Since the pyramid is regular, there is a square at its base, the diagonals of which, at the point of intersection, are divided in half. AO1 = CO1 = AC / 2 = 12/2 = 6 cm.
Consider a right-angled triangle OO1C, and by the Pythagorean theorem we define the leg OO1, which is the height of the pyramid.
OO1 ^ 2 = OC ^ 2 – CO1 ^ 2 = 10 ^ 2 – 6 ^ 2 = 100 – 36 = 64.
OO1 = h = 8 cm.
Knowing the length of the diagonal, we determine the area of the square at the base of the pyramid.
Sax = d ^ 2/2, where d is the length of the diagonal of the square.
Sbn = 122/2 = 72 cm2.
Then the volume of the pyramid is:
V = Sbn * h / 3 = 72 * 8/3 = 192 cm3.
Answer: The volume of the pyramid is 192 cm3.
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