SUBJECT:
Properties of the elements of a right triangle. Bisector property triangle angle.
mathematics teacher at a municipal educational institution
average secondary school №13
KOSTROMA 2009
EXPLANATORY NOTE
When compiling these didactic materials, the following goals were set:
Help the teacher organize educational process when studying the topics “Property of the bisector of an angle of a triangle” and “Property of the height dropped from a vertex right angle to the hypotenuse"
Supplement the geometry textbook on these topics with tasks for independent work students;
Identification of tasks for preparing for the Unified State Exam in mathematics.
These didactic materials help to consolidate the skills of solving tasks on the application of properties arising from the similarity of right triangles. A selection of tasks can be used for current and final control, for independent work, for individual assignment at home, both in the 9th grade and in the 10-11th grades when repeating material and preparing for the Unified State Exam. The materials present 22 problems, half of them are accompanied by solutions. Problems whose solutions are similar to those considered are offered either for independent solution in class, or as a homework. The tasks are arranged in order of increasing difficulty.
Why did I, as a teacher, need a selection of tasks on this particular topic? There are several answers here. Firstly, in the textbook I am working on, there are practically no problems on this topic (only two problems: No. 40 p. 106 and several more problems in the didactic materials), but they are of the same type and generally do not reflect various situations to apply properties. There are no problems at all on applying the properties of the angle bisector of a triangle.
Secondly, this topic has been reflected more than once in Unified State Exam materials, and therefore I consider it necessary to outline this topic in more detail for students. The number of geometry problems in the mathematics exam has increased
Literature:
« Exam questions and answers to 5"
“Handbook for applicants to universities”
Zelensky I. I. “Geometry in problems.” Mathematics Series: “Reboot”
"Collection of problems in geometry"
Ziv A. G. “Geometry problems”
Gusev A.I. " Didactic materials in geometry"
Heading
Property No. 1
The height of a right triangle drawn from the vertex of a right angle is the average proportional between the projections of the legs onto the hypotenuse
Property No. 2
The leg of a right triangle is the mean proportional between the hypotenuse and its projection onto the hypotenuse
Property No. 3
The bisector of a triangle divides the opposite side into segments proportional to the other two sides
Level A
A1 The perimeter of the triangle is 25 cm, and its bisector divides the opposite side into segments equal to 7.5 cm and 2.5 cm. Find the sides of the triangle.
A2 The perimeter of the triangle is 35 cm. Find the segments into which the bisector of the triangle divides the opposite side.
A3 One of the legs of a right triangle is 10 dm, and its projection onto the hypotenuse is 8 dm. Find the second leg and hypotenuse.
A4 Find the legs of a right triangle if their projections to the hypotenuse are 36 cm 64 cm.
A5 Find the altitude of a right triangle drawn from the vertex of a right angle if its base divides the hypotenuse into segments 4 cm and 9 cm.
A6 The altitude of a right triangle drawn from the vertex of the right angle to the hypotenuse is 4. Find the hypotenuse if one of the legs is 8.
Level B
B1 B right triangle the height drawn to the hypotenuse is 36 cm and divides it into segments in the ratio 9:16. Find RAVS
https://pandia.ru/text/78/060/images/image003_197.gif" width="71" height="23">; SK2= AK ∙ HF;
362 = 9x∙16x; 1296 = 144x2; x2 = 9; x = 3
AK=27cm; VK=48cm; AB=75cm.
2) From ∆ AKS according to the Pythagorean theorem: AC= https://pandia.ru/text/78/060/images/image006_144.gif" width="49" height="24 src=">=45 (cm)
From ∆ ABC according to the Pythagorean theorem: BC===60 (cm)
3) P ABC = AC+AB+BC; RABC = 180cm.
Answer 180cm
B2 In a right triangle, the altitude drawn to the hypotenuse divides it into segments in the ratio 16:9. The longest leg of the triangle is 60cm. find the length of this height. (this problem is similar to the previous one and therefore its solution is not considered )
Answer: 36cm
B3 A perpendicular is drawn from a point on the circle to the diameter, which divides the diameter into segments whose lengths are in the ratio 9:4. Find the circumference if the perpendicular length is 24 cm.
https://pandia.ru/text/78/060/images/image010_107.gif" width="12" height="19">AO = 26 cm
3) To find the circumference, apply the formula: L = 2https://pandia.ru/text/78/060/images/image011_97.gif" width="15" height="15 src="> cm
Answer: 52https://pandia.ru/text/78/060/images/image012_89.gif" width="208" height="172 src=">Solution
1) Let us apply the property of the height drawn
from the vertex of the right angle ∆ABC to the hypotenuse AC: VK= https://pandia.ru/text/78/060/images/image014_72.gif" width="83" height="27">cm, AK=4cm, KS =16cm.
