Consider a circle inscribed in a triangle (Fig. 302). Recall that its center O is located at the intersection of the bisectors of the interior angles of the triangle. The segments OA, OB, OC connecting O with the vertices of triangle ABC will split the triangle into three triangles:
AOV, VOS, SOA. The height of each of these triangles is equal to the radius, and therefore their areas will be expressed as
The area of the entire triangle S is equal to the sum of these three areas:
where is the semi-perimeter of the triangle. From here
The radius of the inscribed circle is equal to the ratio of the area of the triangle to its semi-perimeter.
To obtain a formula for the circumradius 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 be the acute angle of the triangle. Let's draw the radii OB, OS of the circle and drop the perpendicular OK from its center O to side BC of the triangle. Note that angle a of a triangle is measured by half of the arc BC, for which angle BOC is the central angle. From this it is clear that . Therefore, from the right triangle RNS we find , or , which is what we needed to prove.
The given fig. 303 and the reasoning refer to the case of an acute angle of a triangle; It would be easy to carry out the proof for the cases of right and obtuse angles (the reader will do this on his own), but you can use the theorem of sines (218.3). Since it must be from where
The sine theorem is also written in. form
and comparison with the notation form (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 incircle and circumcircle have radii respectively
Solution. Let's write formulas expressing the radii of the inscribed and circumscribed circles of a 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
which leads to a quadratic equation with respect to
It has two solutions:
Substituting instead of its expression in any of the equations for or R, we will finally find two answers to our problem:
Exercises
1. The altitude of a right triangle drawn from the vertex of a right angle, dividing the hypotenuse in the ratio Find the ratio of each of the legs to the hypotenuse.
2. The bases of an isosceles trapezoid circumscribed 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 tangent points 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 altitude 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. The side and altitudes of a triangle are known. Find 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?
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How to find the radius of a circle? This question is always relevant for schoolchildren studying planimetry. Below we will look at several examples of how you can cope with this task.
Depending on the conditions of the problem, you can find the radius of the circle like this.
Formula 1: R = 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 = 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 = 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 = S / (P/2), where - S and P are the area and perimeter of the figure, respectively.
Formula 2: R = (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:
The radius of a circle that is inscribed in a rhombus
A circle can be inscribed in any rhombus, both equilateral and unequal.
Formula 1: R = 2 * H, where H is the height of the geometric figure.
Formula 2: R = S / (A*2), where S is and A is the length of its side.
Formula 3: R = √((S * sin A)/4), where S is the area of the rhombus, and sin A is the sine of the acute angle of this geometric figure.
Formula 4: R = B*G/(√(B² + G²), where B and G are the lengths of the diagonals of the 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
If in the problem statement you are given the lengths of all sides of the figure, then first calculate (P), and then the semi-perimeter (p):
P = 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 one, then you can calculate the required radius as follows.
Formula 2: R = S * 2(A + B + C)
Formula 3: R = S/n = S / (A+B+B)/2), where - n is the semi-perimeter of the geometric figure.
Formula 4: R = (n - A) * tan (A/2), where n is the semi-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 = A * √3/6.
The radius of a circle that is inscribed in a right triangle
If the problem gives the lengths of the legs, as well as the hypotenuse, then the radius of the inscribed circle is determined as follows.
Formula 1: R = (A+B-C)/2, where A, B are legs, C is 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 = √(A²+B²).
The radius of a circle that is inscribed in a square
A circle that is inscribed in a square divides all 4 of its sides exactly in half at the points of contact.
Formula 1: R = A/2, where A is the length of the side of the square.
Formula 2: R = S / (P/2), where S and P are the area and perimeter of the square, respectively.
First, let's understand the difference between a circle and a circle. To see this difference, it is enough to consider what both figures are. These are an infinite number of points on the plane, located at an equal distance from a single central point. But, if the circle also consists of internal space, then it does not belong to the circle. It turns out that a circle is both a circle that limits it (circle(r)), and an innumerable number of points that are inside the circle.
For any point L lying on the circle, the equality OL=R applies. (The length of the segment OL is equal to the radius of the circle).
A segment that connects two points on a circle is its chord.
A chord passing directly through the center of a circle is diameter this circle (D). The diameter can be calculated using the formula: D=2R
Circumference calculated by the formula: C=2\pi R
Area of a circle: S=\pi R^(2)
Arc of a circle is called that part of it that is located between its two points. These two points define two arcs of a circle. The chord CD subtends two arcs: CMD and CLD. Identical chords subtend equal arcs.
Central angle An angle that lies between two radii is called.
