Task 8 for the Mathematics exam
3.1. The diagonal of the smaller side face of a rectangular parallelepiped is equal to the larger edge of the base. The height of the parallelepiped is 2 cm, the diagonal of the base is 14 cm. Find the volume of the parallelepiped.
3.2. The base of a straight prism - right triangle with a hypotenuse of 10 cm and a leg of 6 cm. The larger leg of the triangle at the base of the prism equal to the diagonal the smaller of the side faces. Find the height of the prism.
3.3. The base of a straight prism is a rhombus with a side of 12 cm and an angle of 60°. The smaller of the diagonal sections of the prism is a square. Find the volume of the prism.
3.4. At the base of a straight prism lies isosceles trapezoid with an acute angle of 60°; the lateral side and the smaller of the parallel sides of the trapezoid are 4 cm; The diagonal of the prism makes an angle of 30° with the plane of the base. Calculate the volume of the prism.
3.5. The diagonal of a rectangular parallelepiped makes an angle of 45° with the plane of the base, and the diagonal of the side face makes an angle of 60°. The height of a rectangular parallelepiped is 8 cm. Find its volume.
3.6. At the base of a straight prism is a rhombus; the diagonals of the prism make angles of 30° and 60° with the plane of the base; The height of the prism is 6 cm. Find the volume of the prism.
3.7. At the base of a straight prism lies a rhombus with a side of 10 cm. The base side is removed from the two parallel sides of the opposite side face by 5 cm and 13 cm, respectively. Find the volume of the prism.
3.8. The edge of the lower base of a regular quadrangular prism is 10 cm away from the plane of the upper base. The distances between the opposite side edges are 8 cm. Find the volume of the prism.
3.9. At the base of a straight prism lies a trapezoid. The areas of the parallel lateral faces of the prism are 8 cm and 12 cm, and the distance between them is 5 cm. Find the volume of the prism.
3.10. At the base of a straight prism lies a trapezoid. The volume of the prism is 40 cm. The areas of the parallel side faces are 6 cm and 14 cm. Find the distance between them.
3.11. Diagonal of the base of a rectangular parallelepiped
is equal to 10 cm, and the diagonals of the side faces are 2 * / W cm and 2 l / 17 cm. Find the volume of the parallelepiped.
3.12. At the base of a straight prism is a rhombus. The area of the base of the prism is 48 cm, and the area of its diagonal
sections are 30 cm and 40 cm. Find the volume of the prism.
3.13. In a regular quadrangular pyramid, the height is 3 cm, the lateral surface area is 80 cm. Find the volume of the pyramid.
3.14. In a regular quadrangular pyramid, the side of the base is 6 cm, the area of the lateral surface is twice the area of the base. Find the volume of the pyramid.
3.15. The area of the lateral surface of the cone is 60tcm; the distance from the center of the base to the generatrix is 4.8 cm. Find the volume of the cone.
3.16. The base of the inclined prism is a square with a side of 6 cm; one of the diagonal sections of the prism is perpendicular to the plane of the base and is a rhombus with an angle of 60°. Find the volume of the prism.
3.17. At the base inclined parallelepiped- a square with a side of 3 cm. Two opposite side faces perpendicular to the base, the other two form angles of 30° with the plane of the base. The total surface of the parallelepiped is 72 cm. Find the volume of the parallelepiped.
3.18. At the base of the inclined parallelepiped there is a rhombus with a side of 4 cm and an acute angle of 45°; the side edge makes an angle of 60° with the plane of the base; the diagonal of one side face is perpendicular to the plane of the base. Find the volume of the parallelepiped.
3.19. All 9 edges of the inclined prism are equal to 4 cm. The volume of the prism is 24 cm. Find the angle of inclination of the side edge of the prism to the plane of the base.
3.20. In an inclined triangular prism, the distances between the side edges are 5 cm, 12 cm and 13 cm. The area of the smaller side edge is 22 cm. Find the volume of the prism.
3.21. At the base of an inclined prism lies a right triangle with legs of 4 cm and 6 cm. The lateral edge of the prism makes an angle of 60° with the plane of the base. The volume of the prism is 60 cm. Find the length of the side edge of the prism.
3.22. The two lateral faces of an inclined triangular prism form an angle of 60°; the distance from their common edge to the other two ribs is 5 cm; the lateral edge of the prism is 8 cm. Find the lateral surface of the prism.
