Theorems: properties of a trapezoid
\\ [(\\ Large (\\ text (Free trapezoid))) \\]
Definitions
A trapezoid is a convex quadrilateral in which two sides are parallel and the other two are not parallel.
The parallel sides of a trapezoid are called its bases, and the other two sides are called lateral sides.
The height of a trapezoid is a perpendicular dropped from any point on one base to another base.
Theorems: properties of a trapezoid
1) The sum of the angles at the side is \\ (180 ^ \\ circ \\).
2) Diagonals divide the trapezoid into four triangles, two of which are similar, and the other two are equal.
Evidence
1) Because \\ (AD \\ parallel BC \\), then the angles \\ (\\ angle BAD \\) and \\ (\\ angle ABC \\) are one-sided for these lines and the secant \\ (AB \\), therefore, \\ (\\ angle BAD + \\ angle ABC \u003d 180 ^ \\ circ \\).
2) Because \\ (AD \\ parallel BC \\) and \\ (BD \\) is a secant, then \\ (\\ angle DBC \u003d \\ angle BDA \\) as criss-cross.
Also \\ (\\ angle BOC \u003d \\ angle AOD \\) as vertical.
Therefore, in two angles \\ (\\ triangle BOC \\ sim \\ triangle AOD \\).
Let us prove that \\ (S _ (\\ triangle AOB) \u003d S _ (\\ triangle COD) \\)... Let \\ (h \\) be the height of the trapezoid. Then \\ (S _ (\\ triangle ABD) \u003d \\ frac12 \\ cdot h \\ cdot AD \u003d S _ (\\ triangle ACD) \\)... Then: \
Definition
The middle line of a trapezoid is a segment connecting the midpoints of the sides.
Theorem
The middle line of the trapezoid is parallel to the bases and equal to their half-sum.
Evidence*
1) Let us prove parallelism.
Draw a straight line \\ (MN "\\ parallel AD \\) (\\ (N" \\ in CD \\)) through the point \\ (M \\). Then, by Thales' theorem (i.e. \\ (MN "\\ parallel AD \\ parallel BC, AM \u003d MB \\)) the point \\ (N "\\) is the midpoint of the segment \\ (CD \\). So the points \\ (N \\) and \\ (N" \\) will coincide.
2) Let us prove the formula.
Let's run \\ (BB "\\ perp AD, CC" \\ perp AD \\). Let be \\ (BB "\\ cap MN \u003d M", CC "\\ cap MN \u003d N" \\).
Then, by Thales' theorem, \\ (M "\\) and \\ (N" \\) are the midpoints of the segments \\ (BB "\\) and \\ (CC" \\), respectively. So, \\ (MM "\\) is the middle line \\ (\\ triangle ABB" \\), \\ (NN "\\) is the middle line \\ (\\ triangle DCC" \\). Therefore: \
Because \\ (MN \\ parallel AD \\ parallel BC \\) and \\ (BB ", CC" \\ perp AD \\), then \\ (B "M" N "C" \\) and \\ (BM "N" C \\) are rectangles. By Thales' theorem, from \\ (MN \\ parallel AD \\) and \\ (AM \u003d MB \\) it follows that \\ (B "M" \u003d M "B \\). Hence, \\ (B" M "N" C "\\) and \\ (BM "N" C \\) are equal rectangles, therefore \\ (M "N" \u003d B "C" \u003d BC \\).
Thus:
\ \\ [\u003d \\ dfrac12 \\ left (AB "+ B" C "+ BC + C" D \\ right) \u003d \\ dfrac12 \\ left (AD + BC \\ right) \\]
Theorem: property of an arbitrary trapezoid
The midpoints of the bases, the point of intersection of the trapezoid diagonals and the point of intersection of the extensions of the lateral sides lie on one straight line.
Evidence*
It is recommended to read the proof after studying the topic "Similarity of triangles".
1) Let us prove that the points \\ (P \\), \\ (N \\) and \\ (M \\) lie on one straight line.
Draw the line \\ (PN \\) (\\ (P \\) is the intersection point of the extensions of the lateral sides, \\ (N \\) is the middle of \\ (BC \\)). Let it intersect the side \\ (AD \\) at the point \\ (M \\). Let us prove that \\ (M \\) is the midpoint of \\ (AD \\).
Consider \\ (\\ triangle BPN \\) and \\ (\\ triangle APM \\). They are similar in two angles (\\ (\\ angle APM \\) - common, \\ (\\ angle PAM \u003d \\ angle PBN \\) as corresponding for \\ (AD \\ parallel BC \\) and \\ (AB \\) secant). Means: \\ [\\ dfrac (BN) (AM) \u003d \\ dfrac (PN) (PM) \\]
Consider \\ (\\ triangle CPN \\) and \\ (\\ triangle DPM \\). They are similar in two angles (\\ (\\ angle DPM \\) - common, \\ (\\ angle PDM \u003d \\ angle PCN \\) as corresponding with \\ (AD \\ parallel BC \\) and \\ (CD \\) secant). Means: \\ [\\ dfrac (CN) (DM) \u003d \\ dfrac (PN) (PM) \\]
From here \\ (\\ dfrac (BN) (AM) \u003d \\ dfrac (CN) (DM) \\)... But \\ (BN \u003d NC \\), therefore \\ (AM \u003d DM \\).
2) Let us prove that points \\ (N, O, M \\) lie on one straight line.
Let \\ (N \\) be the midpoint of \\ (BC \\), \\ (O \\) be the intersection point of the diagonals. Draw a line \\ (NO \\), it intersects the side \\ (AD \\) at the point \\ (M \\). Let us prove that \\ (M \\) is the midpoint of \\ (AD \\).
\\ (\\ triangle BNO \\ sim \\ triangle DMO \\) in two angles (\\ (\\ angle OBN \u003d \\ angle ODM \\) as crosswise with \\ (BC \\ parallel AD \\) and \\ (BD \\) secant; \\ (\\ angle BON \u003d \\ angle DOM \\) as vertical). Means: \\ [\\ dfrac (BN) (MD) \u003d \\ dfrac (ON) (OM) \\]
Similarly \\ (\\ triangle CON \\ sim \\ triangle AOM \\)... Means: \\ [\\ dfrac (CN) (MA) \u003d \\ dfrac (ON) (OM) \\]
From here \\ (\\ dfrac (BN) (MD) \u003d \\ dfrac (CN) (MA) \\)... But \\ (BN \u003d CN \\), therefore \\ (AM \u003d MD \\).
\\ [(\\ Large (\\ text (isosceles trapezoid))) \\]
Definitions
A trapezoid is called rectangular if one of its corners is straight.
A trapezoid is called isosceles if its sides are equal.
Theorems: properties of an isosceles trapezoid
1) At an isosceles trapezoid, the angles at the base are equal.
2) The diagonals of an isosceles trapezoid are equal.
3) The two triangles formed by the diagonals and the base are isosceles.
Evidence
1) Consider an isosceles trapezoid \\ (ABCD \\).
From the vertices \\ (B \\) and \\ (C \\) we drop the perpendiculars \\ (BM \\) and \\ (CN \\) to the side \\ (AD \\), respectively. Since \\ (BM \\ perp AD \\) and \\ (CN \\ perp AD \\), then \\ (BM \\ parallel CN \u200b\u200b\\); \\ (AD \\ parallel BC \\), then \\ (MBCN \\) is a parallelogram, therefore \\ (BM \u003d CN \\).
Consider right-angled triangles \\ (ABM \\) and \\ (CDN \\). Since their hypotenuses are equal and the leg \\ (BM \\) is equal to the leg \\ (CN \\), these triangles are equal, therefore, \\ (\\ angle DAB \u003d \\ angle CDA \\).
2) 
Because \\ (AB \u003d CD, \\ angle A \u003d \\ angle D, AD \\) - general, then on the first basis. Therefore, \\ (AC \u003d BD \\).
