How to find the total area and the volume of a right circular cone inscribed in a cube

Please help me! I would really appreciate it.

Let cube have side length = x
Then cone has radius = x/2 and h = x
and slant height s = √((x/2)²+x²) = √5/2 x

V = π/3 r²h
V = π/3 (x/2)² x
V = πx³/12

Base of cone has area = πr² = π(x/2)² = πx²/4
and lateral area = πrs = π(x/2)(√5/2 x) = √5πx²/4

SA = πx²/4 + √5πx²/4
SA = (√5+1)πx²/4

You probably know the formulas for the total area and the volume of a right circular cone.

If the cube has side a, the height of the right circular cone inscribed in it is also a.
The radius of the base is a/2. So the volume is (Ï€/3)*(a/2)^2*a = (Ï€/12)a^3,
and the total surface area is π(a/2)^2 + √(a^2+(a/2)^2)*(π(a/2)) = ((1+ √5)/4)πa^2

[√(a^2+(a/2)^2) is, by the Pythagorean theorem, the length of the generator (slant height) of the cone
calculated as the hypotenuse of a right triangle with sides a (height) and a/2 (radius of the base))]

You will have to provide me with the given info. Like the dimensions of the cube or any other info.
Basically the formula for the total surface area of a right circular cone is pi*r*l +pi*r^2(l is the slant height.
Volume of a cone is 1/3 pi* r^3

Go to either Bing.com or Google.com and enter volume of a right circular cone.

You will see many sites that will answer your question.

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