The lateral edge of a regular quadrangular pyramid is 5√3, the diagonal of its base is 10√2.

| July 24, 2021 | education

The lateral edge of a regular quadrangular pyramid is 5√3, the diagonal of its base is 10√2. Find the side of the pyramid, the full surface of the pyramid, the dihedral at the base of the pyramid.

Since there is a square at the base of the pyramid, we determine the length of the side of this square.

AB ^ 2 + AD ^ 2 = BD ^ 2 = 200.

2 * AB ^ 2 = 200.

AB ^ 2 = 100.

AB = 10 cm.

OH is the middle line of the triangle ABD, then OH = BD / 2 = 10/2 = 5 cm.

Point O is the middle of the diagonal ВD, ОВ = ВD / 2 = 5 * √2 cm. In the right-angled triangle МОВ, according to the Pythagorean theorem, we determine the length of the leg MO.

MO ^ 2 = BM ^ 2 – OB ^ 2 = 75 – 50 = 25.

MO = 5 cm.

In the right-angled triangle MOH, we define the angle MOH. tgMHO = OM / OH = 5/5 = 1.

Angle INR = 45.

In the right-angled triangle MOH, we calculate the height MH. MH ^ 2 = MO ^ 2 + OH ^ 2 = 25 + 25 = 50.

MH = 5 * √2 cm.

Determine the area of ​​the triangle MOH. Smon = AB * MH / 2 = 10 * 5 * √2 / 2 = 25 * √2 cm2.

Then S side = 4 * Smon = 4 * 25 * √2 = 100 * √2 cm2.

Determine the area of ​​the base. Sbn = AB ^ 2 = 100 cm2.

Spov = Sside + Sbn = 100 * √2 + 100 = 100 * (1 + √2) cm2.

Answer: The dihedral angle is 45, S side is 100 * √2 cm2, Sпов is equal to 100 * (1 + √2) cm2.



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