The height of a regular quadrangular pyramid is 4 m, and the diagonal of the base is 10 m. Find the area of its lateral surface.
Since the pyramid is regular, there is a square at its base, and all sides at the base are equal.
AB = BC = CD = AD.
Consider a right-angled triangle ACD, according to the Pythagorean theorem AC ^ 2 = AD ^ 2 + D ^ 2 = 2 * AD ^ 2.
AD ^ 2 = AC ^ 2/2 = 100/2 = 50.
AD = 5 * √2 cm.
Let’s draw the height О1Н in the triangle DО1С, which will be equal to half of the side of the base of the pyramid.
О1Н = АD / 2 = 5 * √2 / 2 cm.
Let’s draw an apothem OH and consider a right-angled triangle OO1H.
By the Pythagorean theorem OH ^ 2 = OH1 ^ 2 + OH ^ 2 = 16 + ((5 * √2) / 2) ^ 2 = (64 + 50) / 4 = 114/4.
OH = √114 / 2 cm.
The lateral surface area of a regular pyramid is equal to the product of the half-perimeter and the apothem.
Sside = p * L = (4 * 5 * √2) * (√114 / 2) / 2 = 5 * √288 = 10 * √72 cm2.
Answer: Side = 10 * √72 cm2.
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