The height of a regular quadrangular pyramid is 4 cm and its aporhema forms an elevated angle of 45 degrees.

| March 12, 2021 | education

The height of a regular quadrangular pyramid is 4 cm and its aporhema forms an elevated angle of 45 degrees. Find the area of the base of the pyramid and the side surface of the pyramid.

Since the pyramid is correct, the square ABCD lies at its base. Let’s draw the diagonals of the square and connect the point O with the point H of the apothem. In a right-angled triangle РOН, the angle РНO = 45, then this triangle is isosceles, OH = PO = 4 dm.

Since the diagonals of the square are divided in half at point O, and point H divides the side of CD in half as well, then OH is the middle line of the triangle ACD, then AD = 2 * OH = 2 * 4 = 8 dm.

Determine the area of ​​the base.

Sbn = АD ^ 2 = 8 ^ 2 = 64 dm2.

Let us determine the length of the PH apothem. PH ^ 2 = PO ^ 2 + OH ^ 2 = 16 + 16 = 32. PH = 4 * √2 dm2.

Determine the area of ​​the side face of the pyramid. Sрсд = DC * РН / 2 = 8 * 4 * √2 / 2 = 16 * √2 cm2.

Then Sside = 4 * Srcd = 4 * 16 * √2 = 64 * √2 cm2.

Answer: The base area is 64 dm2, the lateral surface area is 64 * √2 cm2.



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