3.23. The two lateral faces of an inclined triangular prism are perpendicular. The sum of their areas is 70 cm. The length of the side edge is 5 cm. The volume of the prism is 120 cm. Find the distances between the side edges of the prism.
3.24. In a regular triangular pyramid, the height is equal to the side of the base. Find the angle between the side edge and the plane of the base.
3.25. In a regular quadrangular pyramid, the lateral edge forms an angle of 45° with the plane of the base. The side of the base of the pyramid is 6 cm. Find the volume of the pyramid.
3.26. In a regular quadrangular pyramid, the lateral edge forms an angle of 60° with the plane of the base. The height of the pyramid is 3 cm. Find the surface area of the pyramid.
3.27. In a regular quadrangular pyramid, the apothem forms an angle of 60° with the plane of the base. The height of the pyramid is 6 cm. Find the surface area of the pyramid.
3.28. In a regular quadrangular pyramid, the apothem forms an angle of 30° with the plane of the base. The side of the base of the pyramid is 12 cm. Find the surface area of the pyramid.
3.29. The height of a regular quadrangular pyramid is 6 cm and forms an angle of 30° with its side face. Find the volume of the pyramid.
3.30. The height of a regular quadrangular pyramid is 10 cm and forms an angle of 45° with the side edge. Find the volume of the pyramid.
3.31. The height of a regular triangular pyramid is 8 cm, and the side edge is 10 cm. Find the area of the lateral surface of the pyramid.
3.32. The height of a regular quadrangular pyramid is 20 cm, and the side edge is 16 cm. Find the area of the lateral surface of the pyramid.
3.33. The height of a regular hexagonal pyramid is 12 cm, and the side edge is 13 cm. Find the area of the lateral surface of the pyramid.
3.34. In a regular quadrangular pyramid, the side of the base is 8 cm; the dihedral angle at the base of the pyramid is 60°. Find the volume of the pyramid.
3.35. In a regular quadrangular pyramid, the height is 8 cm; the dihedral angle at the base of the pyramid is 30°. Find the volume of the pyramid.
3.36. In a regular quadrangular pyramid, the apothem is 16 cm; the dihedral angle at the base of the pyramid is 45°. Find the volume of the pyramid.
3.37. The side of the base of a regular quadrangular pyramid is 5 cm; the diagonal section is equal to the base. Find the lateral surface area of the pyramid.
3.38. The height of a regular quadrangular pyramid is 10 cm; the diagonal section is equal to the base. Find the side surface of the pyramid.
3.39. The radius of the cylinder is 8 cm, and its height is 12 cm. A straight line is drawn through the middle of the cylinder axis, intersecting the plane of the lower base of the cylinder at a distance of 24 cm from the center of the lower base. In what ways does this line divide the generatrices of the cylinder that intersect it?
3.40. The radius of the cylinder is 6 cm, and its height is 10 cm. A straight line is drawn through the middle of the generatrix of the cylinder, intersecting the axis of the cylinder. This line intersects the lower base of the cylinder at a distance of 3 cm from the center of the lower base. In what ratio does this straight line divide the axis of the cylinder?
3.41. The radius of the cylinder is 8 cm. A straight line is drawn through the middle of the cylinder axis, intersecting the plane containing the lower base of the cylinder, at a distance of 12 cm from the center of the lower base. This straight line intersects the generatrix of the cylinder at a distance of 2 cm from the plane of the lower base. Find the height of the cylinder.
3.42. The height of the cylinder is 12 cm. A straight line is drawn through the middle of the generatrix of the cylinder, intersecting the axis of the cylinder at a distance of 4 cm from the lower base. This line intersects the plane containing the lower base of the cylinder at a distance of 18 cm from the center of the lower base. Find the radius of the base of the cylinder.
3.43. The height of the cone is 20 cm, the distance from the center of the base to the generatrix is 12 cm. Find the volume of the cone.
3.44. The radius of the base of the cone is 20 cm; the distance from the center of the base to the generatrix is 12 cm. Find the area of the lateral surface of the cone.
3.45. At the base of the pyramid lies a right triangle, the hypotenuse of which is 15 cm, and one of the legs is 9 cm. Find the area of the section drawn through the middle of the height of the pyramid parallel to its base.
