In a regular quadrangular pyramid, the height is 4 cm, and the length of the diagonal is 6√2

| September 17, 2021 | education

In a regular quadrangular pyramid, the height is 4 cm, and the length of the diagonal is 6√2. Find the surface area of the pyramid.

Since the pyramid is correct, there is a square at its base, AB = BC = CD = AD.

Then in a right-angled triangle ACD, according to the Pythagorean theorem, AC ^ 2 = AD ^ 2 + CD ^ 2 = 2 * AD ^ 2.

AD ^ 2 = AC ^ 2/2 = 72/2 = 36.

AD = 6 cm.

Determine the area of ​​the base of the pyramid. Sbn = АD ^ 2 = 36 cm2.

The diagonals of the square, at the intersection point, are divided in half, then OD = BD / 2 = (6 * √2) / 2 = 3 * √2 cm.

In a right-angled triangle OKD, KD ^ 2 = KO ^ 2 + OD ^ 2 = 16 * 18 = 34. KD = √34 cm.

Let’s draw an apothem of KH, which in an equilateral triangle KCD is the height and median, then DH = CH =CD / 2 = 6/2 = 3 cm.

Then in a right-angled triangle KHD KH ^ 2 = KD ^ 2 – DH ^ 2 = 34 – 9 = 25.

КН = 5 cm. Let us determine the area of ​​the side face of the pyramid. Sksd = KH * CD / 2 = 5 * 6/2 = 15 cm2.

Then S side = 4 * Sxd = 4 * 15 = 60 cm2.

Sпов = Sside + Sсн = 60 + 36 = 96 cm2.

Answer: The surface area of ​​the pyramid is 96 cm2.



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