Arc length can be found using the formula:
- Using degree measure: CD = \frac(\pi R \alpha ^(\circ))(180^(\circ))
- Using radian measure: CD = \alpha R
The diameter, which is perpendicular to the chord, divides the chord and the arcs contracted by it in half.
If the chords AB and CD of the circle intersect at the point N, then the products of the segments of the chords separated by the point N are equal to each other.
AN\cdot NB = CN\cdot ND
Tangent to a circle
Tangent to a circle It is customary to call a straight line that has one common point with a circle.
If a line has two common points, it is called secant.
If you draw the radius to the tangent point, it will be perpendicular to the tangent to the circle.
Let's draw two tangents from this point to our circle. It turns out that the tangent segments will be equal to one another, and the center of the circle will be located on the bisector of the angle with the vertex at this point.
AC = CB
Now let’s draw a tangent and a secant to the circle from our point. We obtain that the square of the length of the tangent segment will be equal to the product of the entire secant segment and its outer part.
AC^(2) = CD \cdot BC
We can conclude: the product of an entire segment of the first secant and its external part is equal to the product of an entire segment of the second secant and its external part.
AC\cdot BC = EC\cdot DC
Angles in a circle
The degree measures of the central angle and the arc on which it rests are equal.
\angle COD = \cup CD = \alpha ^(\circ)
Inscribed angle is an angle whose vertex is on a circle and whose sides contain chords.
You can calculate it by knowing the size of the arc, since it is equal to half of this arc.
\angle AOB = 2 \angle ADB
Based on a diameter, inscribed angle, right angle.
\angle CBD = \angle CED = \angle CAD = 90^ (\circ)
Inscribed angles that subtend the same arc are identical.
Inscribed angles resting on one chord are identical or their sum is equal to 180^ (\circ) .
\angle ADB + \angle AKB = 180^ (\circ)
\angle ADB = \angle AEB = \angle AFB
On the same circle are the vertices of triangles with identical angles and a given base.
An angle with a vertex inside the circle and located between two chords is identical to half the sum of the angular values of the arcs of the circle that are contained within the given and vertical angles.
\angle DMC = \angle ADM + \angle DAM = \frac(1)(2) \left (\cup DmC + \cup AlB \right)
An angle with a vertex outside the circle and located between two secants is identical to half the difference in the angular values of the arcs of the circle that are contained inside the angle.
\angle M = \angle CBD - \angle ACB = \frac(1)(2) \left (\cup DmC - \cup AlB \right)
Inscribed circle
Inscribed circle is a circle tangent to the sides of a polygon.
At the point where the bisectors of the corners of a polygon intersect, its center is located.
A circle may not be inscribed in every polygon.
The area of a polygon with an inscribed circle is found by the formula:
S = pr,
p is the semi-perimeter of the polygon,
r is the radius of the inscribed circle.
It follows that the radius of the inscribed circle is equal to:
r = \frac(S)(p)
The sums of the lengths of opposite sides will be identical if the circle is inscribed in a convex quadrilateral. And vice versa: a circle fits into a convex quadrilateral if the sums of the lengths of opposite sides are identical.
AB + DC = AD + BC
It is possible to inscribe a circle in any of the triangles. Only one single one. At the point where the bisectors of the internal angles of the figure intersect, the center of this inscribed circle will lie.
The radius of the inscribed circle is calculated by the formula:
r = \frac(S)(p) ,
where p = \frac(a + b + c)(2)
Circumcircle
If a circle passes through each vertex of a polygon, then such a circle is usually called described about a polygon.
At the point of intersection of the perpendicular bisectors of the sides of this figure will be the center of the circumcircle.
The radius can be found by calculating it as the radius of the circle that is circumscribed about the triangle defined by any 3 vertices of the polygon.
There is the following condition: a circle can be described around a quadrilateral only if the sum of its opposite angles is equal to 180^( \circ) .
\angle A + \angle C = \angle B + \angle D = 180^ (\circ)
Around any triangle you can describe a circle, and only one. The center of such a circle will be located at the point where the perpendicular bisectors of the sides of the triangle intersect.
The radius of the circumscribed circle can be calculated using the formulas:
R = \frac(a)(2 \sin A) = \frac(b)(2 \sin B) = \frac(c)(2 \sin C)
R = \frac(abc)(4 S)
a, b, c are the lengths of the sides of the triangle,
S is the area of the triangle.
Ptolemy's theorem
Finally, consider Ptolemy's theorem.
Ptolemy's theorem states that the product of diagonals is identical to the sum of the products of opposite sides of a cyclic quadrilateral.
AC \cdot BD = AB \cdot CD + BC \cdot AD
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