3) Because \\ (\\ triangle ABD \u003d \\ triangle ACD \\)then \\ (\\ angle BDA \u003d \\ angle CAD \\). Therefore, the triangle \\ (\\ triangle AOD \\) is isosceles. Similarly, it is proved that \\ (\\ triangle BOC \\) is isosceles.
Theorems: signs of an isosceles trapezoid
1) If the angles at the base of the trapezoid are equal, then it is isosceles.
2) If the diagonals of the trapezoid are equal, then it is isosceles.
Evidence
Consider a trapezoid \\ (ABCD \\) such that \\ (\\ angle A \u003d \\ angle D \\).
Let's complete the trapezoid to the triangle \\ (AED \\) as shown in the picture. Since \\ (\\ angle 1 \u003d \\ angle 2 \\), the triangle \\ (AED \\) is isosceles and \\ (AE \u003d ED \\). Angles \\ (1 \\) and \\ (3 \\) are equal as corresponding for parallel lines \\ (AD \\) and \\ (BC \\) and secant \\ (AB \\). Similarly, the angles \\ (2 \\) and \\ (4 \\) are equal, but \\ (\\ angle 1 \u003d \\ angle 2 \\), then \\ (\\ angle 3 \u003d \\ angle 1 \u003d \\ angle 2 \u003d \\ angle 4 \\)therefore the triangle \\ (BEC \\) is also isosceles and \\ (BE \u003d EC \\).
Eventually \\ (AB \u003d AE - BE \u003d DE - CE \u003d CD \\), that is, \\ (AB \u003d CD \\), as required.
2) Let \\ (AC \u003d BD \\). Because \\ (\\ triangle AOD \\ sim \\ triangle BOC \\), then we denote their similarity coefficient as \\ (k \\). Then if \\ (BO \u003d x \\), then \\ (OD \u003d kx \\). Likewise \\ (CO \u003d y \\ Rightarrow AO \u003d ky \\).
Because \\ (AC \u003d BD \\), then \\ (x + kx \u003d y + ky \\ Rightarrow x \u003d y \\). So \\ (\\ triangle AOD \\) is isosceles and \\ (\\ angle OAD \u003d \\ angle ODA \\).
Thus, by the first sign \\ (\\ triangle ABD \u003d \\ triangle ACD \\) (\\ (AC \u003d BD, \\ angle OAD \u003d \\ angle ODA, AD \\) - general). So, \\ (AB \u003d CD \\), etc.
"Series" MSU - School "was founded in 1999 by S. M. Sahakyan Geometry. Lesson development. 10-11 grades: C12 Textbook for general education. organizations / S. M. Sahakyan, V. F. Butuzov. - ... "
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3. Discuss orally the solutions of problems 1, 2, 3, given in the textbook.
In connection with the need to carry out continuous work on the development of the oral speech of students, it should be required from them not only to construct sections in the problems under consideration, but also to give an oral story about the construction progress with appropriate justifications.
For brevity of writing decisions, you can use well-known symbology.
More complex tasks for constructing sections of a tetrahedron and a parallelepiped, when the data points through which the section is drawn, lie inside the faces, can be considered in optional lessons and special courses.
For class and homework, you can use tasks 74, 75, 79-87, additional tasks to chapter I.
Problem 105. Draw a tetrahedron DABC and mark points M and N on edges BD and CD and an interior point K of face ABC. Construct a section of the tetrahedron with the MNK plane.
Solution. Let us denote the cutting plane by a letter.
Then M, N, M CDB, N CDB, CDB \u003d MN.
Two cases are possible: 10) MN BC \u003d P; 20) MN BC.
Let's consider them separately.
10) Draw a straight line MN. P, K, P ABC, K ABC, ABC \u003d PK. We draw a straight PK. Let it intersect sides AC and AB at points E and F. Draw segments NE and MF. The desired section is a quadrangle MNEF (Fig. 1.31).
20) Draw EF BC through point K. Draw segments NE and MF. The desired section is a quadrangle MNEF.
Problem 85. Draw a parallelepiped ABCDA1B1C1D1 and construct its section with the BKL plane, where K is the midpoint of edge AA1 and L is the midpoint of CC1.
Prove that the constructed section is a parallelogram.
Solution. Draw the segment KL. According to the A2 axiom, it lies in the section plane.
Since the points K and L are the midpoints of the lateral edges, the segment KL passes through the midpoint of the diagonal AC1, and therefore, according to property 20 of the parallelepiped (item 13), it passes through the midpoint of the diagonal BD1 (point O in Figure 1.32).
B, O, hence BD1. The desired section is the BLD1K quadrangle. Since its diagonals KL and BD1 are bisected by the intersection point, the quadrilateral BLD1K is a parallelogram.
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1. Explain how to construct a section of a DABC tetrahedron with a plane passing through the given points M, N, K.
2. In tasks 1-3, find the perimeter of the section if M, N, K are the midpoints of the edges and each edge of the tetrahedron is equal to a.
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1. Explain how to construct a section of a cube with a plane passing through three given points, which are either the vertices of the cube or the midpoints of its edges (three given points are highlighted in the figures).
2. In problems 1-4 and 6, find the perimeter of the section if the edge of the cube is a. In Problem 5, prove that AE \u003d 1 a.
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1. Explain how to construct a section of a parallelepiped by a plane passing through points B, D and M if M is the midpoint of edge B1C1.
2. Prove that the constructed section is equal to the femoral trapezoid.
3. Find the sides of the trapezoid.
Decision.
1) Let is the cutting plane, ABCD \u003d BD, BCC1B1 \u003d BM, MN BD, the section is the trapezoid BDNM.
2) BB1M \u003d DD1N, BM \u003d DN, BDNM trapezoid is isosceles.
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When solving problems related to the section of a tetrahedron by a certain plane, Menelaus's theorem often turns out to be useful, in some other problems - Cheva's theorem. Therefore, in classes with an in-depth study of mathematics, it is advisable to combine the study of item 14 "Problems of constructing sections" with the study of the theorems of Menelaus and Cheva (items 95 and 96). Let us give an example of such a problem.
Problem 1. In the tetrahedron ABCD on the edges AB, AD and BC, the points K, L and M are taken, respectively, so that AK: KB \u003d 2: 3, AL \u003d LD, BM: MC \u003d 4: 5.
Construct a section of the tetrahedron with the KLM plane and find the ratio in which this plane divides the edge CD.
Decision.
1) Draw the segments KL and KM, and then continue the segments KL and BD, lying in the plane ABD, until the intersection at point E (Fig. 1.33). Points E and M lie in the KLM cut plane and also in the BCD plane.
Having drawn the segment ME, we get the point N at which the cutting plane KLM intersects the edge CD.
The four-sided KLNM is the required section.
2) Find the ratio CN: ND. To this end, we apply Menelaus' theorem to triangles ABD and BCD. Points K and L lie on sides AB and AD of triangle ABD, and point E on the extension of side BD, and points K, L and E lie on one straight line. Therefore, according to Menelaus' theorem, the equality
AK BE DL
= 1.
KB ED LA
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MC BE find the required ratio CN: ND \u003d 15: 8.
In order to use Menelaus' theorem in problem 105 of the textbook, an additional task can be given:
Find the ratio in which the plane MNK divides the edge AB if CN: ND \u003d 2: 1, BM \u003d MD and point K is the midpoint of the median AL of triangle ABC. (Answer: 3: 2.) A similar additional task can be given in Problem 106:
Find the ratio in which the plane MNK divides the edge BC if it divides the edge AB in the ratio 1: 4 (counting from point A), point K is the midpoint of edge CD, and point N lies on the median DL of triangle ACD, for h m DN: NL \u003d 3: 2. (Answer: 4: 3.) On the application of Cheva's theorem, we can consider the following problem:
Problem 2. Points C1, A1, B1 are marked on the edges AB, BC, and CA of the tetrahedron ABCD so that AC1: C1B \u003d 1: 2, BA1: A1C \u003d 2: 3, CB1: B1A \u003d 3: 1.
