The lateral edge of a regular quadrangular pyramid is 4cm and forms 45 degrees with the plane of the pyramid.

| September 8, 2021 | education

The lateral edge of a regular quadrangular pyramid is 4cm and forms 45 degrees with the plane of the pyramid. a) find the height of the pyramid b) find the area of the side and full surface of the pyramids c) find the volume of the pyramid.

The PTO triangle is rectangular and isosceles, since one of its acute angles is 45, then ОМ = ОВ. ОМ ^ 2 + ОВ ^ 2 = BM ^ 2.

2 * OM ^ 2 = 42 = 16.

ОМ = ОВ = √8 = 2 * √2 cm.

Since ABCD is a square, then BD = 2 * OB = 4 * √2 cm.

From a right-angled triangle ABD, AB ^ 2 + AD ^ 2 = BD ^ 2 = 32.

2 * AB ^ 2 = 32.

AB ^ 2 = 16.

AB = 4 cm. Determine the area of ​​the base. Sbn = AB ^ 2 = 16 cm2.

Let’s calculate the volume of the pyramid. V = Sbase * MO / 3 = 16 * 2 * √2 / 3 = 32 * √2 / 3 cm3.

Let’s draw the height in the MAB triangle, and since the MAB is isosceles, then MH and its median, then AH = BH = AB / 2 = 4/2 = 2 cm.

Then MH ^ 2 = MA ^ 2 – AH ^ 2 = 16 – 4 = 12.

MH = 2 * √3 cm.

Smav = AB * MН / 2 = 4 * 2 * √3 / 2 = 4 * √3.

Side = 4 * Smav = 4 * 4 * √3 = 16 * √3 cm2.

Spov = Sb + S side = 16 + 16 * √3 = 16 * (1 + √3) cm2.

Answer: The height of the pyramid is 2 * √2 cm, the lateral surface area is 16 * √3 cm2, the total surface area is 16 * (1 + √3) cm2, the volume of the pyramid is 32 * √2 / 3 cm3.



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