3.46. At a distance of 4 cm from the top of the pyramid, a section parallel to the base was drawn. The cross-sectional area is 10 cm and is equal to the area of the base of the pyramid.
Find the volume of the pyramid.
3.47. The radius of the base of the cone is 6 cm, and the height is 12 cm. A section is drawn in the cone parallel to the base. The radius of the section is 4 cm. In what ratio does the section divide the height of the cone?
3.48. The height of the cone is 12 cm, and the radius of the base is 3 cm. At what distance from the top of the cone should a section parallel to the base be drawn so that its area is equal to cm?
3.49. In a right parallelepiped, a section is drawn through the diagonal of the lower base and the middle of the side edge not in contact with this diagonal. Distance from the section plane to the top of the lower base,
not lying in the section plane is equal to 5 cm. Area
Section 2 is equal to 10 cm. Find the volume of the parallelepiped.
3.50. In a regular quadrangular prism, a section is drawn through the diagonal of the lower base and the end of the non-parallel diagonal of the upper base. The base area of the prism and the cross-sectional area are 20 cm. Find the volume of the prism.
3.51. In a regular triangular prism, a section is drawn through the side of the lower base and the middle of the opposite side edge. The section plane is inclined to the base plane at an angle of 45°; The cross-sectional area is 4 l/6 cm. Find the volume of the prism.
3.52. The height of a regular triangular prism is 12 cm. A section is drawn in the prism through the side of the lower base and the opposite vertex of the upper base. The section plane is inclined to the plane of the base of the prism at an angle of 60°. Find the volume of the prism.
3.53. In a right parallelepiped, a section is drawn through the diagonal of the lower base and the middle of a side edge that does not intersect with this diagonal. The volume of the smaller of the two polyhedra into which the parallelepiped is divided section plane, is equal to 40 cm. Find the volume of the parallelepiped.
3.54. In a triangular prism, a section is drawn through the side of the lower base and the opposite vertex of the upper base. In what ratio does the section plane divide the volume of the prism?
3.55. In a triangular pyramid, a section is drawn through midline the bottom base and the top of the pyramid. In what ratio does the section plane divide the volume of the pyramid?
3.56. In a regular quadrangular pyramid, a section is drawn through the midpoints of two adjacent sides of the base, perpendicular to the base. In what ratio does the section plane divide the volume of the pyramid?
3.57. IN rectangular parallelepiped a section is drawn through the edge of the lower base and the intersection point of the diagonals of the opposite side face. In what ratio does the section plane divide the volume of the parallelepiped?
3.58. The pyramid has a section parallel to the base. The section plane divides the pyramid into parts, the volumes of which are in the ratio 1:26, counting from the top. In what ratio does the cutting plane divide the height of the pyramid?
3.59. The pyramid has a section parallel to the base. The section plane divides the height of the pyramid into parts, the ratio of which is 2:1, counting from the top. In what ratio does the sectional plane divide the volume of the pioamyl?
3.60. The area of the base of the pyramid is 1 m. A plane parallel to the base of the pyramid divides it into two equal parts. Find the cross-sectional area of the pyramid.
3.61. The development of the lateral surface of a regular triangular prism is a rectangle with sides 15 cm and 12 cm. Determine the volume of this prism. Find both solutions.
3.62. The development of the lateral surface of a regular triangular prism is a rectangle with sides 18 cm and 9 cm. Determine the total surface area of this prism. Find both solutions.
3.63. A rectangle with sides 12 cm and 16 cm can be folded in two ways into the lateral surface of a regular quadrangular prism. Compare the volumes of these prisms.
3.64. A rectangle with sides 24 cm and 10 cm can be folded in two ways in the form of the lateral surface of a regular quadrangular prism. Compare the total surface areas of these Prisms.
3.65. A rectangle with sides 12 cm and 8 cm is folded for the first time in the form of the lateral surface of a regular quadrangular prism with a height of 8 cm, and for the second time - a regular triangular prism with the same height. Compare the volumes of these prisms.
3.66. A rectangle with sides 24 cm and 10 cm is folded for the first time in the form of the lateral surface of a regular quadrangular prism with a height of 10 cm, and for the second time - a regular triangular prism with the same height. Compare the total surface areas of these prisms.
