Let BC=2a, angle ABC=30 degrees. Then 2a/AB=cos30 From here we find AB=4a/\sqrt(3), then the radius of the circle R=2a/\sqrt(3) At the same time we find AC=2a/\sqrt(3) Let's move on to finding the height. Desired face SCB Let's draw OE perpendicular to BC (at the same time OE is parallel to AC and is the middle line and therefore equals half of AC, OE=a/\sqrt(3)). According to the theorem on three perpendiculars, SE will also be perpendicular to BC, and therefore the linear angle of the dihedral angle is equal to SEO=45/ Then SO=OE The height is found. Next, we find the volume of the cone using the standard formula.
Related tasks:
Write an expression to solve the problem:
a) The perimeter of a rectangle is 16 cm, one of its sides is m cm. What is the area of the rectangle?
b) The area of a rectangle is 28 m², and one of its sides is a m. What is the perimeter of the rectangle?
c) From two cities, the distance between which is s km, two cars left at the same time towards each other. The speed of one of them is u km/h, and the speed of the other is v 2 km/h. In how many hours will they meet?
d) After what time will the motorcyclist overtake the cyclist if the distance between them is s km, the speed of the cyclist is v 1 km/h, and the speed of the motorcyclist is v 2 km/h?
(Problem-research.) Compare the sum of the lengths of the medians of a triangle with its perimeter.
1) Draw an arbitrary triangle ABC and draw the median BO.
2) On the ray BO, set aside the segment OD = BO and connect point D with points A and C. What is the shape of quadrilateral ABCD?
3) Consider triangle ABD. Compare 2m b with the sum of BC + AB (m b is the median of VO).
4) Write similar inequalities for 2m a and 2m c.
5) Using addition of inequalities, estimate the sum m a + m b + m c .
1. 240 students from Moscow and Orel arrived at the tourist camp. There were 125 boys among the arrivals, of which 65 were Muscovites. Among the students who arrived from Orel, there were 53 girls.
How many students came from Moscow in total?
2. Draw a rectangle whose area is 12 cm and the perimeter is 26 cm.
3. How many times will the area of a square increase if each side is doubled?
4. How many times more number, expressed by four units of the fourth digit, than the number expressed by four of the first digit?
5. The hockey team played three matches, scoring only 3 goals and conceding 1 goal. She won one of the matches, drew another, and lost the third.
What was the score of each match?
6. The sum of two numbers is 715. One number ends with a zero. If this zero is crossed out, then the second number will be obtained. Find those numbers.
7. Arrange the brackets so that the equality is true: 15-35+5:4=5
8. 7 people participated in the chess tournament. Each played one game with each. How many games did they play in total?
Preferably with a solution.
the included angle is 30 degrees. The side face of the pyramid passing through this leg makes an angle of 45 degrees with the base plane. Find the volume of the pyramid
If the base of the pyramid is right triangle, and the pyramid is inscribed in a cone, so this triangle is inscribed in the circle of the base of the cone. And if a triangle has a right angle, then it relies on the diameter of this circle. So one of the faces of the pyramid that goes up from the diagonal is perpendicular to the base.
If the leg is 2a, the angle next to it is 30 degrees, then the second leg is 2a tg 30 = 2a / √3
The angle between the side face and the plane of the base is the angle between the lines 1. perpendicular from the center of the hypotenuse of the base (the center of the circumference of the base of the cone) to the leg 2a and the line from the top of the pyramid to the base of this perpendicular. (need a drawing?)
The perpendicular from the center is equal to half of the second leg, since it is parallel to it and comes out of the center of the hypotenuse (similar to triangles)
those. equals a/√3
If side face inclined at 45 degrees, which means that in a triangle formed by a height perpendicular to the leg and a straight line from the vertex, where one angle is right and the second is 45, the third angle is also 45. So the legs are equal. So the height of the pyramid is equal to the perpendicular a√3.
The height of the pyramid is 1/3 Sbase H
H=
A pyramid is inscribed in a cone if the base of the pyramid is a polygon inscribed in the base of the cone. The top of the pyramid coincides with the top of the cone. The lateral edges of an inscribed pyramid for a cone are generators. Accordingly, in this case, the cone is described near the pyramid.
A pyramid can be inscribed in a cone if a circle can be circumscribed near its base (another option is that a pyramid can be inscribed in a cone if all of its side ribs are equal). The heights of the inscribed pyramid and the cone are the same.
If a triangular pyramid is inscribed in a cone, the location of the center of the circumscribed circle depends on the type of triangle that lies at its base.
If this triangle is acute-angled, the center of the circle circumscribed about the pyramid (as well as the base of the height of the pyramid and the cone) lies inside the triangle, if it is obtuse-angled, outside it. If a rectangular pyramid is inscribed in a cone, the center of the circumscribed circle lies in the middle of the hypotenuse of the base, that is, the radius of the circumscribed cone is equal to half the hypotenuse. In this case, the height of the cone and cylinder coincides with the height of the side face containing the hypotenuse.
A quadrangular pyramid can be inscribed in a cone if the sums of the opposite corners of the quadrilateral at the base are equal to 180º (from parallelograms this condition is satisfied for a rectangle and a square, from trapezoids - only for isosceles).
Find the ratio of the volume of the inscribed pyramid to the volume of the cone.
Here SO=H is the height of the cone and the height of the pyramid, SA=l is the generatrix of the cone, AO=R is the radius of the cone (and the radius of the circumscribed circle near the base of the pyramid).
When a regular hexagonal pyramid is inscribed in a cone, the ratio of the volume of the pyramid to the volume of the cone is:
(Clue, ).
If inscribed in a cone right pyramid, the projection of its apothem onto the plane of the base is the radius of the circle inscribed in the base (in the figures SF is the apothem, OF=r). Thus, depending on the initial data, in the course of solving the problem on a pyramid inscribed in a cone, one can consider a right-angled triangle SOA or SOF (or both).
