In a regular quadrangular pyramid MABCD, the side of the base is 6, and the lateral edge is 5.

| March 6, 2021 | education

In a regular quadrangular pyramid MABCD, the side of the base is 6, and the lateral edge is 5. Find: A) the area of the lateral surface of the pyramid. B) the volume of the pyramid C) the angle of inclination of the lateral face to the plane of the base D) the scalar product of vectors (AD + AB) x AM E) the area of the sphere described near the pyramid E) the angle between BD and the BMC plane

Let’s construct an apothem of MН. Since the triangle CDM is isosceles, then CH = DH = CD / 2 = 6/2 = 3 cm.In a right-angled triangle MDH, according to the Pythagorean theorem, MH ^ 2 = DM ^ 2 – DN ^ 2 = 25 – 9 = 16.

MH = 4 cm.

Sside = 4 * Sdm = 4 * CD * MН / 2 = 2 * 6 * 4 = 48 cm2.

The length of the segment OH = AD / 2, since OH is the middle line of the triangle ACD. In a right-angled triangle MOН CosOHM = OH / MH = 3/4. Angle OHM = arcos (3/4).

ОМ ^ 2 = МН ^ 2 – ОН ^ 2 = 16 – 9 = 7. ОМ = √7 cm

Let’s define the volume of the pyramid. V = S main * ОМ / 3 = 36 * √7 / 3 = 12 * √7 cm3.

Let us determine the length of the AC diagonal. AC ^ 2 = 2 * AB ^ 2 = 2 * 36. AC = 6 * √2 cm, then AO = AC / 2 = 3 * √2 cm.

In a right-angled triangle AOM CosOAM = AO / AM = 3 * √2 / 5.

The sum of vectors ↑ AD + ↑ AB = ↑ AC.

↑ АС * ↑ АМ = | ↑ AC | * | ↑ AM | * CosOAM = 6 * √2 * 5 * 3 * √2 / 5 = 36.

Let us construct a perpendicular DE to the MC, then BE is the projection of BD onto the IUD plane.

Ssdm = 12 = CM * DE / 2. DE = 24/5 = 4.8 cm.

The triangle BED is isosceles, and the angle DE is the angle between BD and the plane of the ВМС.

By the cosine theorem DE ^ = BD ^ 2 + BE ^ 2 – 2 * BD * BE * CosDBE.

23.04 = 72 + 23.4 – 2 * 6 * √2 * 4.8 * CosDBE.

CosDBE = 5 * √2 / 8 = 0.88. Angle DBE ≈ 280.

The radius of the sphere is the radius of the circle circumscribed about the isosceles triangle ACM.

Let us determine the area of ​​the triangle ACM. Sкмн = АС * ОМ / 2 = 6 * √2 * √7 / 2 = 3 * √14 cm2.

R = КМ * МН * КН / 4 * Sкмн = 5 * 5 * 6 * √2 / 12 * √14 = 25/2 * √7 cm.

Sball = 4 * π * R2 = 4 * π * 625/28 = 625 * π / 7 cm2.

Answer: a) 48 cm2, b) 12 * √7 cm3, c) arcos (3/4), d) 36, e) 625 * π / 7 cm2, f) 280.



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