3.67. A square with a side of 12 cm is folded for the first time in the form of the lateral surface of a regular triangular prism, and for the second time - a regular quadrangular prism. Compare the total surface areas of these prisms.
3.68. A square with a side of 24 cm is folded for the first time in the form of the lateral surface of a regular triangular prism, and for the second time - a regular quadrangular prism. Compare the volumes of these prisms.
3.69. A rhombus with a side of 10 cm and an acute angle of 60° rotates around the side. Find the volume of the body of revolution.
3.70. A rhombus with a side of 8 cm and an acute angle of 60° rotates around the side. Find the surface area of the body of rotation.
3.71. A rectangular trapezoid with bases 5 cm and 8 cm and a height of 4 cm rotates around a larger base. Find the volume of the body of rotation.
3.72. A rectangular trapezoid with bases 6 cm and 10 cm and a height of 3 cm rotates around a larger base. Find the surface area of the body of rotation.
3.73. A rectangular trapezoid with bases 10cm and 14cm and a height of 3cm rotates around a smaller base. Find the volume of the body of rotation.
3.74. A rectangular trapezoid with bases 12 cm and 15 CMJ^ and a height of 4 cm rotates around a smaller base. Find the surface area of the body of revolution.
3.75. A rectangular trapezoid with bases 10 cm and 15 cm and a height of 12 cm rotates for the first time around the smaller base, and for the second time around the larger one. Compare the volumes of bodies of revolution.
3.76. A rectangular trapezoid with bases 12 cm and 20 cm and a height of 15 cm rotates the first time around the smaller base, and the second time around the larger one. Compare the surface areas of bodies of revolution.
3.77. An equilateral trapezoid with bases 10 cm and 16 cm and a height of 4 cm rotates around a smaller base. Find the volume of the body of revolution.
3.78. An equilateral trapezoid with bases 10 cm and 18 cm and a height of 3 cm rotates around a smaller base. Find the surface area of the body of revolution.
3.79. An equilateral trapezoid with bases 12 cm and 18 cm and a height of 4 cm rotates around a larger base. Find the volume of the body of revolution.
3.80. An equilateral trapezoid with bases of 15 cm and 25 cm and a height of 12 cm rotates around a larger base. Find the surface area of the body of revolution.
3.81. An equilateral trapezoid with bases 12 cm and 24 cm and a height of 8 cm rotates for the first time around the smaller base, and the second time around the larger one. Compare the volumes of bodies of revolution.
3.82. An equilateral trapezoid with bases 12 cm and 28 cm and a height of 6 cm rotates for the first time around the smaller base, and the second time around the larger one. Compare the surface areas of bodies of revolution.
3.83. A right triangle with a leg of 3 cm and a hypotenuse of 6 cm rotates around an axis passing through the vertex of the right angle parallel to the hypotenuse. Find the volume of the body of revolution.
3.84. A square with a side of 8 cm rotates about a straight line drawn through a vertex parallel to a diagonal that does not pass through this vertex. Find the volume of the body of revolution.
3.85. A regular triangle with a side of 4 cm rotates about an axis drawn through a vertex parallel to the side not passing through this vertex. Find the volume of the body of revolution.
3.86. A right triangle with legs 3 cm and 4 cm rotates about a straight line parallel to the smaller of the legs and passing through the vertex of the smaller of the angles of the triangle. Find the volume of the body of revolution.
3.87. A rhombus with a side of 13 cm and a diagonal of 10 cm rotates about an axis passing through the vertex of an obtuse angle parallel to the diagonal that does not pass through this vertex. Find the volume of the body of revolution.
3.88. A rhombus ABCD with side 10 cm and diagonal AC = 12 cm rotates for the first time about an axis passing through vertex A parallel to diagonal BD, and for the second time through vertex B parallel to diagonal AC. Compare the volumes of bodies of revolution.
3.89. A rectangular trapezoid with bases 10 cm and 18 cm and a height of 6 cm rotates about a straight line passing through the vertex of an acute angle perpendicular to the bases. Find the volume of the body of revolution.
3.90. Three metal cubes with edge a are fused into one ball. What is greater: the surface area of this ball or the total surface area of the cubes?
3.91. Four metal balls of radius a are fused into one cube. What is greater: the surface area of this cube or the total surface area of the balls?