2) From ∆AKV according to the Pythagorean theorem:
3) From ∆VKS according to the Pythagorean theorem:
4) SAVSD =AB ∙ ; S ABCD = 160 cm2
Answer: 160cm2
B6 From the vertices of the opposite corners of the rectangle, perpendiculars are drawn to the diagonal, the distance between the bases of which is 16 cm. Find the area of the rectangle if the lengths of these perpendiculars are 6 cm. (The problem is similar to the previous one, so its solution is not presented)
Answer: 120cm2
Problems B7, B8, B9 can be offered to students either as homework or taken to class. independent decision in class
Q7 The area of a right triangle is 150, one of the legs is 15. Find the length of the height dropped from the vertex of the right angle
Q8 The altitude of a right triangle drawn from the vertex of the right angle to the hypotenuse is equal to Find the hypotenuse if one of the legs is 8.
Q9 The height of a right triangle, lowered to the hypotenuse, is equal to b, and one of the acute angles is 60○. Find the hypotenuse.
B10 The bisector of an acute angle of a right triangle divides a leg of 12 cm and 15 cm. Find the area of the triangle using the segments.
https://pandia.ru/text/78/060/images/image022_49.gif" width="148" height="41">
Let x be the proportionality coefficient, then
5x – side AB, 4x – side AC
2) For ∆ACV we apply the Pythagorean theorem
AB2 = AC2 + BC2;
25x2 = 16x2 +729;
3) Apply the formula for the area of the triangle: S∆ = AC∙BC; AC = 36(cm); Sun = 27(cm)
S∆ASV =486 cm2
Answer: 486 cm2
Q11, Q12 are similar to the previous problem.
B11 The bisector of a right angle of a triangle divides its hypotenuse into segments of 15 cm and 20 cm. Find the area of the triangle.
Answer: 294cm2
Q12 In a right triangle, the bisector of an acute angle divides the opposite leg into segments of length 8 cm and 10 cm. Find the perimeter of this triangle.
Answer: 72cm
B13 The bisector of a right angle of a right triangle divides the hypotenuse into segments of 20 cm and 15 cm. Find the radius of the inscribed circle.
https://pandia.ru/text/78/060/images/image025_41.gif" width="148" height="41">
2) Let x be the proportionality coefficient, then AC -4x, CB-3x
For ∆ASV we apply the Pythagorean theorem:
AB2 = AC2+CB2
x=7 AC=28cm, CB=21cm
3) To find the radius of the inscribed circle, apply the formula: r═;r=cm
Answer: 7cm
B14 The bisector of an acute angle of a right triangle divides the leg into segments of 10 cm and 26 cm. Find the radius of the circle circumscribed about this triangle.
AB - 13x, AC - 5x
3) Let us apply the Pythagorean theorem for ∆ ASV:
AB2= AC2 + BC2
169x2= 1396+25x2https://pandia.ru/text/78/060/images/image030_35.gif">4) Because the center of the circle circumscribed about a right triangle is the midpoint of the hypotenuseR= R=19.5cm
Answer: 19.5 cm
Q15, Q16, Q17 can be assigned at home, followed by testing in the classroom.
Problem No. 15 The bisector of a right angle of a right triangle divides the hypotenuse into segments in a ratio of 4:3. Find these segments if the radius of the inscribed circle is 7.
Answer: 32cm and 24cm
IN 1 6 A bisector drawn from the vertex of a rectangle divides its diagonal into segments of 65 cm and 156 cm. Find the area of the rectangle.
Answer 17340cm2
Q17The length of the circle circumscribed about a right triangle is 39https://pandia.ru/text/78/060/images/image023_47.gif" width="16" height="41">DВ∙DК; ВD - ? DК - ?
2) Let’s find S∆ABC using Heron’s formula: p = 21, S∆ABC = 84.
3) On the other hand, S ∆ABC = AC∙DB AC∙DB = 2S; DВ = ; DB = 12;
4) Let's take AK = x, then SC = 14 – x; Let's apply the property of the angle bisector of a triangle: =https://pandia.ru/text/78/060/images/image036_29.gif" width="21" height="41 src=">.gif" width="20" height= "16 src="> x = 6.5: AK = 6.5
5) DK = AK – AD..gif" width="16" height="41 src=">∙12∙1.5 = 9.