Prove that planes ADA1, BDB1 and CAC1 intersect in a straight line.
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1. Review the main questions of the topic "Parallelism of lines and planes", after hearing the answers of the students. These questions are formulated in credit cards # 1.
2. Conduct mathematical dictation No. 1.1. The dictation is given in the didactic materials.
3. Consider solutions to some of the problems from the credit cards and from the textbook.
The study of the topic "Parallelism of lines and planes" ends with a test No. 1.2 and credit No. 1 on this topic.
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Test work No. 1.2 Option 1
10. Lines a and b lie in parallel planes and. Can these lines be: a) parallel;
20. Through the point O, lying between the parallel planes and, lines l and m are drawn. Line l intersects the planes and at points A1 and A2, respectively, line m - at points B1 and B2. Find the length from cutting A2B2 if A1B1 \u003d 12cm, B1O: OB2 \u003d 3: 4.
3. Draw a parallelepiped ABCDA1B1C1D1 and construct its section by a plane passing through points M, N, and K, which are the midpoints of edges AB, BC, and DD1.
Option 2
10. Lines a and b lie in intersecting planes and. Can these lines be: a) parallel;
b) crossing? Draw a drawing for every possible occasion.
20. Through the point O, not lying between the parallel planes and, lines l and m are drawn. Line l intersects the planes and at points A1 and A2, respectively, line m - at points B1 and B2. Find the length from cutting A1B1 if A2B2 \u003d 15cm, OB1: OB2 \u003d 3: 5.
3. Draw the tetrahedron DABC and construct its section by a plane passing through points M and N, which are the midpoints of edges DC and BC, and a point K such that K DA, AK: KD \u003d 1: 3.
Answer:
Option 2 Option 1
10. Fig. 1.35, a b, a b.
10. Fig. 1.34, a b, a b.
3. Section - trapezoid.
3. Section - pentagon.
Fig. 1.34 Fig. 1.35
Lesson number 24 Exam t number 1. Parallelism of lines and planes Card 1
1. Formulate the axioms A1, A2 and A3 of stereometry.
Formulate and prove the corollaries of the axioms.
2. Prove that through any point in space that does not lie on the given straight line, there is a straight line parallel to the given one, and, moreover, only one.
3. The plane intersects sides AB and AC of triangle ABC at points B1 and C1, respectively. It is known that BC, AB: B1B \u003d 5: 3, AC \u003d 15 cm. Find AC1.
Card 2
1. Formulate the definition of parallel lines and planes. Formulate and prove a theorem expressing the criterion for parallelism of a line and a plane.
2. Prove that if one of two parallel lines intersects this plane, then the other line also intersects this plane.
3. Each edge of the tetrahedron DABC is equal to 2 cm. Construct a section of the tetrahedron with a plane passing through points B, C and the midpoint of the edge AD. Calculate the perimeter of the section.
Card 3
1. Formulate the definition of crossed lines. Formulate and prove a theorem expressing the criterion of intersecting lines.
2. Prove that if two lines are parallel to the third line, then they are parallel.
3. Construct a section of the parallelepiped ABCDA1B1C1D1 with the plane passing through points A, C and M, where M is the midpoint of edge A1D1.
Card 4
1. Formulate the definition of parallel planes. Formulate and prove a theorem expressing the criterion for the parallelism of two planes.
2. Prove that through each of the two crossing lines there is a plane parallel to the other line, and, moreover, only one.
3. ABCDA1B1C1D1 is a cube with an edge of 4 cm. Construct a section of the cube by a plane passing through points A, D1 and M, where M is the midpoint of the edge BC. Calculate the perimeter of the section.
Card 5
1. Prove that the opposite faces of the parallelepiped are parallel and equal.
2. Prove that if the sides of two angles are correspondingly codirectional, then such angles are equal.
3. Parallel planes and intersect the AB side of the BAC angle, respectively, at points A1 and A2, and the AC side of this angle, respectively, at the B1 and B2 points. Find AA1 if A1A2 \u003d 6 cm, AB2: AB1 \u003d 3: 2.
Card 6
1. Prove that the diagonals of the parallelepiped meet at one point and are halved by this point.
2. Prove that if two parallel planes are intersected by the third, then the lines of their intersection are parallel.
3. Point C lies on the segment AB. A plane is drawn through point A, and parallel lines are drawn through points B and C, intersecting this plane at points B1 and C1, respectively. Find the length of the segment BB1 if AC: CB \u003d 4: 3, CC1 \u003d 8 cm.
1. Students are given credit cards containing the main theoretical questions and some typical tasks in advance (about two weeks before the credit).
2. In preparation for the test, students take notes. These notes (perhaps in the form of drafts), which indicate the repetition of the training material and preparation for the test, are shown by the students to the teacher on the day of the test. They can be used for credit. At the same time, on the basis of the conversation and additional questions, the teacher finds out the depth of the mastery of the topic by the students.
3. What is taught by the teacher with the help of the most prepared student - consultants. To do this, the class must be divided into several groups, each of which has 4-5 students. One of them is the teacher's assistant in carrying out the test. In the preceding lessons and at the beginning of the test, the teacher should make sure that the counselors themselves have a good command of the teaching material.
4. During the lesson, the teacher interviews many students. At the end of the lesson, he approves the grades given by the consultants. In some cases, after the lesson, the teacher can check the students' notes made in the lesson, and then give the final grade for the test.
5. The teacher gives the final mark for the six months on the basis of the current marks for independent and control work, as well as the oral answer of the students.
The decisive role in this is played by the credit score.
P E R P E N D I K UL Y R N O S T P R I M Y X
AND PLANES
§ 1. DIRECT PERPENDICULARITY
AND PLANES
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The main objectives of the lesson Introduce the concept of perpendicular lines in space, prove the lemma on the perpendicularity of two parallel lines to the third line, define the perpendicularity of a line and a plane, prove theorems that establish a connection between the parallelism of lines and their perpendicularity to the plane.
1. Recall the concept of the angle between two crossing lines, introduce the concept of perpendicularity of two straight lines in space. Note that perpendicular lines can intersect and can be crossed (see Fig. 43 of the textbook).
2. Prove the lemma: if one of two parallel lines is perpendicular to the third line, then the other line is also perpendicular to this line.
The proof is based on the use of the concept of the angle between lines and can be carried out by the students themselves, based on the text and figure 44 of the textbook.
3. Formulate the definition of the perpendicularity of a line and a plane. Enter designation a. Illustrate the concept of perpendicularity of a line and a plane using Figure 45 and examples from real life.
4. Prove the theorem: if one of two parallel lines is perpendicular to the plane, then the other line is also perpendicular to this plane.
The proof of the theorem is simple. It is based on the definition of the perpendicularity of a straight line and a plane and the lemma considered above and consists of two stages:
1) x, x is an arbitrary straight line. From the condition a it follows (by the definition of the perpendicularity of a straight line and a plane) that a x;
2) since a1 a (by hypothesis) and a x, then (according to the lemma on the perpendicularity of two parallel lines to the third line) a1 x.
So, line a1 is perpendicular to an arbitrary line x lying in the plane. This means that a1.
5. Prove the inverse theorem: if two straight lines are perpendicular to the plane, then they are parallel.
The proof is carried out according to the textbook (see Fig. 47, a, b). You can repeat this proof in the following lessons.
At first glance, it may seem strange why this theorem is called the reverse of the previous theorem.
Indeed, in the previous theorem, the condition was that a a1 and a, and the conclusion of the theorem was: a1. In this theorem, the condition is that a and a1, and the conclusion is that a a1.
