The formula for calculating the length of an arc of a circle. Circle geometry. An example of a problem of medium complexity using this formula

Tasks for finding the area of ​​a circle are a mandatory part of the exam in mathematics. As a rule, several tasks are assigned to this topic at once in the certification test. All high school students should understand the algorithm for finding the circumference and area of ​​a circle, regardless of their level of preparation.

If such planimetric tasks cause you difficulties, we recommend that you turn to the Shkolkovo educational portal. With us you can fill the gaps in knowledge.

The corresponding section of the site contains a large selection of tasks for finding the circumference and area of ​​a circle, similar to those included in the exam. Having learned how to perform them correctly, the graduate will be able to successfully cope with the exam.

Basic moments

Problems that require the use of area formulas can be direct or inverse. In the first case, the parameters of the elements of the figure are known. In this case, the desired value is the area. In the second case, on the contrary, the area is known, and it is necessary to find any element of the figure. The algorithm for calculating the correct answer in such tasks differs only in the order in which the basic formulas are applied. That is why, starting to solve such problems, it is necessary to repeat the theoretical material.

The educational portal "Shkolkovo" provides all the basic information on the topic "Finding the length of a circle or arc and the area of ​​a circle", as well as on other topics, for example, our specialists prepared and presented it in the most accessible form.

After remembering the basic formulas, students can start completing tasks for finding the area of ​​a circle, similar to those included in the exam, online. For each exercise on the site, a detailed solution is presented and the correct answer is given. If necessary, any task can be saved in the "Favorites" section in order to return to it later and discuss it with the teacher.

Task 10 (OGE - 2015)

Points A and B are marked on a circle centered at O ​​so that ∠ AOB = 18°. The length of the smaller arc AB is 5. Find the length of the larger arc of the circle.

Decision

∠AOB = 18°. The whole circle is 360°. So ∠AOB is 18/360 = 1/20 of the circle.

This means that the smaller arc AB is 1/20 of the entire circle, so the larger arc is the rest, i.e. 19/20 circumference.

1/20 of the circle corresponds to an arc length of 5. Then the length of the larger arc is 5*19 = 95.

Task 10 (OGE - 2015)

Points A and B are marked on a circle centered at O ​​so that ∠ AOB = 40°. The length of the smaller arc AB is 50. Find the length of the larger arc of the circle.

Decision

∠AOB = 40°. The whole circle is 360°. So ∠AOB is 40/360 = 1/9 of the circle.

This means that the smaller arc AB is 1/9 of the entire circle, so the larger arc is the rest, i.e. 8/9 circle.

1/9 of the circle corresponds to an arc length of 50. Then the length of the larger arc is 50 * 8 = 400.

Answer: 400.

Task 10 (GIA - 2014)

The chord length of the circle is 72, and the distance from the center of the circle to this chord is 27. Find the diameter of the circle.

Decision

By the Pythagorean theorem, from a right triangle AOB we get:

AO 2 \u003d OB 2 + AB 2,

AO 2 \u003d 27 2 +36 2 \u003d 729 + 1296 \u003d 2025,

Then the diameter is 2R = 2*45 = 90.

Task 10 (GIA - 2014)

Point O is the center of the circle on which points A, B and C lie. It is known that ∠ABC = 134° and ∠OAB = 75°. Find the angle BCO. Give your answer in degrees.

The formula for finding the length of an arc of a circle is quite simple, and very often in important exams such as the USE there are such problems that cannot be solved without using it. You also need to know it to pass international standardized tests, such as SAT and others.

What is the length of the arc of a circle?

The formula looks like this:

l = prα / 180°

What is each of the elements of the formula:

• π - Pi number (constant value equal to ≈ 3.14);
• r is the radius of the given circle;
• α - the value of the angle on which the arc rests (central, not inscribed).

As you can see, in order to solve the problem, r and α must be present in the condition. Without these two quantities, the arc length cannot be found.

How is this formula derived and why does it look like this?

Everything is extremely easy. It will become much clearer if you put 360 ° in the denominator, and add a deuce in the numerator in front. You can also α do not leave it in a fraction, print it and write it with a multiplication sign. This is quite possible to afford, since this element is in the numerator. Then the general view will be like this:

l = (2πr / 360°) × α

Just for convenience, we reduced 2 and 360 °. And now, if you look closely, you can see a very familiar formula for the length of the entire circle, namely - 2pr. The whole circle consists of 360 °, so we divide the resulting measure into 360 parts. Then we multiply by the number α, that is, for the number of "pieces of the pie" that we need. But everyone knows for sure that a number (that is, the length of the entire circle) cannot be divided by a degree. What to do in this case? Usually, as a rule, the degree is reduced with the degree of the central angle, that is, with α. After that, only numbers remain, and as a result, the final answer is obtained.

This can explain why the length of the arc of a circle is found in this way and has this form.

An example of a problem of medium complexity using this formula

Condition: There is a circle with a radius of 10 centimeters. The degree measure of a central angle is 90°. Find the length of the circular arc formed by this angle.

