A cone is given with a base radius of 7.5 dm and a height of 6 dm. Find the area of the section drawn through

A cone is given with a base radius of 7.5 dm and a height of 6 dm. Find the area of the section drawn through the vertex if the distance from it to the center of the base of the cone is 3.6 dm.

Let’s draw an OK perpendicular to NM.
OK = 3.6 dm.

Consider a right-angled triangle ОКМ:
KM² = OM² – OK² = 7.5² – 3.6² = 173.16.
KM = 6.58 dm.

NOM isosceles, NM = 2 * KM = 13.16 in.

Consider a right-angled triangle СON:
CN² = 6² + 7.5² = 92.25.
CN = 9.60 dm.

The area of the NCM is found by Heron’s formula:
p = (a + b + c) / 2 = (9.60 + 13.16 + 9.60) / 2 = 16.18.
S = √ (p * (p – a) * (p – b) * (p – c)) = √ (16.18 * 6.58 * 3.02 * 6.58) = 45.99 dm²