Thus, from the formal point of view, this theorem is not the reverse of the previous one, since the condition and conclusion of this theorem do not coincide, respectively, with the conclusion and condition of the previous theorem. Nevertheless, it is possible to formulate these theorems in such a way that each of them will be the opposite of the other.
Let me give this formulation.
Let the line a be perpendicular to the plane. Then:
if a a1, then a1, and vice versa:
if a1, then a a1.
6. For classroom and homework, you can use tasks 116-118, 120.
Problem 116 a). Given a parallelepiped ABCDA1B1C1D1.
Prove that DC B1C1 and AB A1D1 if BAD \u003d 90 °.
Decision.
1) In a parallelepiped, all faces are parallelograms. Since BAD \u003d 90 ° (by condition), the face ABCD is a rectangle, therefore AB AD and DC BC (Fig. 2.1).
2) B1C1 BC (since BB1C1C is a parallelogram) and BC DC. Hence, by the lemma on the perpendicularity of two parallel lines to the third, B1C1 DC. Fig. 2.1
3) It can be proved similarly that AB A1D1. Indeed, A1D1 AD (since AA1D1D is a parallelogram) and AB AD, therefore AB A1D1.
Problem 120. Through the point O of the intersection of the diagonals of the square with side a, a straight line OK is drawn, perpendicular to the plane of the square.
Find the distance from point K to the vertices of the square, if Fig. 2.2 OK \u003d b.
Decision.
2) Triangles KAO, KBO, KCO and KDO are equal in two legs, whence KA \u003d KB \u003d KC \u003d KD (Fig.2.2).
KAO we get AO \u003d a 2. Since KA \u003d
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Lesson number 26 Lesson topic: The sign of the perpendicularity of a line and a plane The main objectives of the lesson To study a theorem expressing the sign of perpendicularity of a line and a plane; consider the problem of applying this theorem.
Approximate lesson plan
1. Review the theoretical material of the previous lesson by interviewing students.
2. As preparatory work for the study of new material, solve problem 119.
Problem 119. Line OA is perpendicular to plane OBC, and point O is the midpoint of segment AD.
Prove that: a) AB \u003d DB; b) AB \u003d AC if OB \u003d OC;
c) OB \u003d OC if AB \u003d AC.
Decision.
a) OA OBC by condition, therefore, OA OB by definition of the perpendicularity of a straight line to a plane.
OA \u003d OD by the problem statement; therefore, the line OB is the median perpendicular to the segment AD, and, therefore, AB \u003d DB (Fig. 2.3).
b) Since by the condition OA OBC, then OA OC. If OB \u003d OC, then right-angled triangles AOC and AOB are equal in two legs, and, therefore, their hypotenses are equal, ie AB \u003d AC.
c) If AB \u003d AC, then rectangular triangles AOC and Fig. 2.3 AOB are equal in leg and hypotenuse, from which it follows that OB \u003d OC.
3. Prove a theorem expressing the linearity of a straight line and plane: if a straight line is perpendicular to two intersecting straight lines lying in a plane, then it is perpendicular to this plane.
In the process of proving the theorem, the following stages are distinguished:
1) First, we consider the case when the straight line a passes through the point O of intersection of the straight lines p and q lying on the plane. We prove that line a is perpendicular to any line lying in the plane and passing through the point O.
2) Using the lemma on the perpendicularity of two parallel lines to the third, we conclude that the line a is perpendicular to any line lying in the plane. This means that a.
3) We now consider the case when the line a does not pass through the point O of the intersection of p and q. In this case, draw a straight line a1 through point O and parallel to the straight line a. By the above lemma, a1 p and a1 q, and therefore, according to what was proved in the first case, a1. Hence, by the first theorem in Section 16, it follows that a. This completes the proof of the theorem.
4. Due to the fact that the proof of the theorem consists of several stages, you can invite students to write the outline of the proof in accordance with the content of slide 2.1.
The slide can be used to summarize this lesson and in the next lesson.
5. For classroom and homework, you can use tasks 121, 124, 126, 128.
Problem 128. Through the intersection point O of the diagonals of the parallelogram ABCD, a line OM is drawn so that MA \u003d MC, MB \u003d MD. Prove that line OM is perpendicular to the plane of the parallelogram.
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1. Formulate the definition of perpendicularity of a line and a plane.
2. Theorem. If a straight line is perpendicular to two intersecting straight lines lying in a plane, then it is perpendicular to this plane.
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Decision.
1) Since MA \u003d MC (by condition) and AO \u003d OC (the diagonals of the parallelogram are halved by the point of intersection), then the segment MO - the median is equal to the femoral triangle AMC (Fig. 2.4).
Therefore, MO is also the height of this triangle, i.e.
2) It is proved similarly, Fig. 2.4 what is MO BD.
3) Since MO AC and MO BD, then according to the perpendicularity of the straight line and the plane MO ABCD.
Lesson number 27 Lesson topic: Theorem about a straight line perpendicular to a plane The main objectives of the lesson Repeat the proof of the theorem, which expresses at the sign of perpendicularity of a straight line and a plane; consider the theorem from item 18: through any point in space there is a straight line perpendicular to the given plane, and moreover only one.
Approximate lesson plan
1. Repeat the proof of the theorem expressing the criterion for the perpendicularity of a line and a plane.
2. Check selectively problem solving from homework.
3. Formulate the theorem: through any point in space there is a straight line perpendicular to the given plane, and moreover, only one.
The assertion of the theorem seems to be quite obvious, but its rigorous proof is not simple.
Students with an increased interest in mathematics can be encouraged to parse the proof at home using the textbook themselves. In this case, their attention should be paid to the fact that in the first part of the proof a plane is introduced into consideration, passing through a given point M and perpendicular to a given line a.
The existence of such a plane was proved in the problem with a solution given in Sec. 17, and the uniqueness of such a plane was proved in Problem 133, which is also given with a solution. Thus, the complete proof of this theorem is very cumbersome, and therefore the teacher, at his own discretion, can present it with varying degrees of completeness, depending on the level of the class. Some fragments of the proof (problem from p. 17, task 133) can be considered in lessons no. 28-30, devoted to repetition of the theory and solving problems on the topic.
4. Conduct a frontal survey of students using slide 2.2.
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5. For classroom and homework, you can use tasks 122, 123, 125, 127.
Problem 122. Line CD is perpendicular to the plane of a regular triangle ABC. Line OK, parallel to line CD, is drawn through the center O of this triangle. It is known that AB \u003d 16 3 cm, OK \u003d 12 cm, CD \u003d 16 cm. Find the distances from points D and K to the apexes A and B of the triangle.
Decision.
1) By the condition of the problem, OK CD, therefore, OK ABC (Fig. 2.5).
2) Point O is the center of a regular triangle ABC, therefore, OA \u003d OB \u003d OC \u003d AB \u003d 16 cm.
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Lessons 28-30 Lesson topic: Solving problems on the perpendicularity of a line and a plane. Repetition of questions of theory The main tasks of the lessons To develop the skills of solving the main types of problems on the perpendicularity of a straight line and a plane, to repeat the questions of theory.
1. Repeat the theory questions during the survey of students (pp. 15-18).
2. Solve selectively tasks 129-137, use questions 1-9 to chapter II.
3. Consider partially or completely the proof of the theorem from item 18.
4. You can use tasks from didactic materials.
5. You can conduct a mathematical dictation (No. 2 in didactic materials).
6. Useful work in the lesson with slide 2.3.
In the lesson number 30 independent work is carried out.
Independent work No. 2.1 Option 1
10. Given: AB, M and K are arbitrary points of the plane. Prove that AB MK.
2. Triangle ABC is regular, point O is its center. Line OM is perpendicular to plane ABC.
a) 0 Prove that MA \u003d MB \u003d MC.
b) Find MA if AB \u003d 6 cm, MO \u003d 2 cm.