3.92. How many balls with a diameter of 2 cm can be cast from a metal cube with an edge of 4 cm?
3.93. How many cubes with an edge of 2 cm can be cast from a metal ball with a diameter of 4 cm?
3.94. A cylinder is inscribed in a regular quadrangular prism. The volume of the cylinder is V. Find the volume of the prism.
3.95. A cylinder is inscribed in a regular triangular prism. The lateral surface area of the prism is S. Find the lateral surface area of the cylinder.
3.96. The correct triangular prism. The lateral surface area of the prism is 5. Find the lateral surface area of the cylinder.
3.97. A cone is inscribed in a regular triangular pyramid. The volume of the cone is V. Find the volume of the pyramid.
3.98. A regular quadrangular pyramid is inscribed in a cone. The volume of the pyramid is V. Find the volume of the cone.
3.99. A ball is inscribed in a cube. Find the ratio of the surface areas of the cube and the sphere.
3.100. A cube is inscribed in a sphere. Find the ratio of the volumes of a sphere and a cube.
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During the lesson, everyone will be able to get an idea of the topic "Polyhedra. Prism. Prism problems." In this lesson we will review the basic information about polyhedra. We will pay special attention to the definition of a prism. Let us recall the theorem on the area of the lateral surface of a straight prism. Then we will solve several problems on this topic.
Topic: Polyhedra
Lesson: Polyhedra. Prism. Prism problems
In this lesson we will review the basic information about polyhedra. We will pay special attention to the definition of a prism. Let us recall the theorem on the area of the lateral surface of a straight prism.
Figure 1 shows a prism ABCDFA 1 B 1 C 1 D 1 F 1, its foundations ABCDF And A 1 B 1 C 1 D 1 F 1. Pentagons ABCDF And A 1 B 1 C 1 D 1 F 1 are equal and lie in parallel planes.
Rice. 1
Prism bases- these are two faces that are equal polygons that lie in parallel planes.
Lateral The faces are all the faces of the prism, except the bases. Each side face is a parallelogram.
The common sides of the lateral faces are called lateral ribs.
Let's return to Figure 1. In a pentagon ABCDFA 1 B 1 C 1 D 1 F 1:
ABCDF And A 1 B 1 C 1 D 1 F 1- base of the prism.
The side faces are the faces AA 1 B 1 B, BB 1 C 1 C,CC 1 D 1 D, DD 1 F 1 F, FF 1 A 1 A. And the side ribs - AA 1, BB 1, SS 1, DD 1 , FF 1 .
Definition. If the lateral edge of a prism is perpendicular to the plane of its base, then such a prism is called straight.
Consider a pentagonal prism ABCDFA 1 B 1 C 1 D 1 F 1(Fig. 2).
Let the side edge AA 1 perpendicular to the plane of the base. This means that this prism is straight. Since the edge AA 1 perpendicular to the plane ABC, then this side edge is perpendicular to any straight line from the plane of the base ABC, including direct A.F.. This means that the side face is a rectangle.
Rice. 2
Consider a parallelepiped ABCDA 1 B 1 C 1 D 1-(Fig. 3) is a special case of a prism. The bases of the prism are parallelograms ABCD And A 1 B 1 C 1 D 1.
Rice. 3
If the side edge is perpendicular to the plane of the base, then such a parallelepiped will be called a right parallelepiped.
Rice. 4
Consider a parallelepiped ABCDA 1 B 1 C 1 D 1-(Fig. 4). If the edge AA 1 perpendicular to the plane ABCD, then the parallelepiped ABCDA 1 B 1 C 1 D 1- straight.
If a rectangle lies at the base of a right parallelepiped, then such a parallelepiped is called rectangular. Designation: ABCDA 1 B 1 C 1 D 1- or briefly AC 1.
Definition. Correct n-An angle prism is a straight prism that has a regular base at its base. n-gon.
Theorem. The lateral surface area of a straight prism is equal to the product of the perimeter of the base and the height of the prism.
Let's consider this theorem using the example of a triangular right prism ABCA 1 B 1 C 1(Fig. 5) . Prism ABCA 1 B 1 C 1-- straight, which means that all side edges are perpendicular to the plane of the base.
Given: ABCA 1 IN 1 WITH 1- straight prism, i.e. AA 1 ⊥ ABC.
AA 1 = h.
Prove: S side = P main ∙ h.
