuse will be:
The answer must be written:
Answer: 4
Task 8.
In triangle ABC, AC = 4, BC = 3, angle C is equal to 90 0 . Find the radius of the inscribed circle.
Let's use the formula for the radius of a circle inscribed in a triangle:
where a, b, c are the sides of the triangle
S is the area of the triangle
Two sides are known (these are legs), we can calculate the third (hypotenuse), we can also calculate the area.
According to the Pythagorean theorem:
Let's find the area:
Thus:
Answer: 1
Task 9.
The sides of an isosceles triangle are 5, the base is 6. Find the radius of the inscribed circle.
Let's use the formula for the radius of a circle inscribed in a triangle:
where a, b, c are the sides of the triangle
S is the area of the triangle
All sides are known, and the area is calculated. We can find it using Heron's formula:
Then
Very often, when solving geometric problems, you have to perform actions with auxiliary figures. For example, find the radius of an inscribed or circumscribed circle, etc. This article will show you how to find the radius of a circle circumscribing a triangle. Or, in other words, the radius of the circle in which the triangle is inscribed.
How to find the radius of a circle circumscribed about a triangle - the general formula
The general formula is as follows: R = abc/4√p(p - a)(p - b)(p - c), where R is the radius of the circumscribed circle, p is the perimeter of the triangle divided by 2 (half-perimeter). a, b, c are the sides of the triangle.
Find the radius of the circumcircle of the triangle if a = 3, b = 6, c = 7.
Thus, based on the above formula, we calculate the semi-perimeter:
p = (a + b + c)/2 = 3 + 6 + 7 = 16. => 16/2 = 8.
Substitute the values in the formula and get:
R = 3 × 6 × 7/4√8(8 – 3)(8 – 6)(8 – 7) = 126/4√(8 × 5 × 2 × 1) = 126/4√80 = 126/16 √5.
Answer: R = 126/16√5
How to find the radius of a circle circumscribed about an equilateral triangle
To find the radius of a circle circumscribed about an equilateral triangle, there is a fairly simple formula: R = a/√3, where a is the size of its side.
Example: The side of an equilateral triangle is 5. Find the radius of the circumscribed circle.
Since all sides of an equilateral triangle are equal, to solve the problem, you just need to enter its value in the formula. We get: R = 5/√3.
Answer: R = 5/√3.
How to find the radius of a circle circumscribed about a right triangle
The formula looks like this: R = 1/2 × √(a² + b²) = c/2, where a and b are legs and c is the hypotenuse. If we fold the squares of the legs into right triangle, we get the square of the hypotenuse. As can be seen from the formula, this expression is under the root. By calculating the root of the square of the hypotenuse, we get the length itself. Multiplying the resulting expression by 1/2 eventually leads us to the expression 1/2 × c = c/2.
Example: Calculate the radius of the circumscribed circle if the legs of the triangle are 3 and 4. Substitute the values into the formula. We get: R = 1/2 × √(3² + 4²) = 1/2 × √25 = 1/2 × 5 = 2.5.
In this expression, 5 is the length of the hypotenuse.
Answer: R = 2.5.
How to find the radius of a circle circumscribed about an isosceles triangle
The formula looks like this: R = a² / √ (4a² - b²), where a is the length of the thigh of the triangle and b is the length of the base.
Example: Calculate the radius of a circle if its hip = 7 and its base = 8.
Solution: We substitute these values \u200b\u200binto the formula and get: R \u003d 7² / √ (4 × 7² - 8²).
R = 49/√(196 - 64) = 49/√132. The answer can be written directly like this.
Answer: R = 49/√132
Online Resources for Calculating the Radius of a Circle
It is very easy to get confused in all these formulas. Therefore, if necessary, you can use online calculators, which will help you in solving problems on finding the radius. The principle of operation of such mini-programs is very simple. Substitute the value of the side in the appropriate field and get a ready-made answer. You can choose several options for rounding the answer: to decimals, hundredths, thousandths, etc.
