Find the area of the side face and the base of a regular 4-angular pyramid with a side edge of 8 and a height of 4.

| March 1, 2021 | education

At the base of the pyramid lies the square ABCD. Let us construct the diagonals of the squares AC and BD.

Point O is the midpoint of the diagonals and the projection of point P onto the base of the pyramid. Then the triangle COP is rectangular, in which, according to the Pythagorean theorem, we determine the length of the leg OC. OC ^ 2 = CP ^ 2 – OP ^ 2 = 64 – 16 = 48.

OS = 4 * √3 cm.

Triangle OCD is rectangular and isosceles, CD is its hypotenuse, then CD ^ 2 = 2 * OC ^ 2 = 2 * 48 = 96.

CD = AD = 4 * √6 cm.

In a right-angled triangle ACD, the OH segment is the middle line, then OH = AD / 2 = 4 * √6 / 2 = 2 * √6 cm.

From the right-angled triangle РOН, by the Pythagorean theorem, we define the hypotenuse РН.

PH ^ 2 = OP ^ 2 + OH ^ 2 = 16 + 24 = 40.

PH = 2 * √10 cm.

Determine the area of ​​the side face of the pyramid.

Sрсд = СD * РН / 2 = 4 * √6 * 2 * √10 / 2 = 4 * √60 = 8 * √15 cm2.

Answer: The area of ​​the side face is 8 * √15 cm2, the side of the base is 4 * √6 cm.



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