The lateral edge of a regular quadrangular pyramid = 8 cm and forms an angle of 45 degrees

| May 7, 2021 | education

The lateral edge of a regular quadrangular pyramid = 8 cm and forms an angle of 45 degrees with the plane of the base of the pyramid. a) find the height of the pyramid. b) find the area of the lateral surface.

Consider a right-angled triangle MOB in which one of the acute angles is 45, then this triangle is also isosceles. OB = OM, then MB ^ 2 = OB ^ 2 + OM ^ 2 = 2 * OB ^ 2.

2 * OB ^ 2 = 64.

ОМ ^ 2 = 64/2 = 32.

ОМ = ОВ = √32 = 4 * √2 cm.

Diagonal ВD is divided in half at point O, then ВD = 2 * ОВ = 2 * 4 * √2 = 8 * √2 cm.

In a right-angled triangle ABD AD = AB as the sides of a square, then 2 * AB ^ 2 = BD ^ 2 = 32.

AB ^ 2 = 32/2 = 16.

AB = 4 cm.

OH = AB / 2 = 4/2 = 2 cm, since OH is the middle line of the triangle ABD.

In a right-angled triangle MOH, MH ^ 2 = OM ^ 2 + OH ^ 2 = 32 + 4 = 36.

MH = 6 cm.

Determine the area of ​​the triangle ABM.

Sawm = AB * MH / 2 = 4 * 6/2 = 12 cm2.

Then S side = 4 * Sawm = 4 * 12 = 48 cm2.

Answer: The height of the pyramid is 4 * √2 cm, the lateral surface area is 48 cm2.



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