Solution: l = 10π × 90° / 180° = 10π × 1 / 2=5π

Answer: l = 5π

It is also possible that instead of the degree measure, the radian measure of the angle would be given. In no case should you be scared, because this time the task has become much easier. To convert a radian measure to a degree measure, you need to multiply this number by 180 ° / π. So now we can substitute instead α the following combination: m × 180° / π. Where m is the radian value. And then 180 and the number π are reduced and a completely simplified formula is obtained, which looks like this:

• m is the radian measure of the angle;
• r is the radius of the given circle.

Let's first understand the difference between a circle and a circle. To see this difference, it is enough to consider what both figures are. This is an infinite number of points in 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 bounds it (o-circle (g)ness), and an uncountable 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 line segment that connects two points on a circle is 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 called that part of it, which is located between two of its points. These two points define two arcs of a circle. The chord CD subtends two arcs: CMD and CLD. The same chords subtend the same arcs.

Central corner is the angle between two radii.

arc length can be found using the formula:

1. Using degrees: CD = \frac(\pi R \alpha ^(\circ))(180^(\circ))
2. Using a radian measure: CD = \alpha R

The diameter that is perpendicular to the chord bisects the chord and the arcs it spans.

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 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 points in common, it is called secant.

If you draw a radius at the point of contact, 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 segments of the tangents 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 we draw a tangent and a secant to the circle from our point. We get that the square of the length of the tangent segment will be equal to the product of the entire secant segment by its outer part.

AC^(2) = CD \cdot BC

We can conclude: the product of an integer segment of the first secant by its outer part is equal to the product of an integer segment of the second secant by its outer 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 diameter, inscribed angle, straight.

\angle CBD = \angle CED = \angle CAD = 90^ (\circ)

Inscribed angles that lean on the same arc are identical.

The inscribed angles based on the same chord are identical or their sum equals 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 inside 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 magnitudes of the arcs of a circle that are 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 the polygon.

At the point where the bisectors of the angles of the 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 semiperimeter of the polygon,

r is the radius of the inscribed circle.

It follows that the radius of the inscribed circle is:

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 is inscribed in a convex quadrilateral if the sums of the lengths of opposite sides in it are identical.

AB+DC=AD+BC

It is possible to inscribe a circle in any of the triangles. Only one single. At the point where the bisectors of the inner 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)

Circumscribed circle

If a circle passes through every vertex of a polygon, then such a circle is called circumscribed about a polygon.

At the point of intersection of the perpendicular bisectors of the sides of this figure will be the center of the circumscribed circle.

The radius can be found by calculating it as the radius of a circle that is circumscribed about a triangle defined by any 3 vertices of the polygon.

There is the following condition: a circle can be circumscribed 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)

Near any triangle it is possible to describe a circle, and one 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 by the formulas:

R = \frac(a)(2 \sin A) = \frac(b)(2 \sin B) = \frac(c)(2 \sin C)

R = \frac(abc)(4S)

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 the diagonals is identical to the sum of the products of the opposite sides of an inscribed quadrilateral.

AC \cdot BD = AB \cdot CD + BC \cdot AD

The part of a figure that forms a circle whose points are equidistant is called an arc. If, from the point of the center of the circle, the rays are drawn to points coinciding with the ends of the arc, its central angle will be formed.

Determining the length of an arc

Produced according to the following formula:

where L is the desired length of the arc, π = 3.14, r is the radius of the circle, α is the central angle.

The length of the arc of a circle is 14.82 centimeters.

In elementary geometry, an arc is understood as a subset of a circle located between two points located on it. In practice, solve problems definition her length engineers and architects often have to, since this geometric element is widespread in a wide variety of designs.

Perhaps the first to face this task were the ancient architects, who one way or another had to determine this parameter for the construction of vaults, widely used to bridge the gaps between supports in round, polygonal or elliptical buildings. If you look closely at the masterpieces of ancient Greek, ancient Roman and especially Arabic architecture that have survived to this day, you will notice that arcs and vaults are extremely common in their designs. The creations of modern architects are not so rich in them, but these geometric elements are present, of course, in them.

Length various arcs it is necessary to calculate during the construction of roads and railways, as well as autodromes, and in many cases, traffic safety largely depends on the correctness and accuracy of calculations. The fact is that many turns of highways from the point of view of geometry are precisely arcs, and various physical forces act on transport along them. The parameters of their resultant are largely determined by the length of the arc, as well as its central angle and radius.

Designers of machines and mechanisms have to calculate the lengths of various arcs for the correct and accurate layout of the components of various units. In this case, errors in calculations are fraught with the fact that important and critical parts will interact incorrectly with each other and the mechanism simply will not be able to function as its creators plan. Examples of designs rich in geometric elements such as arcs include internal combustion engines, gearboxes, wood and metalworking equipment, body parts for cars and trucks, etc.

arcs quite widely found in medicine, in particular, in dentistry. For example, they are used to correct malocclusion. Corrective elements, called braces (or bracket systems) and having the appropriate shape, are made from special alloys and are installed in such a way as to change the position of the teeth. It goes without saying that in order for the treatment to be successful, these arcs must be very precisely calculated. In addition, arcs are widely used in traumatology, and perhaps the most striking example of this is the famous Ilizarov apparatus, invented by a Russian doctor in 1951 and extremely successfully used to this day. Its integral parts are metal arcs, equipped with holes through which special knitting needles are threaded, and which are the main supports of the entire structure.