Question

в правильной 4 угольной пирамиде МАВСД с вершиной М стороны основания=3,а боковые ребра =8,Найти площадь сечения пирамиды плоскостью,проходящей через В и середину МД,ПАРАЛЛЕЛЬНО ас

Answer 1

искомое сечение -  симметричный четырехугольник  BPKL

диагонали  PL , BK  пересекаются под углом 90 град

по условию

стороны основания  AB=BC=CD=AD =3

боковые ребра  MA=MB=MC=MD =8

точка К - середина ребра MD ;  KD = MD /2 = 8/2=4

ABCD -квадрат

диагональ  AC = BD =  3√2

пересечение диагоналей  точка  F  :  BF =FD = BD/2 =3√2 /2 =1.5√2

BK - медиана треугольника  MBD

длина медианы  BK = 1/2 √(2 BM^2 +2 BD^2  - MD^2 ) =1/2 √(2*8^2 +2*(3√2)^2  - 8^2 ) =5

по теореме косинусов

cos KBD = ( KD^2 - (BK^2+BD^2) )/ (-2*BK*BD)= ( 4^2 - (5^2+(3√2)^2) )/ (-2*5*3√2)= 9/(10√2)

MF - высота

треугольник  EBF - прямоугольный

BE = BF / cos KBD = 1.5√2 / [ 9/(10√2)] = 10/3

по теореме Пифагора EF =√(BE^2 - BF^2) =√( (10/3)^2 - (1.5√2)^2) =√238/6

MF - высота

треугольник  MFB - прямоугольный

по теореме Пифагора MF =√( MB^2 -BF^2) =√( 8^2- (1.5√2)^2 ) =√238/2

ME =MF -EF =√238/2- √238/6= √238/3

треугольники  MPL  ~ MCA    подобные

PL / AC = ME /MF ; PL = AC * ME /MF = 3√2 * √238/3 /√238/2 =2√2

площадь   сечения(четырехугольника  BPKL)     

Sс = PL*BK *sin<BEP /2 = 2√2*5*sin90 /2 = 5√2         

Ответ  5√2

Answer 2

Problem Statement

We are given a right quadrangular pyramid MAVSD with the vertex M and the base sides equal to 3. The length of the lateral edges is 8. We need to find the area of the section of the pyramid by a plane passing through B and the midpoint of MD, parallel to AS.

Solution

To find the area of the section of the pyramid, we need to determine the shape of the section and then calculate its area.

Let's start by visualizing the pyramid and the given information:

- The pyramid is right quadrangular, which means it has a rectangular base. - The vertex of the pyramid is M. - The sides of the base are equal to 3. - The length of the lateral edges is 8.

To find the shape of the section, we need to determine the intersection of the plane passing through B and the midpoint of MD, parallel to AS, with the pyramid.

Since the plane is parallel to AS, it will intersect the pyramid in a rectangle. Let's call the intersection points of the plane with the lateral edges of the pyramid P and Q.

To find the coordinates of P and Q, we can use similar triangles. The triangle BMD is similar to the triangle BPQ. Therefore, we can set up the following proportion:

(BM / BP) = (MD / PQ)

Since BM = 8 (the length of the lateral edges) and MD = 3 (the length of the base sides), we can solve for PQ:

(8 / BP) = (3 / PQ)

Cross-multiplying, we get:

8 * PQ = 3 * BP

PQ = (3 * BP) / 8

Now, let's find the length of BP. Since the plane passes through the midpoint of MD, we can find the coordinates of the midpoint using the formula:

Midpoint = ( (x1 + x2) / 2, (y1 + y2) / 2, (z1 + z2) / 2 )

In this case, the coordinates of M are (0, 0, 0) and the coordinates of D are (3, 0, 0). Therefore, the coordinates of the midpoint are:

Midpoint = ( (0 + 3) / 2, (0 + 0) / 2, (0 + 0) / 2 ) = (1.5, 0, 0)

Now, let's find the coordinates of B. Since the plane passes through B and the midpoint of MD, the x-coordinate of B will be the same as the x-coordinate of the midpoint, which is 1.5. The y-coordinate and z-coordinate of B will be the same as the y-coordinate and z-coordinate of M, which are both 0. Therefore, the coordinates of B are (1.5, 0, 0).

Now, we can substitute the coordinates of B into the equation for PQ:

PQ = (3 * BP) / 8 = (3 * 1.5) / 8 = 0.5625

Therefore, the length of PQ is 0.5625.

Now that we have the dimensions of the rectangle PQRS, we can calculate its area. The area of a rectangle is given by the formula:

Area = length * width

In this case, the length of the rectangle is PQ, which is 0.5625, and the width of the rectangle is the same as the length of the base sides, which is 3. Therefore, the area of the section of the pyramid is:

Area = 0.5625 * 3 = 1.6875 square units.

Therefore, the area of the section of the pyramid, formed by a plane passing through B and the midpoint of MD, parallel to AS, is 1.6875 square units.

Conclusion

In a right quadrangular pyramid MAVSD with a vertex M and base sides equal to 3, and lateral edges equal to 8, the area of the section of the pyramid formed by a plane passing through B and the midpoint of MD, parallel to AS, is 1.6875 square units.