Triangle - a polygon with three sides, or a closed broken line with three links, or a figure formed by three segments connecting three points that do not lie on the same straight line (see Fig. 1).
Basic elements of triangle abc
Peaks – points A, B, and C;
Parties – segments a = BC, b = AC and c = AB connecting the vertices;
Angles – α, β, γ formed by three pairs of sides. Angles are often designated in the same way as vertices, with the letters A, B, and C.
The angle formed by the sides of a triangle and lying in its interior area is called an interior angle, and the one adjacent to it is the adjacent angle of the triangle (2, p. 534).
Heights, medians, bisectors and midlines of a triangle
In addition to the main elements in a triangle, other segments with interesting properties are also considered: heights, medians, bisectors and midlines.
Height
Triangle heights- these are perpendiculars dropped from the vertices of the triangle to opposite sides.
To plot the height, you must perform the following steps:
1) draw a straight line containing one of the sides of the triangle (if the height is drawn from the vertex of an acute angle in an obtuse triangle);
2) from the vertex lying opposite the drawn line, draw a segment from the point to this line, making an angle of 90 degrees with it.
The point where the altitude intersects the side of the triangle is called height base (see Fig. 2).
Properties of triangle altitudes
In a right triangle, the altitude drawn from the vertex of the right angle splits it into two triangles similar to the original triangle.
In an acute triangle, its two altitudes cut off similar triangles from it.
If the triangle is acute, then all the bases of the altitudes belong to the sides of the triangle, and obtuse triangle two heights fall on the continuation of the sides.
Three heights in acute triangle intersect at one point and this point is called orthocenter triangle.
Median
Medians(from Latin mediana – “middle”) - these are segments connecting the vertices of the triangle with the midpoints of the opposite sides (see Fig. 3).
To construct the median, you must perform the following steps:
1) find the middle of the side;
2) connect the point that is the middle of the side of the triangle with the opposite vertex with a segment.
Properties of triangle medians
The median divides a triangle into two triangles of equal area.
The medians of a triangle intersect at one point, which divides each of them in a ratio of 2:1, counting from the vertex. This point is called center of gravity triangle.
The entire triangle is divided by its medians into six equal triangles.
Bisector
Bisectors(from Latin bis - twice and seko - cut) are the straight line segments enclosed inside a triangle that bisect its angles (see Fig. 4).
To construct a bisector, you must perform the following steps:
1) construct a ray coming out from the vertex of the angle and dividing it into two equal parts (the bisector of the angle);
2) find the point of intersection of the bisector of the angle of the triangle with the opposite side;
3) select a segment connecting the vertex of the triangle with the intersection point on the opposite side.
Properties of triangle bisectors
The bisector of an angle of a triangle divides the opposite side in the ratio equal to the ratio two adjacent sides.
Bisectors internal corners triangles intersect at one point. This point is called the center of the inscribed circle.
The bisectors of the internal and external angles are perpendicular.
If the bisector external corner triangle intersects the continuation of the opposite side, then ADBD=ACBC.
The bisectors of one internal and two external angles of a triangle intersect at one point. This point is the center of one of the three excircles this triangle.
The bases of the bisectors of two internal and one external angles of a triangle lie on the same straight line if the bisector of the external angle is not parallel to the opposite side of the triangle.
If the bisectors of the external angles of a triangle are not parallel to opposite sides, then their bases lie on the same straight line.
Hello, dear readers! Today we will start solving problems onproperties of the bisector and median of a triangle.
First, let's remember what a bisector and a median are.
Bisector - this is the segment CD that extends from the vertex of the angle of the triangle, bisects an angle and ends on the opposite side.
Median is a segment of CM, which connects vertex of the triangle With the middle of the opposite side.
Since a triangle has three vertices and three sides, it will also have three median bisectors.
Task 1. Given rectangular triangle ABC. The median AD and bisector AM are drawn from vertex A to side BC. The angle between the median and the bisector is 17°. Find sharp corners triangle.
Solution: Since AM is a bisector, then angle BAM equal to angle MAC and they are equal to 45°. But the angle DAM is 17°. Hence, the angle VAD is equal to the difference between the angles VAM and LAM, or 45-17 = 28°.
We know that The median drawn from the vertex of the right angle of a right triangle divides this triangle into 2 isosceles triangles. Namely triangles АВД and АДС.
And now, since the triangle ABC is isosceles, the angles at its base are equal, i.e. angle VAD is equal to angle AAD and they are both equal to 28°.
This means that in a right triangle, angle B is 28°.