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Option 2
10. Given: line MA is perpendicular to the plane of triangle ABC. Prove that MA BC.
2. The four-gon ABCD is a square, point O is its center. Line OM is perpendicular to the plane of the square.
a) 0 Prove that MA \u003d MB \u003d MC \u003d MD.
b) Find MA if AB \u003d 4 cm, OM \u003d 1 cm.
Answers:
Option 1.
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Problem 129. Line AM is perpendicular to the plane of the square ABCD, the diagonals of which meet at point O. Prove that:
a) straight line BD is perpendicular to the plane AMO;
Decision.
a) MA ABCD, therefore, MA BD by the definition of the perpendicularity of a straight line and a plane, BD AC by the property of the diagonals of a square (Fig. 2.7).
So, BD AO and BD AM, therefore, BD AMO on the basis of the perpendicularity of a straight line and a plane.
b) Since BD is MOA, then the line BD is perpendicular to any line lying in the plane MOA, in particular, BD MO.
Problem 134. Prove that all lines passing through a given point M of line a and perpendicular to this line lie in a plane passing through point M and perpendicular to line a.
Solution. Let us denote by a letter the plane passing through point M by line a and perpendicular to this line, and consider an arbitrary line b, also passing through point M and perpendicular to line a.
It is required to prove that b (Fig. 2.8). Let's say it isn't. Then the plane passing through lines a and b intersects the plane along some straight line b1 passing through point M and different from line b. Since a and b1, then a b1. We got that in the plane through the point M there are two straight lines (b and b1) perpendicular to the straight line a, which is impossible. Hence, the assumption is incorrect and the line b lies in the plane.
Fig. 2.7 Fig. 2.8
Problem 136. Prove that if a point X is equidistant from the ends of a given segment AB, then it lies in a plane passing through the midpoint of segment AB and perpendicular to line AB.
Solution. Let us denote by a letter the plane passing through the midpoint O of the segment AB and Fig. 2.9 perpendicular to the line AB (Fig. 2.9). Let the point X be equidistant from the ends of the segment AB, that is, XA \u003d XB. It is required to prove that X.
If point X lies on line AB, then it coincides with point O, and therefore X.
If the point X does not lie on the line AB, then the segment XO is the median of the isosceles triangle AXB and, therefore, the height of this triangle, i.e.
Thus, line XO passes through point O of line AB and is perpendicular to line AB. Hence it follows (see Problem 134) that the line XO lies in the plane, and therefore X.
Problem 137. Prove that through each of the two mutually perpendicular crossing lines there is a plane perpendicular to the other line.
Solution. Let a and b be mutually perpendicular intersecting lines. Let us prove that a plane passes through line a and is perpendicular to line b.
1) Through an arbitrary point O of the line a draw a line b1 parallel to the line b. Then a b1, since by the condition a b (Fig. 2.10).
2) Let us denote by a letter the plane passing through the intersecting straight lines a and b1, and draw a straight line c through the point O, perpendicular to the plane. Then c b1, and since b b1, then c b.
3) Let us denote by a letter the plane passing through the intersecting straight lines a and c. Since b a (by condition) and b c, Fig. 2.10 then b (based on the perpendicularity of a straight line and a plane). So, through line a there is a plane perpendicular to line b.
It is proved in a similar way that a plane perpendicular to the line a passes through line b.
§ 2. PERPENDICULAR AND INCLINED.
ANGLE BETWEEN STRAIGHT AND PLANE
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The main objectives of the lesson Introduce the concept of distance from a point to a plane, prove the theorem about three perpendiculars, show the application of this theorem in solving problems.
Approximate lesson plan
1. Using Figure 51 of the textbook, introduce the concept of a perpendicular to a plane, inclined, projection on an inclined plane. Considering the right-angled triangle AMH (see Fig. 51), prove that the perpendicular drawn from a given point to the plane is less than any oblique drawn from the same point to this plane. The length of a perpendicular drawn from a point to a plane is called the distance from that point to the plane.
2. Pay attention to remarks 1, 2, 3 in item 19 of the textbook, in which the concept of the distance between parallel planes parallel to a straight line and a plane crossing straight lines are introduced. It is useful to carry out the pictures and justify the validity of the statements given in the remarks.
Remark 1. If two planes are parallel, then all points of one plane are equally distant from the other plane.
Let, A, M. We carry out AA0 and MM0, then Fig. 2.11 AA0 MM0 (fig. 2.11), therefore AA0 \u003d MM0 (as segments of parallel lines, enclosed between parallel planes).
So, the distances from two arbitrary points A and M of the plane to the plane are equal to each other. The same applies to the distances from plane points to plane.
The distance from an arbitrary point of one of the parallel planes to another plane is called the distance between the parallel planes.
Remark 2. If the line and the plane are parallel, then all points of the line are equidistant from this plane.
The proof of the statement is given in the solution to problem 144, students can read it on their own.
You can offer another version of the proof.
Let a, A a, B a. Let's draw AA1 and BB1 (fig. 2.12). Then AA1 BB1. Let us prove that AA1 \u003d BB1.
The plane passing through parallel lines AA1 and BB1 intersects with the plane along straight line A1B1 and contains line AB. It is clear that AB A1B1 (if these lines intersect, then line AB (i.e., line a) would intersect with the plane, which contradicts condition a).
So AA1 BB1 and AB A1B1. Therefore, the quadrilateral ABB1A1 is a parallelogram, and therefore AA1 \u003d BB1.
Thus, the distances from two arbitrary points A and B of the straight line a to the plane parallel to it are equal to each other.
If a straight line and a plane are parallel, then the distance between the straight line and the plane is the distance from an arbitrary point of the straight line to this plane.
Remark 3. If two straight lines are intersecting, then the distance between them is the distance between one of them and a plane passing through another straight line parallel to the first straight line.
It is advisable to recall how a plane is constructed containing one of the intersecting straight lines and parallel to another straight line (Fig. 2.13).
Fig. 2.12 Fig. 2.13
Let a b. Draw a line a1 parallel to a through an arbitrary point M of line b. The intersecting straight lines a1 and b define a certain plane parallel to the straight line a.
From an arbitrary point A on a straight line a draw the perpendicular AA1 to the plane. The length of this perpendicular is the distance between crossing straight lines a and b.
Later, in the process of solving problems, it is possible to show how to construct a common perpendicular to two given intersecting lines a and b, i.e., a segment perpendicular to lines a and b, the ends of which lie on these lines.
3. Prove the theorem on three perpendiculars and its converse. In this case, you can use figure 53 of the textbook or slide 2.4.
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4. For classroom and homework, you can use tasks 138-145, 153.
Problem 143. The distance from point M to each of the vertices of a regular triangle ABC is 4 cm. Find the distance from point M to plane ABC if AB \u003d 6 cm.
Decision.
1) By the condition MA \u003d MB \u003d MC \u003d 4. Let MO ABC (Fig. 2.14), then OA \u003d OB \u003d OC (as projections of equal oblique, see problem 139). This means that point O is the center of a circle circumscribed about triangle ABC,
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and OA is the radius of this circle. It is known that a3 \u003d R 3, where a3 \u003d AB, R \u003d AO, therefore AO \u003d 6 \u003d 2 3.
2) From MAO we get MO \u003d MA2 - AO2, MO \u003d 16 - 12 \u003d 4 \u003d 2.
Answer: 2 cm.
Problem 145. Through the vertex A of a rectangular triangle ABC with a right angle C, a straight line AD is drawn, perpendicular to the plane of the triangle.
a) Prove that triangle CBD is right-angled.
b) Find BD if BC \u003d a, DC \u003d b.
Decision.
a) Segment AC - projection of inclined DC on the plane of triangle ABC (Fig. 2.15). BC AC by hypothesis, therefore, BC DC by the three perpendicular theorem and therefore triangle CBD is rectangular.
b) BC \u003d a, DC \u003d b. From BCD we get BD \u003d BC2 + CD2, BD \u003d a2 + b2.