Rice. 5
Proof.
Triangular prism ABCA 1 IN 1 WITH 1- straight, which means side edges AA 1 B 1 B, AA 1 C 1 C, BB 1 C 1 C - rectangles. And all the side edges of the prism are equal to the height of the prism.
Let's find the area of the lateral surface as the sum of the areas of the rectangles AA 1 B 1 B, AA 1 C 1 C, BB 1 C 1 C:
S side = AB∙ AA 1 + BC∙ BB 1 + CA∙ SS 1 = AB∙ h + BC∙ h + CA∙ h = (AB + BC + CA) ∙ h = P main ∙ h.
We get S side = P main ∙ h, Q.E.D.
In the right n- the side of the base of a coal prism is equal to a and the height is h. Calculate the area of the lateral and total surface of the prism if n = 3, h= 15 cm, a= 10 cm. See fig. 6.
Given: ABCA 1 IN 1 WITH 1- prism,
AA 1 ⊥ ABC,
h =AA 1 = 15cm ,
AB=BC=CA=a= 10 cm.
Find: S side, S full.
Rice. 6
Solution:
By condition the prism is straight. So the rib AA 1 perpendicular to the plane of the base and equal to the height of the prism.
The lateral surface area of a straight prism is equal to the product of the perimeter of the base of the prism and its height. Let's find the area of the lateral surface.
S side = P main ∙ h = P ABC ∙ AA 1 = 3 ∙ AB ∙ h = 3∙ 10 ∙ 15 = 450 (cm 2).
At the base of the prism lies a regular triangle ABC. Let's find its area.
The total surface area of a prism is the area of all its faces, that is, the area of the lateral surface plus the areas of the two bases. Means:
Answer: (cm 2).
The lateral edge of an inclined quadrangular prism is 12 cm. The perpendicular section is a rhombus with a side of 5 cm. Find the area of the lateral surface.
Given: prism ABCDA 1 B 1 C 1 D 1(Fig. 7) ,
AA 1 = 12 cm,
perpendicular section - rhombus with side 5 cm.
Find: Sside
Rice. 7
Solution:
We proved in the last lesson that the area of the lateral surface of an inclined prism is equal to the product of the perpendicular section perimeter and the lateral edge.
According to the condition, the perpendicular section is a rhombus with a side of 5 cm. All sides of the rhombus are equal. This means that the perimeter of the perpendicular section is equal to 
cm.
Now let's calculate the lateral surface area:
(cm 2).
Answer: 240 cm 2 .
The base of a straight prism is an isosceles trapezoid with bases 25 cm and 9 cm and a height of 8 cm. Find the dihedral angles at the lateral edges of the prism. See fig. 8.
Given:ABCDA 1 B 1 C 1 D 1- prism,
AA 1 ⊥ ABC,
AB∥CD, CB = AD,
AB = 9 cm , CD = 25 cm,
hladder= 8 cm.
Find: dihedral angles at the lateral edges of the prism.
Rice. 8
Solution:
Let's remember what a dihedral angle is. Let us have two half-planes α and β that intersect in a straight line CC 1(Fig. 9). Then they form a dihedral angle with the edge CC 1. A dihedral angle is measured by its linear angle.
How is a linear angle constructed? An arbitrary point is taken M on the edge, and two perpendiculars are drawn: one perpendicular in the β plane - perpendicular b, the second perpendicular in the α plane is perpendicular a. Then the angle between the lines a And b and will be the linear angle of the dihedral angle.
Rice. 9
Let's find the linear angle at the edge SS 1. Since the edge CC 1 perpendicular to the entire plane ABC, then the edge CC 1 perpendicular to any straight line from this plane, including straight lines B.C. And CD. Then the angle between the lines B.C. And CD, namely the angle DCB, is the linear angle of the dihedral angle at the edge CC 1.
In a similar way, we find that the linear angle at the edge AA 1- this is the angle INAD, at edge DD 1 - ∠ADC, at edge BB 1 - ∠ABC. All these angles are trapezoid angles ABCD. Let's find their degree measure.
Consider a trapezoid ABCD(Fig. 10) . Let's trace the heights AN And KV. According to the condition, the height of the trapezoid is 8 cm. This means AN = KV= 8 cm.