But the sum of the acute angles in a right triangle is 90°. Hence, angle C will be equal to 90 - 28 = 62°.
Answer: The acute angles in a right triangle are 28° and 62°.
Task 2. Prove that the bisectors adjacent corners perpendicular.
Solution: We know the property of measuring angles, which states that if rays are drawn inside an angle, they will split it into several angles and the sum of the degree measures of these angles will be equal to degree measure original angle.
Therefore we have: α+α+β+β = 180°.
Or 2α+2β = 180°.
We shorten the right and left side equation by 2, we get: α + β = 90°.
Those. angle DVK between bisectors VD and VK of adjacent angles ALWAYS equal to 90° regardless of the size of adjacent angles.
Task 3. Given a trapezoid ABCD. The bisectors of angles A and B intersect at point M.
Find AB if AM = 24, BM = 18.
Solution: From previous task we found out that bisectors of adjacent angles always form an angle of 90°.
Bisectors drawn from the corners of a trapezoid adjacent to the side also form an angle of 90°.
In fact: angles A and B of a trapezoid add up to 180°, as are one-sided angles with parallel lines AD and BC and secant AB.
This means that the halves of these angles will add up to 90°.
And if in a triangle 2 angles add up to 90°, then the third angle will be equal to 90°, because the sum of the interior angles of a triangle is 180°.
So this is a right-angled triangle. We know that it has 2 legs; we can find the hypotenuse using the Pythagorean theorem.
AB² = AM² + BM² = 24² + 18² = 900. Hence, AB = 30.
Answer: AB = 30.
Among the numerous subjects of secondary school there is one such as “geometry”. It is traditionally believed that the founders of this systematic science are the Greeks. Today, Greek geometry is called elementary, since it was she who began the study of the simplest forms: planes, straight lines, and triangles. We will focus our attention on the latter, or rather on the bisector of this figure. For those who have already forgotten, the bisector of a triangle is a segment of the bisector of one of the corners of the triangle, which divides it in half and connects the vertex with a point located on the opposite side.
The bisector of a triangle has a number of properties that you need to know when solving certain problems:
- The angle bisector is locus points removed by equal distances from the sides adjacent to the corner.
- The bisector in a triangle divides the side opposite the angle into segments that are proportional to the adjacent sides. For example, given a triangle MKB, where a bisector emerges from angle K, connecting the vertex of this angle with point A on the opposite side MB. Having analyzed this property and our triangle, we have MA/AB=MK/KB.
- The point at which the bisectors of all three angles of a triangle intersect is the center of a circle that is inscribed in the same triangle.
- The bases of the bisectors of one external and two internal angles are on the same straight line, provided that the bisector of the external angle is not parallel to the opposite side of the triangle.
- If two bisectors of one then this
It should be noted that if three bisectors are given, then constructing a triangle from them, even with the help of a compass, is impossible.
Very often, when solving problems, the bisector of a triangle is unknown, but it is necessary to determine its length. To solve this problem, you need to know the angle that is bisected by the bisector and the sides adjacent to this angle. In this case, the required length is defined as the ratio of twice the product of the sides adjacent to the corner and the cosine of the angle divided in half to the sum of the sides adjacent to the corner. For example, given the same triangle MKB. The bisector emerges from angle K and intersects the opposite side of the MV at point A. The angle from which the bisector emerges is denoted by y. Now let’s write down everything that is said in words in the form of a formula: KA = (2*MK*KB*cos y/2) / (MK+KB).
If the value of the angle from which the bisector of a triangle emerges is unknown, but all its sides are known, then to calculate the length of the bisector we will use an additional variable, which we will call the semi-perimeter and denote by the letter P: P=1/2*(MK+KB+MB). After this, we will make some changes to the previous formula by which the length of the bisector was determined, namely, in the numerator of the fraction we put double the product of the lengths of the sides adjacent to the corner by the semi-perimeter and the quotient, where the length of the third side is subtracted from the semi-perimeter. We will leave the denominator unchanged. In the form of a formula, it will look like this: KA=2*√(MK*KB*P*(P-MB)) / (MK+KB).
Bisector isosceles triangle together with general properties has several of its own. Let's remember what kind of triangle this is. Such a triangle has two equal sides and equal angles adjacent to the base. It follows that the bisectors that descend to sides isosceles triangle, equal to each other. In addition, the bisector lowered to the base is both the height and the median.
You will need
- - right triangle;
- - known length of legs;
- - known length of the hypotenuse;
- - known angles and one of the parties;
- - the known lengths of the parts into which the bisector divides the hypotenuse.