Answer: a2 + b2.
In the future, in the process of solving problems, it is important to draw students' attention to the generalized theorem on three perpendiculars, when the straight line a1 is perpendicular to the inclined projection, but does not pass through the base of the inclined one.
Lesson № Lesson topic: The angle between a straight line and a plane The main objectives of the lesson Introduce the concept of the angle between a straight line and a plane;
consider the tasks that use this concept.
Approximate lesson plan
1. Check selectively solving problems from homework. Solutions to problems like 138-142 and the proof of the three perpendicular theorem can be discussed orally, using ready-made drawings and slides.
2. Introduce the concept of projection of a point onto a plane, projection of a figure onto a plane. Prove that the projection of a straight line onto a plane not perpendicular to this plane is a straight line. In this case, figures 54, 55 of the textbook are used.
3. Enter the definition of the angle between a straight line and a plane.
4. Analyze the solution to problem 162 given in the textbook. Prove that the angle between a given straight line and a plane is the smallest of all the angles that this straight line forms with straight lines drawn in the plane through the point of intersection of the straight line with the plane.
It is helpful for students to take a short note of the evidence in Slide 2.5.
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5. For classroom and homework, you can use tasks 163-165, 146-148.
Problem 165. From point A, remote from the plane at a distance d, inclined AB and AC are drawn to this plane at an angle of 30 ° to the plane. Their projections onto the plane form an angle of 120 °. Find BC.
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Lessons 33-36 Lesson topic: Theory revision. Solving problems on the application of the theorem on three perpendiculars, at the angle between a straight line and a plane. The main tasks of the lessons. Repeat the proof of the theorem on three perpendiculars, the concept of the angle between a straight line and a plane, to strengthen the skills of solving problems.
Approximate lesson plan
1. At each lesson number 33-35, repeat the theory questions by interviewing students.
2. In the process of solving problems, repeat the relations between the elements of a right-angled triangle, the theorem of sines and cosines.
3. Pay special attention to the solution of some typical problems that will be used in the future in calculating the areas of surfaces and volumes of polyhedra. Such tasks include, for example, tasks 147, 151, 158, 161. It is useful to use slide 2.6 below, which is intended for frontal work with students, discussion of approaches to solving problems from the textbook.
4. At the lesson number 36 it is advisable to carry out independent work of a controlling nature.
Independent work No. 2.2
Option 1 From point M draw a perpendicular MB equal to 4 cm to the plane of the rectangle ABCD. The oblique MA and MC form angles of 45 ° and 30 ° with the plane of the rectangle, respectively.
a) 0 Prove that triangles MAD and MCD are rectangular.
b) 0 Find the sides of the rectangle.
c) Prove that triangle BDC is the projection of triangle MDC onto the plane of a rectangle, and find its area.
Option 2 From point M, draw a perpendicular MD equal to 6 cm to the plane of the square ABCD. The oblique MB makes an angle of 60 ° with the plane of the square.
a) 0 Prove that triangles MAB and MCB are right-angled.
b) 0 Find the side of the square.
c) Prove that the triangle ABD is the projection of the triangle MAB onto the plane of the square and find its area.
Answers:
b) AB \u003d 4 cm, BC \u003d 4 3 cm; c) 8 3 cm2.
Option 1.
b) 6 cm; c) 3 cm2.
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Problem 147. From point M draw n perpendicular MB to the plane of rectangle ABCD. Prove that AMD and MCD triangles are rectangular.
Decision.
1) By the condition of the problem, the segment MB is the perpendicular to the plane of the rectangle, therefore, the segment AB is the projection of the inclined MA onto the plane of the rectangle (Fig. 2.17). AD AB (since ABCD is a rectangle), therefore, AD MA by the three perpendicular theorem. Thus, the angle MAD is right and, therefore, the triangle AMD is right-angled.
2) Similarly, since DC BC, DC MC and triangle MCD are rectangular.
Problem 151. Line CD is perpendicular to the plane of triangle ABC. Prove that: a) triangle ABC is the projection of triangle ABD onto plane ABC;
b) if CH is the height of triangle ABC, then DH is the height of triangle ABD.
Decision.
a) By the condition of the problem, the segment DC is the perpendicular to the plane ABC, therefore, point C is the projection of point D onto the plane ABC, the segment CB is the projection onto the inclined DB, and the segment CA is the projection of the inclined DA onto the plane ABC (Fig.2.18).
All points of the segment AB lie in the plane ABC, therefore the projection of the segment AB onto the plane ABC is this segment itself.
Thus, the projections of the sides of the triangle ABD onto the plane ABC are the corresponding sides of the triangle ABC.
It is also obvious that the projection M1 of any interior point M of triangle ABD lies inside triangle ABC and vice versa: any interior point M1 of triangle ABC is the projection onto plane ABC of some interior point M of triangle ABD. This means that the triangle ABC is the projection of the triangle ABD onto the plane ABC.
b) AB CH by hypothesis, therefore, AB DH by the three-perpendicular theorem, that is, DH is the height of triangle ABD.
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Problem 158. Through the vertex B of the rhombus ABCD, a line BM is drawn, perpendicular to its plane. Find the distance from point M to the straight lines containing the sides of the rhombus, if AB \u003d 25 cm, BAD \u003d 60 °, BM \u003d 12.5 cm.
Decision.
1) Let's carry out BK AD (fig.2.19). The segment BK is the projection of the inclined MK onto the rhombus plane, AD BK, therefore, AD MK according to the theorem of three perpendiculars. The length of the segment MK is equal to the distance from point M to line AD.
Similarly ME — distance from point M to line DC.
ABK we get BK \u003d AB sin 60 °, BK \u003d 25 3.
3) Triangle MBK is rectangular as MB ABC. We have
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4) BK \u003d BE (as rhombus heights). Rectangular triangles MBK and MBE are equal in two legs, therefore ME \u003d MK \u003d 25 cm.
5) Distances from point M to straight lines AB and BC are equal to the length of the perpendicular MB, i.e. equal to 12.5 cm.
Answer: 25 cm, 25 cm, 12.5 cm, 12.5 cm.
Problem 161. Beam BA does not lie in the plane of the undeveloped angle CBD. Prove that if ABC \u003d ABD, and ABC is 90 °, then the projection of the ray BA onto the plane CBD is the bisector of the angle CBD.
Decision.
1) Let AE CBD. Draw perpendicular AM to line BC in plane ABC, and perpendicular AK to line BD in plane ABD. Since ABC is 90 °, point M lies on the BC ray (and not on the extension of this ray). Similarly, since ABD is 90 °, point K lies on the BD ray (Fig. 2.20).
Since BC AM, then BC EM (by the theorem converse to the three-perpendicular theorem). It can be proved similarly that BD EK.
2) Right-angled triangles ABK and ABM are equal in hypotenuse (AB - common hypotenuse) and acute angle (ABC \u003d ABD), therefore, BM \u003d BK.
3) Right-angled triangles BME and BKE are equal in hypotenuse (BE - common hypotenuse) and leg (BM \u003d BK), therefore, EM \u003d EK.
4) Point E is equidistant from the sides of angle CBD, therefore, it lies on the bisector of this angle, i.e., ray BE is the bisector of angle CBD.
§ 3. DOUBLE ANGLE.
PERPENDICULARITY OF THE PLANES
Lesson number 37 Lesson topic: Dihedral angle The main objectives of the lesson Introduce the concept of a dihedral angle and its linear angle, consider the tasks for the application of these concepts.Approximate lesson plan
1. Introduce the concept of a dihedral angle using figure 58 of the textbook.
2. Introduce the concept of a linear angle of a dihedral angle.
Prove that all linear angles of a dihedral angle are equal to each other (see Fig. 59, a, b).