Rice. 10
We'll find NK. Direct AN And HF perpendicular to the same line DC. So it's straight AN And HF parallel. Because AN = HF, That ANKV- parallelogram. Means, NK = AB= 9 cm.
Since trapezoid ABCD isosceles, then see
Consider a triangle DHA. It is rectangular because AN ⊥ DC and isosceles, since AN = D.H.. Means, ∠ HAD = ∠ HDA= 45° degrees.
Since trapezoid ABCD isosceles, then ∠ DCB = ∠ WITHD.A.= 45°, ∠ DAB = ∠ ABC= 180° - 45° = 135°.
Answer: 45°, 45°, 135°, 135°.
Bibliography
Homework
Test No. 3 on the topic “Polyhedra. Surface area of a prism, pyramid"
I level
Card No. 1
2. The base of a straight prism is a rhombus with a side of 5 cm and obtuse angle 120°. Side surface The prism has an area of 240 cm2. Find the cross-sectional area of the prism passing through the side edge and the smaller diagonal of the base.
3. The side of a regular triangular pyramid is 6 cm, and the height
Card No. 2
2. The base of a straight prism is a rhombus with an acute angle of 60°. The lateral edge of the prism is 10 cm, and the lateral surface area is 240 cm2. Find the cross-sectional area of the prism passing through the side edge and the smaller diagonal of the base.
3. The lateral edge of a regular triangular pyramid is 5 cm, and the height√13 cm. Find the lateral surface area of the pyramid.
II level
Card No. 1
1. Regular polyhedra.
2. The base of a right parallelepiped is a rhombus. Find the area of the lateral surface of the parallelepiped if the areas of its diagonal sections are P andQ.
3. The base of the pyramid is a right triangle with a leg of 4√3 cm and an opposite angle of 60°. All lateral edges of the pyramid are inclined to the plane of the base at an angle of 45°. Find the lateral surface area of the pyramid.
Card No. 2
1. The lateral surface area of a regular truncated pyramid.
2. The diagonal section of a regular quadrangular prism has an areaQ. Find the lateral surface area of the prism.
3. The base of the pyramid is a right triangle with an acute angle of 30°. The height of the pyramid is 4 cm and forms angles of 45° with all lateral edges. Find the lateral surface area of the pyramid.
Level III
Card No. 1
1. Prism. The lateral surface area of a straight prism.
2. In a straight prism ABCA1B1C1 AB = 13, BC = 21, AC = 20. The diagonal of the side face A1C makes an angle of 30° with the plane of the face CC1B1B. Find the total surface area of the prism.
3. In a regular quadrangular pyramid, the side of the base is equal to a, the angle between adjacent side faces is 120°. Find the lateral surface area of the pyramid.
Card No. 2
1. Pyramid. The lateral surface area of a regular pyramid.
2. In a right parallelepipedABCDA1 B1 C1 D1 AD= 17, DC= 28, AC = 39. Diagonal of the side faceA1 Dcomposes with the plane of the side faceDD1 C1 Cangle 45°. Find the total surface area of the parallelepiped.
3. In a regular triangular pyramid, the side of the base is equal tom. The angle between adjacent side faces is 120°. Find the lateral surface area of the pyramid.
Solutions
Level I (card 1)
№ 1. Given:ABCDA1 B1 C1 D1 - straight prism.ABCD-rhombus. AD = 5cm; ∠ B=120° ; S6OK. = 240 cm2.
Find:Sslaughter.
BB1 D1 D. BB1 D1 D- rectangle.Ssec. =BD· DD1. A.A.= 180° - 120° = 60°, sinceABDC- rhombus, then ΔABD- equilateral andBD= AD= 5 cm.(Answer: 60 cm2.)
№ 2. Given:DABC- regular triangular pyramid AB = BC = AC = 6 cm.DO- height;DO= √3.
Find:Sside.
Solution: Since the pyramid is regular, then O is the center of the circle circumscribed and inscribed in ΔABC.
Whereha- apothem of the lateral face. Rosn. = 3 6 = 18 cm. Consider ΔAA1C:
(Answer:Sside. = 36 cm2.)
Level I (card 2)
№ 1. Given:ABCDA1 B1 C1 D1 - straight prism.ABCD- rhombus∠ A= 60°.A.A.1 = 10 cm.Sside. = 240 cm2.
Find:Sslaughter.