Instructions
Use the following theorem: the relations of the legs and the relations of adjacent segments on which there is a direct angle divides the hypotenuse are equal. That is, divide the legs into each other and equate them to the ratio x/(c-x). At the same time, make sure that the numerator contains the leg adjacent to x. Solve the resulting equation and find x.
Having found out the length of the segments for which the bisector of a straight line angle divided the hypotenuse, find the length of the hypotenuse itself using the theorem of sines. You know the angle between the leg and the bisector - 45⁰, two sides inner triangle Same.
Substitute the data into the sine theorem: x/sin45⁰=l/sinα. Simplifying the expression, you get l=2xsinα/√2. Substitute the found x: l=2c*cosα*sinα/√2(sinα+cosα)=c*sin2α/2cos(45⁰-α). This is the bisector of the line angle, expressed through the hypotenuse.
If you are given legs, you have two options: either find the length of the hypotenuse using the Pythagorean theorem, according to which the sum of the squares of the legs is equal to the square of the hypotenuse and solve in the above way. Or use the following ready-made formula: l=√2*ab/(a+b), where a and b are the lengths of the legs.
Sources:
- how to find the length of a straight line
Dividing an angle in half and calculating the length of a line drawn from its top to the opposite side is something that cutters, surveyors, installers and people of some other professions need to be able to do.
You will need
- Tools Pencil Ruler Protractor Sine and Cosine Tables Mathematical formulas and concepts: Definition of a bisector Theorems of sines and cosines Bisector theorem
Instructions
Construct a triangle of the required size, depending on what is given to you? dfe sides and the angle between them, three sides or two angles and the side located between them.
Label the vertices of the corners and sides with the traditional Latin letters A, B and C. The vertices of the corners denote, opposing sides- lowercase. Label the corners Greek letters?,? And?
Using the theorems of sines and cosines, calculate the angles and sides triangle.
Remember bisectors. Bisector - dividing an angle in half. Angle bisector triangle divides the opposite into two segments, which are equal to the ratio of the two adjacent sides triangle.
Draw the bisectors of the angles. Label the resulting segments with the names of the angles written lowercase letters, with subscript l. Side c is divided into segments a and b with indices l.
Calculate the lengths of the resulting segments using the law of sines.
Video on the topic
note
The length of the segment, which is simultaneously the side of the triangle formed by one of the sides of the original triangle, the bisector and the segment itself, is calculated using the law of sines. In order to calculate the length of another segment of the same side, use the ratio of the resulting segments and the adjacent sides of the original triangle.
To avoid confusion, draw bisectors different angles different colors.
Tip 3: How to find the bisector in a right triangle
A bisector is a ray that divides an angle in half. The bisector, in addition to this, has many more properties and functions. And in order to calculate its length in rectangular triangle, you will need the formulas and instructions below.
You will need
- - calculator
Instructions
Multiply side a, side b, the semi-perimeter of the triangle p and the number four 4*a*b. Next, the resulting amount must be multiplied by the difference between the half-perimeter p and side c 4*a*b*(p-c). Extract the root of what you got earlier. SQR(4*a*b*(p-c)). And divide the result by the sum of sides a and b. Thus, we have obtained one of the formulas for finding the bisector using Stewart's theorem. It can be interpreted in a different way, presenting it this way: SQR(a*b*(a+b+c)(a+b-c)). There are several more options for this formula, obtained on the basis of the same theorem.
Multiply side a by side b. From the result, subtract the lengths of the segments e and d into which the bisector l divides side c. The results look like this: a*b-e*d. Next, you need to extract the root of the presented difference SQR (a*b-e*d). This is another method for the length of the bisector in triangles. Do all calculations carefully, repeating at least 2 times for possible errors.
Multiply two by sides a and b, plus the cosine of angle c divided in half. Next, the resulting product must be divided by the sum of sides a and b. Provided the cosines are known, this method of calculation will be the most convenient for you.
Subtract the cosine of angle b from the cosine of angle a. Then divide the resulting difference in half. The divisor that we will need later has been calculated. Now all that remains is to divide the height drawn to side c by the previously calculated number. Now another calculation method has been demonstrated for finding the bisector in a rectangular triangle. The choice of method for finding the numbers you need is up to you, and also depends on what is provided in the conditions for this or that geometric figure.
Video on the topic
Let two intersecting lines given by their equations be given. It is required to find the equation of a line that, passing through the point of intersection of these two lines, would exactly bisect the angle between them, that is, would be a bisector.