3. Give the definition of the degree measure of the dihedral angle.
Consider examples of acute, right and obtuse dihedral angles using figure 60 of the textbook. A right dihedral can be shown at the intersection of two walls in a classroom, as well as a wall and ceiling or floor.
4. For classroom and homework, you can selectively use tasks 166-170.
Students should pay attention to the designation of dihedral angles. A dihedral angle with an edge AB, on different faces of which points C and D are marked, is called a dihedral angle CABD.
Problem 167. In the tetrahedron DABC, all edges are equal, point M is the midpoint of the edge AC. Prove that DMB is the linear angle of the BACD.
Fig. 2.21 Fig. 2.22
Solution. The medians BM and DM are simultaneously the heights of regular triangles ABC and ADC (Fig. 2.21). Therefore, BM AC and DM AC, and, therefore, DMB is the linear angle of the dihedral angle at the edge AC of the pyramid base.
Problem 170. From vertex B of triangle ABC, side AC of which lies in the plane, draw perpendicular BB1 to this plane. Find the distance from point B to line AC and to the plane if AB \u003d 2 cm, BAC \u003d 150 ° and the dihedral angle BACB1 is 45 °.
Decision.
1) The triangle BAC is obtuse with an obtuse angle A, therefore the base of the height BK drawn from the vertex B lies on the extension of the side AC. The distances from point B to line AC and to the plane are equal to BK and BB1, respectively (Fig. 2.22).
2) Since AC BK, then AC KB1 by a theorem converse to the three perpendicular theorem. Therefore, BKB1 is the linear angle of the dihedral angle BACB1. By the problem statement, BKB1 \u003d 45 °.
3) From BAK we have A \u003d 30 °, BK \u003d BA sin 30 °, BK \u003d 1.
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Lesson number 38 Lesson topic: Sign of perpendicularity of two planes The main objectives of the lesson Introduce the concept of the angle between the planes; give a definition of perpendicular planes; prove a theorem expressing the sign of perpendicularity of two planes; show the application of this theorem in solving problems.
Approximate lesson plan
1. Check selectively solving problems from homework. It is advisable to use slides with ready-made drawings.
2. Draw the students' attention to the fact that when two planes intersect, four dihedral angles are formed. If is the value of that of the four angles that does not exceed each of the others, then they say that the angle between the intersecting planes is equal. It is clear that 0 ° 90 °. If \u003d 90 °, then the planes are called perpendicular. In this case, each of the four dihedral angles formed by intersecting planes is straight.
3. Prove a theorem expressing the criterion for perpendicularity of two planes. The proof of the theorem can be carried out orally through the text of the textbook using Figure 62. The traditional proof given in the textbook is usually successfully assimilated by students.
4. It is important to draw the attention of students to the following two facts that are often used in solving problems:
a) A plane perpendicular to the edge of a dihedral angle is perpendicular to its faces. (This statement in a slightly different formulation is given in section 23 of the textbook as a corollary to the theorem.)
b) A perpendicular drawn from any point of one of two mutually perpendicular planes to the line of their intersection is a perpendicular to the other plane.
(This statement is proved in the solution to problem 178 given in the textbook.)
5. For classroom and homework, you can use tasks 171-180.
Problem 171. The hypotenuse of a right-angled isosceles triangle lies in a plane, and the leg is inclined to this plane at an angle of 30 °. Find the angle between the plane and the plane of the triangle.
Decision.
1) Let ABC be a given triangle, AB, CO. Then the segment OB is the projection of the leg CB onto the plane. By the condition of the problem, CBO \u003d 30 ° (Fig. 2.23).
2) Let COB CO \u003d a in the triangle, then CB \u003d 2a.
3) Draw CD AB, then AB DO by the inverse theorem on three perpendiculars, and CDO is the linear angle of the dihedral angle formed at the intersection of the plane with the plane of the triangle. Let be
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CDO \u003d x. This is the desired angle between the plane and the plane of the triangle.
4) From CDB we get CBD \u003d 45 °, since by the condition the triangle ACB is isosceles and rectangle
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whence \u003d 45 °, i.e. the dihedral angle DABC is 45 °.
5) Since BC is DC and AC DC, ACB is the linear angle of the dihedral angle of BDCA.
Since ACB \u003d 60 °, the dihedral angle BDCA is 60 °.
Answer: 90 °, 45 °, 60 °.
Problem 174. Find the dihedral angle ABCD of the tetrahedron ABCD if the angles DAB, DAC and ACB are straight, AC \u003d CB \u003d 5, DB \u003d 5 5.
Decision.
1) By the condition of the problem, the angles DAB and DAC are straight, therefore, DA AB and DA AC (Fig. 2.25). Hence it follows that the segment DA is the perpendicular to the plane ABC, and, therefore, the segment AC is the projection of the inclined DC onto the plane ABC. Fig. 2.25
2) By the hypothesis of the problem, the angle ACB is a straight line, that is, BC AC, and hence BC DC by the theorem on three perpendiculars. Thus, ACD is the linear angle of the dihedral angle ABCD.
3) From DCB: DC \u003d DB2 - BC2, DC \u003d 25 5 - 25 \u003d 10.
4) From DAC we get ACD \u003d x, cos x \u003d AC, cos x \u003d 5,
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The main objectives of the lesson Introduce the concept of a rectangular parallelepiped, consider the properties of its faces, dihedral angles, diagonal.
Approximate lesson plan
1. Formulate the definition of a rectangular parallelepiped. Prove that all six faces of a rectangular parallelepiped are rectangles.
2. Prove that all dihedral angles of a rectangular parallelepiped are right.
3. Prove the theorem: the square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions.
Note the analogy with the property of the diagonal rectangle. It can also be noted that this theorem is one of the versions of the spatial Pythagorean theorem.
Consider the corollary of the theorem: the diagonals of a rectangular parallelepiped are equal.
4. For classroom and homework, you can selectively use tasks 187-192.
Fig. 2.26 Fig. 2.27
Problem 191. Given a cube ABCDA1B1C1D1. Prove that planes ABC1 and A1B1D are perpendicular.
Decision.
1) BC1 B1C by the property of the square diagonals (Fig. 2.26). DC BCC1, therefore DC BC1, as BC1 BCC1.
Thus, line BC1 is perpendicular to two intersecting lines DC and CB1 lying in the plane A1B1D. Consequently, line BC1 is perpendicular to plane A1B1D in terms of the line and plane perpendicularity.
2) Plane ABC1 passes through line BC1, perpendicular to plane A1B1D, therefore, ABC1 A1B1D by the sign of perpendicularity of two planes.
Problem 192. Find the tangent of the angle between the diagonal of the cube and the plane of one of its faces.
Decision.
1) Let the edge of the cube ABCDA1B1C1D1 be equal to a. Then BD \u003d a 2 (Fig. 2.27). Since D1D is ABC, straight line BD is the projection of straight line BD1 onto the plane of face ABCD, and therefore the angle between these straight lines is the angle between diagonal BD1 and face ABCD. Thus, it is required to find the tangent of the angle D1BD, the value of which we denote.
2) From D1DB we obtain tg \u003d 1, tg \u003d a, tg \u003d 2.
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Lesson № 40 Lesson topic: Solving problems on a rectangular parallelepiped Main tasks of the lesson Repeat the properties of a rectangular parallelepiped, solve a number of problems on a rectangular parallelepiped.
Approximate lesson plan
1. Review the theory questions by interviewing students.
2. Check selectively the solution of tasks from homework, using ready-made drawings, slides.
3. For classroom and homework, you can use problems 193-196.
Problem 195. Find the dimensions of the rectangular parallelepiped ABCDA1B1C1D1 if AC1 \u003d 12 cm and the diagonal BD1 makes an angle of 30 ° with the face plane AA1D1D, and an angle of 45 ° with the edge DD1.
Decision.