Solution: Section passing through the side rib and the smaller diagonal of the baseBB1
D1
D.
BB1
D1
D- rectangle.Ssec. =BD·
DD1.
AB =
DC= AC (by condition). AB = 24/4 = 6 cm. Consider ΔABD, because∠
A = 60°, then ΔABD- equilateral.BD= 6 cm.Ssection = 6 10 = 60 cm. (Answer: 60 cm.)
№ 2. Given:DABC- regular triangular pyramidDC= D.B.= AD= 5 cm.DO- height;DO= √ 1 3 cm.
Find:Sside.
Solution:
Whereha– apothem of the lateral face. Consider ΔAOD:
So,ha = 4 (cm). Let's consider ΔABC - equilateral.
(Answer:Sside = 36 cm2.)
Level II (card 1)
№ 1. Given:ABCDA1 B1 C1 D1 - straight parallelepiped.ABCD- rhombusS.A.C.1 C.A. = R;S.B.1 D1 D.B. = Q.
Find:Sside.
Solution:
2)
3) The diagonals of a rhombus, intersecting, are divided in half and are mutually perpendicular.
(Answer:
)
№ 2. Given:DABC - pyramid∠ C = 90 ° ; SA= 4√3 (cm);∠ B = 60 ° ; ∠ DBO = ∠ DAO = ∠ DCO = 45 ° .
Find:Sside.
Solution: Since the edges of the pyramid are inclined at the same angle, then OA = OB = CO. Point O is the center of the circle described around ΔABC and is the midpoint of the hypotenuse.
1) Consider ΔA.D.B.:
ΔDAO– isosceles (∠
DAO= 45°). Hence,A.O. =
DO. AO = 1/2AB. Let's determine AB from ΔABC.
2) Consider ΔCDA:
DMwe determine from ΔDOM.
Let us determine OM from ΔАВС. OM = 1/2BC. BC = 1/2AB (leg against an angle of 30°). BC = 4 cm MO= 2
cm.
3) Consider ΔCDB:
(Answer:)
IIlevel (card 2)
№ 1. Given:ABCDA1 B1 C1 D1 - regular quadrangular prism.ABCD- square.SACA1 C1 = Q.
Find:Sside.
Solution:
Consider ΔADC:
A.C.2 =
AD2 +
DC2, becauseABCD- square, thenA.C.2= 2
AD2.
(Answer:)
№ 2. Given;DABC- pyramid. ΔАВС - rectangular;∠ WITH = 90 ° ; ∠ IN = 30 ° ; DO- height;DO= 4 cm.∠ ADO= ∠ BDO= ∠ CDO
Find:Sside.
Solution: ΔADO= Δ DBO= Δ CDO(by leg and sharp corner). Therefore, AO = OB= OS. This means that point O is the center of a circle circumscribed about ΔABC and, therefore, the middle of the hypotenuse. From the equality of triangles it follows AO = OB = OC =O.D.(isosceles, rectangular). AO = 4 cm. AB = 8 cm. Consider ΔABC:
1. ConsiderΔADB:
2. ConsiderΔADC:
(Answer:)
Level III (card 1)
№ 1. Given:ABCA1 B1 C1 - straight prism.AB= 13, BC = 21, AC = 20;∠ AFM = 30 ° .
Find:Sfull
Solution: The angle between A1C and plane BB1C1C is 30°. This is the angle between the straight line A1C and its projection onto the plane BB1C1C. A1M⊥ B1C1, MS - projection of A1C onto the planeBB1 CC1. ∠ ACM= 30°.Let's consider ΔA1MC: A1M - height andLet's considerΔA1 M.C.: (that is∠ A1 C.M.= 30°); A1C = 24 and(Answer:)
№ 2. Given:MABCD- regular quadrangular pyramid.D.A.= a;∠ BKD= 120°.
Find:Sside.
Solution:
Angle between facesBMS andDMCequals
120°;DK⊥
M.C.;
since ΔBMC =
Δ
DMC,
ThatB.K.⊥
M.C.And∠
BKD -
linear angle of dihedral angle with edge MC;
ha =
MN;
BD =
a√2 (diagonal of the square);
ΔBKD-isosceles.
Hence,∠
OKD= 60°, a∠
ODK= 30° and
Consider ΔDMC:
orFromΔDKC:
From ΔMNC:
(Answer:)