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«TRAINING) Surname First name Patronymic Course_ Faculty of Communications and Law Group No. _ Review results (the column is filled in by the teacher) _ _ _ _ _ _ Teacher _ Minsk 2014 CONTENTS SHORT METHODOLOGICAL INSTRUCTIONS SECTION 1. GENERAL TEACHING ABOUT CRIMINALISM SUBJECT 1.1 SUBJECT, SUBJECT, AND AND OBJECTIVES. HISTORY AND METHODOLOGY ... "
« Contents Annotation ... 1. The goals of students' independent work. 2. Tasks of independent work of students ... 5 3. Recommendations for independent study disciplines..5 4. Types of independent work of students..5 5. Requirements for the minimum content of the discipline in accordance with the Federal State Educational Standard ... 6. Content of independent work on the topics of the discipline. 7. Tasks for independent work of students 7.1. Themes of essays and creative work in the discipline ... 8 ... "
« Plan of outreach seminars and webinars First half of 2015-2016 academic year October Participation is free. All participants (registration required) issued on October 16, 2015 16.00-17.00 (Moscow time) certificates of participation in the Webinar “Methodological principles for developing tasks for seminars and webinars. International competition "PONY® visiting Pythagoras" for pupils of grades 2-4 and criteria for their assessment. " The webinar analyzes the goals of conducting intellectual competitions, ... "
« RUSSIAN FEDERATION MINISTRY OF EDUCATION AND SCIENCE State educational institution of higher professional education TYUMEN STATE UNIVERSITY "APPROVED": Vice-Rector for Academic Affairs L.M. Volosnikov 08.07. 2011 Organization of speech therapy work in preschool educational institutions Educational-methodical complex. Work program for students of the direction of training 050700.62 Special (defectological) education, training profile Speech therapy, form ... "
« State budgetary professional educational institution of the city of Moscow "First Moscow Educational Complex" Methodological recommendations on practical work On the professional module PM 02. Designing of garments MDK 02.02. Methods of constructive modeling of garments, 3rd year of training 262019 Design, modeling and technology of garments advanced training (name of the training profile) Moscow BBK G1 APPROVED Developed on ... "
« MINISTRY OF EDUCATION AND SCIENCE OF THE RUSSIAN FEDERATION ITMO UNIVERSITY E.P. Suchkova, M.S. Belozerova METHODS OF STUDYING MILK AND DAIRY PRODUCTS Teaching aid St. Petersburg UDC 637.1 / 3 Suchkova E.P., Belozerova M.S. Research methods of milk and dairy products: Textbook-method. allowance. - SPb .: ITMO University; IHiBT, 2015 .-- 47 p. The laboratory works on the discipline "Methods of research of milk and dairy products" are presented. The works are devoted to the study of modern methods ... "
« Contents 1. General ... 2. Characteristics of the direction of training ... 3. Characteristics of the professional activity of graduates 3.1. Region professional activity of a graduate of EP VO.3.2 Objects of professional activity of a graduate of EP VO.3.3 Types of professional activity of a graduate of EP VO.3.4 Generalized labor functions of graduates in accordance with professional standards..8 4. Results of mastering the educational program .. 5. Structure of the educational program .. . "
« FEDERAL SERVICE FOR SUPERVISION IN THE FIELD OF PROTECTION OF CONSUMER RIGHTS AND HUMAN WELL-BEING FBUN "Federal Scientific Center for Medical Preventive Technologies public health risk management "FGBOU HPE" Perm State National Research University "CURRENT DIRECTIONS OF DEVELOPMENT OF SOCIAL AND HYGIENIC MONITORING AND HEALTH RISK ANALYSIS Materials of the All-Russian Scientific and Practical Conference with International Participation (15-17 May 2013) Edited by Academician ... "
« WORKING PROGRAM ON THE SUBJECT "TECHNOLOGY" FOR GRADE 1 "F" Compiled by: primary school teacher Natalia Sergeevna Tambovtseva Moscow, 2014-2015 academic year Explanatory note. The work program on technology is based on the requirements of the Federal State Standard for Primary General Education in the educational field "Technology" and is developed in accordance with the Model Program for Primary General Education, the work program of N.I. Rogovtseva, S.V. Anashenkova ... "
« M. S. Soloveichik N. S. Kuzmenko RUSSIAN LANGUAGE METHODOLOGICAL RECOMMENDATIONS for a textbook for the 2nd grade of general education organizations A manual for teachers Edition 7th, revised Smolensk Association XXI century UDC 372.881.116.11.046. BBK 74.268.1Rus S PAY ATTENTION! Use caution when using tutorials from other publishers! If any of the authors of this textbook are not listed as an editor, consultant or reviewer, the manual may not ... "
« SPECIALIZED STRUCTURAL EDUCATIONAL DIVISION OF THE EMBASSY OF THE RUSSIAN FEDERATION IN THE REPUBLIC OF MADAGASCAR IS THE BASIC GENERAL EDUCATIONAL SCHOOL AT THE EMBASSY OF RUSSIA IN MADAGASCAR WORKING PROGRAM of the training course (literature) 5th CLASS 2014-2015 academic year teacher: Egorova I.V. Explanatory note The work program was drawn up in accordance with the normative documents and methodological materials: The federal component of the state educational standard of the main general ... "
« Considered at the meeting of the Ministry of Defense Protocol No. from 08.24.2015. "Checked" "Approved" _ Deputy Director for Internal Affairs, Director of MBOU "Lyceum" MOK No. 2 "Samofalova Yu.V._ Sverdlov V.Ya. Work program for extracurricular activities Course "School of speech development" 2015-2016 academic year Teacher Asoyan O.I., Bavykina I.E., Ledeneva G.A., Ivashkina N.V., Savvina O.Yu., Sverdlova L. IN. Class 4 "A", "B", "C", "D", "D", "E" Subject "Course" SPEECH. Young smart men and smart girls. School of speech development "(34 hours; 1 hour per week) ..."
« Ministry of Education and Science of the Russian Federation Amur State University E.V. Pshenichnikova BASES FOR DESIGNING CLOTHES FOR INDIVIDUAL CONSUMER Textbook Recommended by the Far Eastern Regional Educational and Methodological Center (DV RUMTs) as a textbook for students studying in the direction of training bachelors 262000.62 "Technology of light industry products", 100100.62 "Service" of universities in the region Blagoveshchensk Publishing house of AmSU BBK 37. 24-2 I 73 P 93 ... "
« PROTECTING CHILDREN AGAINST DISCRIMINATION Interdisciplinary Study Guide CREAN PROTECTING CHILDREN AGAINST DISCRIMINATION PROTECTING CHILDREN FROM DISCRIMINATION Interdisciplinary Study Guide manual Edited by Dagmar Kutsar and Hannah Warming Editor of the Russian translation Vera Ivanovna Zabotkina Dr. of Philology. Sci., prof., Vice-Rector for Innovative International Projects Russian State University for the Humanities European consortium of universities offering master's programs in children's rights within the framework of ... "
« Contents Section 1. The list of planned learning outcomes in the discipline, correlated with the planned results of mastering the educational program .. 4 1.1 List of planned learning outcomes for the discipline. 4 1.2 Planned results of mastering the educational program. 4 Section 2. Place of discipline in the structure of the educational program. 6 Section 3. Scope of the discipline .. 6 Section 4. Structure and content of the discipline. 7 Section 5. The list of educational and methodological support for ... "
« CONTENTS Requirements for the results of mastering the discipline 1. 4 Place of the discipline in the structure of OBOP 2. 5 Structure and content of the discipline 3. 6 Structure of the discipline 3.1. 6 Content of the course 3.2. 7 The list of educational and methodological support for independent work 4. 9 students in the discipline Educational technologies 5. 9 Forms of control of the development of the discipline 6. 9 The list of assessment tools for the current control of the development of the discipline 6.1. 9 Composition of the fund of evaluation funds for conducting ... "
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